/sci/ - Science & Math

<div class="math">\int e^{x^2}dx</div>
<div class="math">\int x^{x^x}dx</div>

>>6556896
/thread

"prove that you did it by providing the computations leading to your result". reopen thread!

>>6556881
f(x) = 1 if x is rational
=0 if x is irrational
<span class="math"> \int f(x) \,\mathrm{d} x [/spoiler]

<span class="math">\int \int \int \pi dV
Good luck OP[/spoiler]

>>6556911

0, that was easy.

>>6556954
Why would it be 0? Also it's not a definite integral.

>>6556881
<span class="math">∫_{0}^{10} x(1+sin^2(e^{x^x})) dx[/spoiler]

>>6556962

The set of rationals have measure 0. There's a bunch of proofs of this, so I won't bother going into detail, but no matter what interval you pick the integral will always be 0, so the antiderivative is 0.

>>6556881
I always hate doing induction on binary trees ask him a question about that

Try this one.

<span class="math"> \int^{1}_{0}\frac{\ln(1-x)}{x} dx [/spoiler]

>>6556965
>countable sets have (Lebesgue-) measure 0
>rationals are countable
>QED

>>6556881
The integral from 0 to 1 of 1/(3^x -x) dx

>>6556881
>Give me the most difficult integral you have ever performed that even a top 10 university mathematics professor can't do in five minutes.

Any triple integral involving anything more complex than two first order trigonometric function.

>>6556983
>do my homework for me pls

>>6557000
>Implying it can be solved without using a computer anyways

<div class="math">\int\limits_{\mathbf{R}^n} \! k_1^{-b_1|x_1|^{a_1}} k_2^{-b_2|x_2|^{a_2}} \cdots k_n^{-b_n|x_n|^{a_n}} \, \mathrm{d}x_1 \, \mathrm{d}x_2 \, \ldots \, \mathrm{d}x_n</div>

For all <span class="math">k \in \mathbf{R}[/spoiler] and <span class="math">b > 0[/spoiler], <span class="math">a > 1[/spoiler]

>>6557027
Answer follows from differentiation under the integral sign, a few change of variables, and recognising the Beta function.

Answer is

<div class="math">\prod_{h=1}^{n} \Gamma\left(\frac{1}{\alpha_h} +1\right)(2b_h\log k_h)^{-1/\alpha_h}</div>

\int {sin (x)/x} dx
but find the indefinite integral.

>>6557120
I can't give a closed form.

<span class="math"> \int\frac{\sin(x)}{x} dx [/spoiler]

Apply Euler's product formula for sinc(x).

<span class="math"> \int\frac{\sin(x)}{x} dx = \int\left(\prod^{\infty}_{n=1}\left(1-\frac{x^2}{n^{2}\pi^2}\right)\right)dx [/spoiler]

Apply the Weierstrass expansion theorem / factorization theorem.

<span class="math"> \int\left(\prod^{\infty}_{n=1}\left(1-\frac{x^2}{n^{2}\pi^2}\right)\right)dx = \int\left(1-\left(\frac{1}{\pi^2}+\frac{1}{4\pi^2}+\frac{1}{9\pi^2}+\cdots\right)x^{2}+\left(\frac{1}{\pi^2}+\frac{1}{4\pi^2}+\frac{1}{9\pi^2}+\cdots\right)x^{4}-\cdots\right)dx [/spoiler]

Monotone convergence theorem fails... shit. (hurrdurr, should have seen that coming)

I think a good approximation might be yielded by stopping on any odd large value of <span class="math"> h [/spoiler] though (or maybe any large even value, I can't really think straight tonight):

<span class="math"> \int\left(1-\left(\frac{1}{\pi^2}+\frac{1}{4\pi^2}+\frac{1}{9\pi^2}+\cdots\right)x^{2}+\left(\frac{1}{\pi^2}+\frac{1}{4\pi^2}+\frac{1}{9\pi^2}+\cdots\right)x^{4}-\cdots\right)dx=x-\int\frac{1}{\pi^2}\sum^{h}_{k=1}(-1)^{k}\left(x^{2^k}\zeta(2)\right)dx+c_1 [/spoiler]

<span class="math"> x-\int\frac{1}{\pi^2}\sum^{h}_{k=1}(-1)^{k}\left(x^{2^k}\zeta(2)\right)dx+c_1=x-\frac{1}{6}\sum^{h}_{k=1}\left(\frac{x^{2^{k}+1}}{2^k+1}\right)+c_1+c_2 [/spoiler]

<span class="math"> x-\frac{1}{6}\sum^{h}_{k=1}\left(\frac{x^{2^{k}+1}}{2^k+1}\right)+c [/spoiler]

That doesn't seem right for some reason though...

>>6557318
I still find my first series answer to be questionable so I'm trying again:

Note that <span class="math"> \frac{\sin(x)}{x}=\sum^{\infty}_{k=0}\frac{(-1)^{k}x^{2k}}{(2k+1)!} [/spoiler] (where this series is centered around x=0).

I am going to assume it can be treated similar to the way the error function is generated from <span class="math"> \int^{x}_{0} e^{t^{2}} dt [/spoiler]

<span class="math"> \int\left(\sum^{\infty}_{k=0}\frac{(-1)^{k}x^{2k}}{(2k+1)!}\right)dx [/spoiler]

<span class="math"> \int\left(\sum^{\infty}_{k=0}\frac{(-1)^{k}x^{2k}}{(2k+1)!}\right)dx =c+ \sum^{\infty}_{k=0}\frac{(-1)^{k}x^{2k+1}}{(2k+1)(2k+1)!} [/spoiler]

Do some algebras.

<span class="math"> c+\sum^{\infty}_{k=0}\frac{(-1)^{k}x^{2k+1}}{(2k+1)(2k+1)!}=\sum^{\infty}_{k=1}\left(\frac{(-1)^{k-1}x^{2k-1}}{(2k-1)(2k-1)!}\right)+c [/spoiler]

That might be closer.

>>6557320
>>6557318
I don't get it, half my latex always loads for only me...

>>6557318
>>6557326
Maybe I'm wrong about the monotone convergence theorem there, as it can be broken up into a posive series and a negative series. If so then I'd assume that <span class="math"> h [/spoiler] can be replaced with <span class="math"> \infty [/spoiler]

>>6556965
Is it zero if you go from -infinity to infinity?

>>6557326
Oh, I know why it didn't seem right. I fucking left the <span class="math"> (-1)^k [/spoiler] out of the sum in the final answer for my first attempt. Fuck.

>>6557027
>>6557031
Is this right? I tried it myself and got a similar answer but with the 2 outside of the <span class="math">(\cdot)^{-1/a_i}[/spoiler] term.
My steps:
1. Factorize the integral into a product of n integrals on R, by symmetry each integral i is (suppressing subscripts)
<div class="math"> 2\int_0^\infty k^{-bx^a} dx</div>
2. Rewrite k = <span class="math">e^{\ln k}[/spoiler] substitute u = <span class="math">(b\ln k)x^a[/spoiler] to get the gamma function times <span class="math">(b\ln k)^{-1/a}[/spoiler].

I'm flagrantly disregarding the ranges -- assuming (b ln k) > 0 because I can't into complex analysis and ln k makes no sense to me otherwise -- but surely the answer should agree with >>6557031 in this special case?

>>6557318
So the last two lines should have been:

<span class="math"> x-\int\frac{1}{\pi^2}\sum^{h}_{k=1}(-1)^{k}\left(x^{2^k}\zeta(2)\right)dx+c_1=x-\frac{1}{6}\sum^{h}_{k=1}(-1)^{k}\left(\frac{x^{2^{k}+1}}{2^k+1}\right)+c_1+c_2 [/spoiler]

<span class="math"> x-\frac{1}{6}\sum^{h}_{k=1}(-1)^{k}\left(\frac{x^{2^{k}+1}}{2^k+1}\right)+c [/spoiler]

>>6558039
This is most annoying, testing on a different browser.

<span class="math"> x-\int\frac{1}{\pi^2}\sum^{h}_{k=1}(-1)^{k}\left(x^{2^k}\zeta(2)\right)dx+c_1=x-\frac{1}{6}\sum^{h}_{k=1}(-1)^{k}\left(\frac{x^{2^{k}+1}}{2^k+1}\right)+c_1+c_2 [/spoiler]

<span class="math"> x-\frac{1}{6}\sum^{h}_{k=1}(-1)^{k}\left(\frac{x^{2^{k}+1}}{2^k+1}\right)+c [/spoiler]

>>6557326
>>6557318
Last tow lines should be this:

<span class="math"> x-\int\frac{1}{\pi^2}\sum^{h}_{k=1}(-1)^{k}\left(x^{2^k}\zeta(2)\right )dx+c_1=x-\frac{1}{6}\sum^{h}_{k=1}(-1)^{k}\left(\frac{x^{2^{k}+1}}{2^k+1}\right)+c_1+c_2 [/spoiler]

<span class="math"> x-\frac{1}{6}\sum^{h}_{k=1}(-1)^{k}\left(\frac{x^{2^{k}+1}}{2^k+1}\right)+c [/spoiler]

>>6558062
Oh wtf...

<span class="math"> x-\int\frac{1}{\pi^2}\sum^{h}_{k=1}(-1)^{k}\left(x^{2^k}\zeta(2)\right )dx+c_1=x-\frac{1}{6}\sum^{h}_{k=1}(-1)^{k}\left(\frac{x^{2^{k}+1}}{2^k+1}\right)+c_1+c_2 [/spoiler]

<span class="math"> x-\frac{1}{6}\sum^{h}_{k=1}(-1)^{k}\left(\frac{x^{2^{k}+1}}{2^k+1}\right)+c [/spoiler]

>>6556972
<span class="math">\int^{1}_{0}\frac{\ln(1-x)}{x} dx[/spoiler]
Let <span class="math">u = 1-x[/spoiler]. Then <span class="math"> x = 1-u [/spoiler] and <span class="math">du = -dx[/spoiler].
We now have <span class="math">\int^{0}_{1}\frac{\ln(u)}{1-u} du[/spoiler].
Let <span class="math">w = \ln(u)[/spoiler] and <span class="math">dv = 1-u[/spoiler].
Then <span class="math">dw = \frac{1}{u} du[/spoiler] and <span class="math">v = u - \frac{u^2}{2} = u(1-\frac{u}{2})[/spoiler].
Integrating by parts we get <span class="math">u(\ln(u))(1-\frac{u}{2}) - \int(1-\frac{u}{2}) du[/spoiler].
Unless my (quick) math is off, simple integration and arithmetic yields <span class="math">-\frac{3}{4}[/spoiler].

I did both the math and the latex quickly so if it's a trainwreck just ignore this post.

>>6557371

the antiderivative is f(x) = 0, and yes, f(inf)-f(-inf) = 0 - 0 = 0

>>6556972
>>6558171
Quick correction:
>Integrating by parts we get u(ln(u))(1−2u)−(1−2u)du .
should go from 1 to 0.
Kinda guessing the Latex syntax but...
\left [ u(\ln(u))(1-\frac{u}{2}) - \int(1-\frac{u}{2}) du \right ] |^{0}_{1}

>>6558174
shit forgot math tags
<span class="math">\left [ u(\ln(u))(1-\frac{u}{2}) - \int(1-\frac{u}{2}) du \right ] |^{0}_{1}[/spoiler]

>>6557326
There are spaces showing up in your keywords for some reason. Instead of "\right" it's "\rig ht" etc...

>>6558171
>>6558176
someone wanna check this shit

>>6556896
e^x^2 = sum x^(2k)/k! from 0->inf
integral e^x^2 = sum x^(2k+1)/(k!(2k+1!))
newb

>>6558253
It's wrong. the answer is -pi^2 /6. Just use a series expansion for the log.

>>6558284
but my math looks correct

>>6556881

>>6558328
Not the same anon but...
From 0 to 1, ln(1-x)/x dx
ln(1-x) = -x-x^2/2 -x^3/3-x^4/4-x^5/5
ln(1-x)/x = -1-x/2-x^2/3-x^3/4-x^4/5...
Integrate ln(1-x)/x = -x-x^2/4-x^3/9-x^4/16-x^5/25
Integral ln(1-x)/x = -sum x^(n+1)/(n+1)^2 from 0-> infinity the polylog(x)-x
=-1-1/4-1/9-1/16-1/25.... which converges to -pi^2/6

>>6558328
I can spot at least two mistakes in >>6558171

First when you let u = 1-x, this flips the limits from 1 to 0 (instead of 0 to 1), which the minus sign in the du (= -dx) flips back again. So the limits should remain as 0 to 1 instead of 1 to 0.
(sign check: the integrand of ln(1-x)/x and ln u/(1-u) are both negative on [0, 1])
Secondly for the integration by parts, you should have used dv = 1/(1-u) instead of dv = 1-u.

>>6558171
Wrong. http://math.ucsb.edu/~cmart07/Evaluating%20Integrals.pdf

>>6558406
Right.

>>6558181
Yeah I know, I haven't been able to figure out why that is happening yet. I've tried working on different browsers too.

>>6558065
>>6558062
Looks like this for those that can't see.

So I was playing around with this integral
<div class="math">I(s) = \int_0^{\pi/2} \frac{dx}{ 1+(\tan x)^s}</div>
which is known to = <span class="math">\pi/4[/spoiler] for all s > 0. The trick is to substitute u = <span class="math">\pi/2 - x[/spoiler] and use the trig identity tan(pi/2 - u) = cot(u) to get
<div class="math">I(s) = \int_0^{\pi/2} {(\tan u)^s du}{ 1+(\tan u)^s}</div>
and add them together to get 2I(s) = pi/2.

I'm trying to get this answer another way: since this is bounded, I should be able to differentiate under the integral sign and get I'(s) = 0. Unfortunately the resulting integral isn't especially friendly, even after substituting t = tan x:
<div class="math">\int_0^\infty -\frac{t^s ln t dt}{(1+t)^2 (1+t^s)^2} = 0?</div>
I suspect there's a complex-analysis method of doing this, but I've never taken the subject myself. Any suggestions?

>>6558640
Never mind, figured it out -- the substitution v = ln t turns it into an integral of an odd function over (-infty, infty).

>>6558427
thanks for the clarifications.

http://math.stackexchange.com/questions/562694/integral-int-11-frac1x-sqrt-frac1x1-x-ln-left-frac2-x22-x1/565626#565626


That shit is other-worldly. I couldn't believe someone solved the thing analytically, but it's there (and long).

>>6558828
Yes, I just saw that today as well.
Some fucking people man.

>>6558640
are you a math student? if so, what year?

>>6558828
Haha wow. That's awesome. Why does he have the dt in such a weird place, and sometimes he just forgets it all together. Are those typos? I'm assuming they are.

>>6558835
>>6558828

Well, look at this then:
http://math.stackexchange.com/questions/523027/a-math-contest-problem-int-01-ln-left1-frac-ln2x4-pi2-right-frac

The smart guy who solved it is an animu faggot as well. Pig disgusting.

>>6558839
I wish I could solve problems like that. At the moment I'm only an undergrad in applied math, so maybe some day.

>>6558852
> in applied math
lel, I don't think you'll ever be capable of it then, except if it's your only hobby to solve complicated integrals.
Besides, knowing how to approximate them is more useful to you as an applied mathematician.

Integrate sin(x^2)

>>6558855
>knowing how to approximate them is more useful to you as an applied mathematician
True.

I guess that's why the ones I usually tackle in these types of threads are the ones that can only be approximated. (>>6558624, >>6557318)

Any of the integrals here:
https://math.stackexchange.com/users/97378/cleo?tab=answers&sort=votesJust

Actually if your professor can do them in few minutes maybe he can provide the steps for one of these.
Relevant thread:
>>6558662

>>6556985

>impyling double/triple...etc integration is difficult

>>6558928
What it OP's professor is Cleo?

>>6556881
I'm not sure if i can say it like this on here, but... triple integrals

>>6558931
Triple integrals are a well-known meme on /sci/.

>>6556881
>prove that you did it
>wanting proof on 4chan
you so funnee

>>6556896

>I haven't taken calc 2 yet

keep dreamin freshman

>>6556911

Yeah, 0 is correct. The rationals are denumerable, while the irrationals aren't, and removing a denumerable set from a non-denumerable sets still yields a non-denumerable set as the result.

>>6558973
solve it then if you're so smart.

>>6558984

>>6558266 already solved it

I can't into latex but you just use the Taylor series approximation of e^x, substitute x^2 for x, and integrate from there. ez.

>>6558266
can't read it
<div class="math"> \exp(x^2) = \sum_{k=0}^\infty \frac{x^{2k}}{k!} </div>
<div class="math"> \int \exp(x^2) = \sum_{k=0}^\infty \frac{x^{2k+1}}{k!(2k+1!)} </div>
I don't remember any of Calc 2, but where's the plus C?

>>6559001
It's a waste of time to add the constant of integration every time you write an integral. It's implied.

>>6558973
>calc 2
>series

>>6559004
series were part of calc 2 at my school.
I just don't remember them

>>6557794
Yeah, you're right, I fucked the TeX.

<div class="math">2\Gamma\left(\frac{1}{\alpha} +1\right)(b\log k)^{-1/\alpha}</div>

Is the right answer

>>6558615
if you don't use spaces in your code jsmath forces them in. put spaces before each command to be safe.
>>6558976
>>6556965

Can you explain like I'm a generic STEM instead of a math major?
every rational is surrounded by irrationals and so every f(x) next to a dx is 0 or something?

what about the opposite?
f(x) = 1 if x is irrational
=0 if x is rational
<span class="math"> \int f(x) \, \mathrm{d} x [/spoiler]

>>6558969
Mathematical proof anon. Why are you even on /sci/?

>>6559122
>if you don't use spaces in your code jsmath forces them in. put spaces before each command to be safe
I'll give that a try then.
<span class="math"> \int \left( \prod^{ \infty}_{n=1} \left(1- \frac{x^2}{n^{2} \pi^2} \right) \right)dx = \int \left(1- \left( \frac{1}{ \pi^2}+ \frac{1}{4 \pi^2}+ \frac{1}{9 \pi^2}+ \cdots \right)x^{2}+ \left( \frac{1}{ \pi^2}+ \frac{1}{4 \pi^2}+ \frac{1}{9 \pi^2}+ \cdots \right)x^{4}- \cdots \right)dx [/spoiler]

>>6559123
Thanks anon, I luv u!

<span class="math"> \int \frac{ \sin(x)}{x} dx = \int \left( \prod^{ \infty}_{n=1} \left(1- \frac{x^2}{n^{2} \pi^2} \right) \right)dx[/spoiler]

>>6559136
I feel like this post was supposed to link me and I feel sad about missing out on the love.
If you use \displaystyle or equation tags you're sums and products won't look so shitty
<span class="math"> \displaystyle \sum^\infty_{n=1} n = [/spoiler] something
or
<div class="math"> \prod^\infty_{n=0} n =0 </div>

>>6559156
sorry [.eqn] tags

>>6559156
<div class="math">\sum n = -\frac{1}{12}</div>

>>6559166
dubs speaks truth

>>6559156
<span class="math"> \displaystyle \sqrt{2 \sqrt{2 \sqrt{2 \sqrt{2 \sqrt{2 \cdots}}}}} = \prod^{ \infty}_{n=1}2^{ \frac{1}{2^{n}}}=2^{ \sum^{ \infty}_{n=1} \frac{1}{2^n}} = 2^{ \frac{ \frac{1}{2}}{1- \frac{1}{2}}} = 2^{1} = 2 [/spoiler]

Thanks, these are much better looking.

>>6559179
>square root of two root of two root of two root of two rooty too rooty too rooty too rooty too...

>>6559007
For me the they put multivariable calc in calc 2 and let series come later.

<div class="math">\int \! \exp(a_nx^n+a^_{n-1}x^{n-1} + \cdots + a_1 x+a_0) </div>

is this calculus?

>>6559179
<span class="math">\sqrt{ 2 x } = x[/spoiler]
<span class="math">x = \left \{ 0, 2 \right \}[/spoiler]
we can rule out x=0
<span class="math">x = 2[/spoiler].

>>6559196
I'm aware of the recursive method of solution, it's pretty.

It can also be used to solve:

<span class="math"> \sqrt{2+ \sqrt{2+ \sqrt{2+ \sqrt{2+ \sqrt{2+ \cdots}}}}} = \sqrt{2}^{ \sqrt{2}^{ \sqrt{2}^{ \sqrt{2}^{ \sqrt{2}^{ \sqrt{2}^{ \cdot^{ \cdot^{ \cdot}}}}}}}} [/spoiler]

>>6558400
isn't that the gaussian integral? Call it the error function and there's your answer. That's the trouble with the OP's question: Integrals aren't, in general, computable.

<span class="math"> \displaystyle \int^{1}_{0} \sqrt{e^{ \pi}+3x \sqrt{e^{ \pi}+3x \sqrt{e^{ \pi}+3x \sqrt{e^{ \pi}+3x \codots}}}} dx [/spoiler]

<span class="math"> \displaystyle \int^{1}_{0} \sqrt{e^{ \pi}+3x \sqrt{e^{ \pi}+3x \sqrt{e^{ \pi}+3x \sqrt{e^{ \pi}+3x \cdots}}}} dx [/spoiler]

>>6558931
>implying they are easily completed in 5 mins
Sure they might not always be difficult, but they aren't procedural often times and are tedious to complete

>>6559227
>Integrals aren't, in general, computable.
Be careful with that word. It's pretty easy to computer integrals by using repeated subdivisions. There's only a few kinds that are like, non-convergent for every point.

>>6559318
That's a blank statement though. I don't think OP's prof was talking about taking more than 5 minutes to apply the procedure, more like taking more than 5 minutes to figure out the method to solve it.

Otherwise, just ask the integral of a*b on a [0;1], where a is a bigass 100-digits number and b is also a bigass 100-digits number. It's gonna take him 10,000 digit-by-digit multiplications to do it, not counting the additions, and that's definitely gonna take more then 5 minutes by hand, but that doesn't mean it took him more than 5 minute to know what the answer looked like.

>>6559337
It's an especially poorly chosen word since computability is a well-defined characteristic of some functions (the typical example of an actually uncomputable function is the busy beaver function).

>>6558834
Unfortunately no, I'm an economics undergrad. That means I study bits and pieces of real analysis with a modicum of rigour.

To get back on OP's topic, I was actually trying to solve an integral that appeared in a /sci/ thread yesterday, which died before I could post my solution. The integral was:

<div class="math">\int_0^{\pi} \ln (1 + s\cos^2x) dx,\; s \geq 0</div>

which can be solved using a similar method to the one I outlined above:
1. By symmetry, we need only integrate from 0 to pi/2, then multiply by 2.
2. Write that I(s) and differentiate under the integral sign. The indefinite integral I'(s) is fairly horrible but yields to the 'Weierstrass' tangent substitution t = tan x, and evaluating at the endpoints produces <span class="math">I'(s) = \frac{\pi }{2(s+1 + \sqrt{s+1})}[/spoiler].
3. This one is easy to do: <span class="math">I(s) = \pi\ln(1 + \sqrt{s+1}) + C[/spoiler] with constant <span class="math">C = I(0) - \pi\ln 2[/spoiler], so the final answer to the original problem is <span class="math">2\pi \ln[ (1 + \sqrt{s+1})/ 2][/spoiler].

Good luck, this is Ramanujan shit.

<div class="math">\int_0^\infty \frac {\cos {\pi x}} {e^{2\pi \sqrt x} - 1} \mathrm d x = \dfrac {2 - \sqrt 2} {8}<div class="math"></div></div>

Good luck, this is Ramanujan shit

<div class="math">\int_0^\infty \frac {\cos {\pi x}} {e^{2\pi \sqrt x} - 1} \mathrm d x <div class="math"></div></div>

Good luck, this is Ramanujan shit

<div class="math">\int_0^\infty \frac {\cos {\pi x}} {e^{2\pi \sqrt x} - 1} \, \mathrm d x <div class="math"></div></div>

>>6559630
This may be wrong; I cannot check it.

<div class="math">\Re \sum_{k=0}^{\infty} \frac{\mathrm{Li}_{2k+2}(1)}{k!}\Gamma(2k+2)\left(\frac{i\pi}{4}\right)^k </div>

>>6559630
This may be wrong; I cannot check it.

>>6559708
That sum looks divergent. Did you interchange a sum and integral without justification at some point?

The final answer is simple. It can be written in the form

<div class="math">\frac {a - \sqrt b} {c} <div class="math">

where <span class="math">a,b,c<span class="math"> are integers.[/spoiler][/spoiler]</div></div>

>>6558989
Anyone? How?

>>6559204
Why can you do this? Whats the name of the method/teorem?

>>6559730
It's just recursion. Look up nested radicals.

>>6559630
What is the exact value?

>>6558266
okay you smart people, i'm a complete and utter idiot when it comes to math. hell, my grammar is equally as horrible. but i find it funny that you brainy types spend your lives doing this stuff when there's no practical uses for it.

prove me wrong weirdos. protip, you won't.

>>6559757

>>6559760
no, i'm being serious here. what practical reasons would that equation be used for. is it a part of the schemes used by quants to goad people into thinking they're purposeful....what's it used for?

>>6559767

>>6559772
could you stop being facetious for a moment and tell me what you use this stuff for. or is it simply a mental exercise, nothing more?

>>6559774
The first function you pointed to is the error function (>>6559001 also shown here) it is used in determining approximations of area under the bell curve. It is useful in statistics and anything where Gaussian type distributions can be applied.

P.S. People are only going to answer serious questions, cut the shit and stop being such a cunt.

>>6559778
It's also useful in probability. http://www.youtube.com/watch?v=yT7eWXYlZaE

>>6559778
>>6559780
thank you. i'm going to do research.

>>6559774
>>6559767
Actually, this kind of stuff is what allows to discover new methods of solving, which leads to open new branches of mathematics! So, in the future, we can find specific situations in the nature that needs this "stuff".

Im not supposing, that is what really happened with (for example) many differential equations which the resolution were known, but not its pratical use. But later, it appeared in some area of physics. See Fermats Theorem, apparently with no pratical use, but the evolution of people trying to solve it allowed the discover of a bunch new stuff.

You can't think like that or else you will never envolve.

>>6559783
Why are you replying to me? I already know this. Did you mean to reply to (>>6559774)?

>>6559785
Yeah, sorry! It was only for >>6559774

>>6559783
>>6559785
okay so the process of trying to solve hard stuff simply enables you to solve the stuff around you when you eventually try to do so?


that's remarkable.

On /g/ several months ago. An anon said that the process of trying to solve and master C makes you a smarter person, even if you never, ever use C or even understand the fundamentals beyond a rudimentary appreciation.

by me trying to understand what's going on. i'm gonna be a better person because of it.

>>6559772
>>6559778
>>6559780
another question for you. could i master OP's equation by simply doing it over and over again. like how some one said of Fortran that they learned by simply observing arguments and did those repeatedly until their minds became familiar with what was going on.

would that work for me? i say that because if i was to try the fundamentals of cal...i'll get bored and leave it again for another twenty years.

i'm incredibly old.

>>6559791
>okay so the process of trying to solve hard stuff simply enables you to solve the stuff around you when you eventually try to do so?

Yeah, but in two senses of helping. The first is that it helps you become a better logical thinker. And the second is that the result that can be yielded from the question you are solving can be applied to real life systems many years from now.

>>6559793
>could i master OP's equation by simply doing it over and over again
I suggest you by a book on basic derivative and integral calculus if you would like to start doing these things. I am happy to hear that you want to learn though. I'm always happy to hear that.

>>6559794
I'm not as smart as I need to be, and logically I'm as sophisticated as a turtle. But whenever I muster the will power and forgo slothfulness that starves purposeful thoughts and actions, I copy equations, formula and bits and pieces from various books hoping that something leaps within my mind that drives me to find the solutions and purposes of each.


>>6559800
Derivative and Integral Calculus. I'm going to focus on this for now, and only this. If i even look at anything else i'll lose focus and get side tracked.

>>6559803
>I'm going to focus on this for now, and only this. If i even look at anything else i'll lose focus and get side tracked
Also be sure to touch up on your algebra and trigonometry skills when needed.

>>6559803
I would suggest visiting https://www.khanacademy.org/
They're pretty good for that, you can also review your algebra and trig there (as another anon said, you will need these for calc).

Very nice to follow together with a book or a lecture (you can find uni lectures on youtube and stuff).

>>6559804
damn i hope i can. i have to fight through either laziness or something that happens whenever i try to do math. you have no idea how hard it is.

>>6559806
thank you much anon. very, very much.

>>6558828
>>6558839
Oh god. Why'd you have to show me this /sci/....
Self esteem destroyed

>>6559818
>that second one
The guy solved it and wrote that in under an hour.

Holy fucking shit.

>>6558173
You know you're talking bullshit?

>>6559824

If you read his blog from time to time (as I do), you'll find he hangs out on an internet forum dedicated to solving wacky integrals and series: http://integralsandseries.prophpbb.com

Dude is massively hardcore at solving stuff.

I also like the dude on stackexchange that gives single word answers to integral problems with no solutions.

>>6559882
>I also like the dude on stackexchange that gives single word answers to integral problems with no solutions.

Here he is: https://math.stackexchange.com/users/97378/cleo

>>6559882
That website is pretty cool.

>>6559231
<span class="math"> \displaystyle \int^{1}_{0} \sqrt{e^{ \pi}+3x \sqrt{e^{ \pi}+3x \sqrt{e^{ \pi}+3x \sqrt{e^{ \pi}+3x \cdots}}}} dx > \frac{3}{2} [/spoiler]

I know that much so far.

>>6559948
Actually <span class="math"> \displaystyle \int^{1}_{0} \sqrt{e^{ \pi}+3x \sqrt{e^{ \pi}+3x \sqrt{e^{ \pi}+3x \sqrt{e^{ \pi}+3x \cdots}}}} dx > \frac{3}{2} + \frac{ \sqrt{4e^{ \pi}}}{2} [/spoiler]

I have no idea how these magic formulas work, I can't even begin to understand what they could represent, nor find an extremely simple layman's definition

Can anyone help me out here?

Integral from 0 to infinity of F(x)dx, where F(x) is the following piecewise function:

F(x) = 1 if floor(x) is the minimum Godel number of a proof, in ZFC, of the most difficult theorem I've ever proved in ZFC.
F(x) = 0 otherwise.

Answer: The integral is 1. I established this by proving the most difficult theorem I've ever proved in ZFC.

>>6560083
<span class="math"> \displaystyle \int^{1}_{0} \sqrt{e^{ \pi}+3x \sqrt{e^{ \pi}+3x \sqrt{e^{ \pi}+3x \sqrt{e^{ \pi}+3x \cdots}}}} dx [/spoiler]

Use <span class="math"> y = \displaystyle \sqrt{e^{ \pi}+3xy} [/spoiler]

Apply <span class="math"> \displaystyle z = \sqrt{a+bz} \rightarrow z = \frac{b+\sqrt{4a+b^2}}{2} [/spoiler] to the above equation.

So <span class="math"> \displaystyle \int^{1}_{0} \sqrt{e^{ \pi}+3x \sqrt{e^{ \pi}+3x \sqrt{e^{ \pi}+3x \sqrt{e^{ \pi}+3x \cdots}}}} dx = \int^{1}_{0} \frac{3x+ \sqrt{4e^{ \pi}+(3x)^2}}{2} dx [/spoiler]

<span class="math"> \displaystyle \int^{1}_{0} \frac{3x+ \sqrt{4e^{ \pi}+(3x)^2}}{2} dx = \frac{3}{2}+ \frac{1}{2} \int^{1}_{0} \sqrt{4e^{ \pi}+(3x)^2} dx [/spoiler]

<span class="math"> \displacestyle \frac{3}{2}+ \frac{1}{2} \int^{1}_{0} \sqrt{4e^{ \pi}+(3x)^2} dx = \frac{3}{2}+ \frac{ \sqrt{4e^{ \pi}}}{2} \int^{1}_{0} \sqrt{1+ \left( \frac{3x}{ \sqrt{4e^{ \pi}} \right)^{2}} dx [/spoiler]

Use the series formula <span class="math"> \displaystyle \sqrt{1+w} = \sum^{ \infty}_{n=0} \frac{(-1)^{n}(2n)!}{(1-2n)(n!)^{2}(4^{n})}w^{n} [/spoiler]. It is important to note that this formula only works for <span class="math"> |w| \leq 1 [/spoiler], fortunately this is true for all values <span class="math"> 0 \leq x \leq 1 [/spoiler] on <span class="math"> w = \left( \frac{3x}{ \sqrt{4e^{ \pi}}} \right)^{2} [/spoiler]

<span class="math"> \displaystyle \frac{3}{2}+ \frac{ \sqrt{4e^{ \pi}}}{2} \int^{1}_{0} \sqrt{1+ \left( \frac{3x}{ \sqrt{4e^{ \pi}} \right)^{2}} dx = \frac{3}{2}+ \frac{ \sqrt{4e^{ \pi}}}{2} \int^{1}_{0} \sum^{ \infty}_{n=0} \frac{(-1)^{n}(2n)!}{(1-2n)(n!)^{2}(4^{n})}x^{n} dx [/spoiler]

<span class="math"> \displaystyle \frac{3}{2}+ \frac{ \sqrt{4e^{ \pi}}}{2} \int^{1}_{0} \sum^{ \infty}_{n=0} \frac{(-1)^{n}(2n)!}{(1-2n)(n!)^{2}(4^{n})}x^{n} dx = \frac{3}{2}+ \frac{ \sqrt{4e^{ \pi}}}{2} \sum^{ \infty}_{n=0} \frac{(-1)^{n}(2n)!}{(1-2n)(n+1)(n!)^{2}(4^{n})} dx [/spoiler]

I'll post a closed form in a bit.

>>6560083
<span class="math"> \displaystyle \int^{1}_{0} \sqrt{e^{ \pi}+3x \sqrt{e^{ \pi}+3x \sqrt{e^{ \pi}+3x \sqrt{e^{ \pi}+3x \cdots}}}} dx [/spoiler]

Use <span class="math"> \displaystyle y = \sqrt{e^{ \pi}+3xy} [/spoiler]

Apply <span class="math"> \displaystyle z = \sqrt{a+bz} \rightarrow z = \frac{b+\sqrt{4a+b^2}}{2} [/spoiler] to the above equation

So <span class="math"> \displaystyle \int^{1}_{0} \sqrt{e^{ \pi}+3x \sqrt{e^{ \pi}+3x \sqrt{e^{ \pi}+3x \sqrt{e^{ \pi}+3x \cdots}}}} dx = \int^{1}_{0} \frac{3x+ \sqrt{4e^{ \pi}+(3x)^2}}{2} dx [/spoiler]

<span class="math"> \displaystyle \int^{1}_{0} \frac{3x+ \sqrt{4e^{ \pi}+(3x)^2}}{2} dx = \frac{3}{2}+ \frac{1}{2} \int^{1}_{0} \sqrt{4e^{ \pi}+(3x)^2} dx [/spoiler]

<span class="math"> \displaystyle \frac{3}{2}+ \frac{1}{2} \int^{1}_{0} \sqrt{4e^{ \pi}+(3x)^2} dx = \frac{3}{2}+ \frac{ \sqrt{4e^{ \pi}}}{2} \int^{1}_{0} \sqrt{1+ \left( \frac{3x}{ \sqrt{4e^{ \pi}} \right)^{2}} dx [/spoiler]

Use the series formula <span class="math"> \displaystyle \sqrt{1+w} = \sum^{ \infty}_{n=0} \frac{(-1)^{n}(2n)!}{(1-2n)(n!)^{2} (4^{n})}w^{n} [/spoiler]. It is important to note that this formula only works for <span class="math"> |w| \leq 1 [/spoiler], fortunately this is true for all values <span class="math"> 0 \leq x \leq 1 [/spoiler] on <span class="math"> w = \left( \frac{3x}{ \sqrt{4e^{ \pi}}} \right)^{2} [/spoiler]

<span class="math"> \displaystyle \frac{3}{2}+ \frac{ \sqrt{4e^{ \pi}}}{2} \int^{1}_{0} \sqrt{1+ \left( \frac{3x}{ \sqrt{4e^{ \pi}} \right)^{2}}} dx = \frac{3}{2}+ \frac{ \sqrt{4e^{ \pi}}}{2} \int^{1}_{0} \sum^{ \infty}_{n=0} \frac{(-1)^{n}(2n)!}{(1-2n)(n!)^{2} (4^{n})}x^{n} dx [/spoiler]

<span class="math"> \displaystyle \frac{3}{2}+ \frac{ \sqrt{4e^{ \pi}}}{2} \int^{1}_{0} \sum^{ \infty}_{n=0} \frac{(-1)^{n}(2n)!}{(1-2n)(n!)^{2} (4^{n})}x^{n} dx = \frac{3}{2}+ \frac{ \sqrt{4e^{ \pi}}}{2} \sum^{ \infty}_{n=0} \frac{(-1)^{n}(2n)!}{(1-2n)(n+1)(n! )^{2}(4^{n})} [/spoiler]

I'll post a closed form in a bit.

>>6560083
<span class="math"> \displaystyle \int^{1}_{0} \sqrt{e^{ \pi}+3x \sqrt{e^{ \pi}+3x \sqrt{e^{ \pi}+3x \sqrt{e^{ \pi}+3x \cdots}}}} dx [/spoiler]

Use <span class="math"> \displaystyle y = \sqrt{e^{ \pi}+3xy} [/spoiler]

Apply <span class="math"> \displaystyle z = \sqrt{a+bz} \rightarrow z = \frac{b+\sqrt{4a+b^2}}{2} [/spoiler] to the above equation

So <span class="math"> \displaystyle \int^{1}_{0} \sqrt{e^{ \pi}+3x \sqrt{e^{ \pi}+3x \sqrt{e^{ \pi}+3x \sqrt{e^{ \pi}+3x \cdots}}}} dx = \int^{1}_{0} \frac{3x+ \sqrt{4e^{ \pi}+(3x)^2}}{2} dx [/spoiler]

<span class="math"> \displaystyle \int^{1}_{0} \frac{3x+ \sqrt{4e^{ \pi}+(3x)^2}}{2} dx = \frac{3}{2}+ \frac{1}{2} \int^{1}_{0} \sqrt{4e^{ \pi}+(3x)^2} dx [/spoiler]

<span class="math"> \displaystyle \frac{3}{2}+ \frac{1}{2} \int^{1}_{0} \sqrt{4e^{ \pi}+(3x)^2} dx = \frac{3}{2}+ \frac{4e^{ \pi}}{2} \int^{1}_{0} \sqrt{1+ \left( \frac{3x}{ \sqrt{4e^{ \pi}}} \right)^{2}} dx [/spoiler]

Use the series formula <span class="math"> \displaystyle \sqrt{1+w} = \sum^{ \infty}_{n=0} \frac{(-1)^{n}(2n)!}{(1-2n)(n!)^{2} (4^{n})}w^{n} [/spoiler]. It is important to note that this formula only works for <span class="math"> |w| \leq 1 [/spoiler], fortunately this is true for all values <span class="math"> 0 \leq x \leq 1 [/spoiler] on <span class="math"> w = \left( \frac{3x}{ \sqrt{4e^{ \pi}}} \right)^{2} [/spoiler]

<span class="math"> \displaystyle \frac{3}{2}+ \frac{4e^{ \pi}}{2} \int^{1}_{0} \sqrt{1+ \left( \frac{3x}{ \sqrt{4e^{ \pi}}} \right)^{2}} dx = \frac{3}{2}+ \frac{ \sqrt{4e^{ \pi}}}{2} \int^{1}_{0} \sum^{ \infty}_{n=0} \frac{(-1)^{n}(2n)!}{(1-2n)(n!)^{2} (4^{n})}x^{n} dx [/spoiler]

<span class="math"> \displaystyle \frac{3}{2}+ \frac{ \sqrt{4e^{ \pi}}}{2} \int^{1}_{0} \sum^{ \infty}_{n=0} \frac{(-1)^{n}(2n)!}{(1-2n)(n!)^{2} (4^{n})}x^{n} dx = \frac{3}{2}+ \frac{ \sqrt{4e^{ \pi}}}{2} \sum^{ \infty}_{n=0} \frac{(-1)^{n}(2n)!}{(1-2n)(n+1)(n! )^{2}(4^{n})} [/spoiler]

I'll post a closed form later, maybe. Typing <span class="math"> \int \sqrt{1+t^{2}} dx [/spoiler] will tell you what I'd be posting though (with all steps included), so I don't really feel the need to post it.

>>6560083
Apply <span class="math"> \displaystyle z = \sqrt{a+bz} \rightarrow z = \frac{b+\sqrt{4a+b^2}}{2} [/spoiler] to the above equation

So <span class="math"> \displaystyle \int^{1}_{0} \sqrt{e^{ \pi}+3x \sqrt{e^{ \pi}+3x \sqrt{e^{ \pi}+3x \sqrt{e^{ \pi}+3x \cdots}}}} dx = \int^{1}_{0} \frac{3x+ \sqrt{4e^{ \pi}+(3x)^2}}{2} dx [/spoiler]

<span class="math"> \displaystyle \int^{1}_{0} \frac{3x+ \sqrt{4e^{ \pi}+(3x)^2}}{2} dx = \frac{3}{2}+ \frac{1}{2} \int^{1}_{0} \sqrt{4e^{ \pi}+(3x)^2} dx [/spoiler]

<span class="math"> \displaystyle \frac{3}{2}+ \frac{1}{2} \int^{1}_{0} \sqrt{4e^{ \pi}+(3x)^2} dx = \frac{3}{2}+ \frac{4e^{ \pi}}{2} \int^{1}_{0} \sqrt{1+ \left( \frac{3x}{ \sqrt{4e^{ \pi}}} \right)^{2}} dx [/spoiler]

Use the series formula <span class="math"> \displaystyle \sqrt{1+w} = \sum^{ \infty}_{n=0} \frac{(-1)^{n}(2n)!}{(1-2n)(n!)^{2} (4^{n})}w^{n} [/spoiler]. It is important to note that this formula only works for <span class="math"> |w| \leq 1 [/spoiler], fortunately this is true for all values <span class="math"> 0 \leq x \leq 1 [/spoiler] on <span class="math"> w = \left( \frac{3x}{ \sqrt{4e^{ \pi}}} \right)^{2} [/spoiler]

<span class="math"> \displaystyle \frac{3}{2}+ \frac{4e^{ \pi}}{2} \int^{1}_{0} \sqrt{1+ \left( \frac{3x}{ \sqrt{4e^{ \pi}}} \right)^{2}} dx = \frac{3}{2}+ \frac{ \sqrt{4e^{ \pi}}}{2} \int^{1}_{0} \sum^{ \infty}_{n=0} \frac{(-1)^{n}(2n)!}{(1-2n)(n!)^{2} (4^{n})}x^{n} dx [/spoiler]

<span class="math"> \displaystyle \frac{3}{2}+ \frac{ \sqrt{4e^{ \pi}}}{2} \int^{1}_{0} \sum^{ \infty}_{n=0} \frac{(-1)^{n}(2n)!}{(1-2n)(n!)^{2} (4^{n})}x^{n} dx = \frac{3}{2}+ \frac{ \sqrt{4e^{ \pi}}}{2} \sum^{ \infty}_{n=0} \frac{(-1)^{n}(2n)!}{(1-2n)(n+1)(n! )^{2}(4^{n})} [/spoiler]

I'll post a closed form later, maybe. Typing <span class="math"> \int \sqrt{1+t^{2}} dt [/spoiler] int Wolfram Alpha will tell you what I'd be posting though (with all steps included), so I don't really feel the need to post it.

>>6560220
>int Wolfram Alpha
into Wolfram Alpha*

>>6560220
> <div class="math"> = \int^{1}_{0} \frac{3x+ \sqrt{4e^{ \pi}+(3x)^2}}{2} dx </div>
><div class="math"> = \frac{3}{2}+ \frac{1}{2} \int^{1}_{0} \sqrt{4e^{ \pi}+(3x)^2} dx [\eqn]
<div class="math"> \int_0^1 \frac{3x}{2} dx = \frac{3}{4} \neq \frac{3}{2} </div></div>

>>6560226
protip latex doesnt greentext
<div class="math">= \int^{1}_{0} \frac{3x+ \sqrt{4e^{ \pi}+(3x)^2}}{2} dx </div>
<div class="math"> = \frac{3}{2}+ \frac{1}{2} \int^{1}_{0} \sqrt{4e^{ \pi}+(3x)^2} dx [\eqn]
<div class="math">\int_0^1 \frac{3x}{2} dx = \frac{3}{4} \neq \frac{3}{2} </div></div>

>>6560226
It's called a mistake and I intend to fix it.

>>6560220
<div class="math"> = \int^{1}_{0} \frac{3x+ \sqrt{4e^{ \pi}+(3x)^2}}{2} dx </div>
<div class="math"> = \frac{3}{2}+ \frac{1}{2} \int^{1}_{0} \sqrt{4e^{ \pi}+(3x)^2} dx </div>
<div class="math"> \int_0^1 \frac{3x}{2} dx = \frac{3}{4} \neq \frac{3}{2} </div>

>>6560236
just making sure you saw it

also, this
>>6560238
is in reference to my failed latex attempts

>>6560238
Wow, I didn't even see that a made a mistake that trivial. There is another mistake in there I have to fix too.

>>6560220
This is wrong.

>>6559231
<span class="math"> \displaystyle \int^{1}_{0} \sqrt{e^{ \pi}+3x \sqrt{e^{ \pi}+3x \sqrt{e^{ \pi}+3x \sqrt

{e^{ \pi}+3x \cdots}}}} dx [/spoiler]

Use <span class="math"> \displaystyle y = \sqrt{e^{ \pi}+3xy} [/spoiler]

Apply <span class="math"> \displaystyle z = \sqrt{a+bz} \rightarrow z = \frac{b+\sqrt{4a+b^2}}{2} [/spoiler] to

the above equation

So <span class="math"> \displaystyle \int^{1}_{0} \sqrt{e^{ \pi}+3x \sqrt{e^{ \pi}+3x \sqrt{e^{ \pi}+3x

\sqrt{e^{ \pi}+3x \cdots}}}} dx = \int^{1}_{0} \frac{3x+ \sqrt{4e^{ \pi}+(3x)^2}}{2} dx [/spoiler]

<span class="math"> \displaystyle \int^{1}_{0} \frac{3x+ \sqrt{4e^{ \pi}+(3x)^2}}{2} dx = \frac{3}{2}+

\frac{1}{2} \int^{1}_{0} \sqrt{4e^{ \pi}+(3x)^2} dx [/spoiler]

>>6560261
Gog fucking dammit.

>>6559231


<span class="math"> \displaystyle \int^{1}_{0} \sqrt{e^{ \pi}+3x \sqrt{e^{ \pi}+3x \sqrt{e^{ \pi}+3x \sqrt

{e^{ \pi}+3x \cdots}}}} dx [/spoiler]

Use <span class="math"> \displaystyle y = \sqrt{e^{ \pi}+3xy} [/spoiler]

Apply <span class="math"> \displaystyle z = \sqrt{a+bz} \rightarrow z = \frac{b+\sqrt{4a+b^2}}{2} [/spoiler] to

the above equation

So <span class="math"> \displaystyle \int^{1}_{0} \sqrt{e^{ \pi}+3x \sqrt{e^{ \pi}+3x \sqrt{e^{ \pi}+3x

\sqrt{e^{ \pi}+3x \cdots}}}} dx = \int^{1}_{0} \frac{3x+ \sqrt{4e^{ \pi}+(3x)^2}}{2} dx [/spoiler]

<span class="math"> \displaystyle \int^{1}_{0} \frac{3x+ \sqrt{4e^{ \pi}+(3x)^2}}{2} dx = \frac{3}{4}+

\frac{1}{2} \int^{1}_{0} \sqrt{4e^{ \pi}+(3x)^2} dx [/spoiler]

<span class="math"> \displaystyle \frac{3}{4}+ \frac{1}{2} \int^{1}_{0} \sqrt{4e^{ \pi}+(3x)^2} dx = \frac

{3}{4}+ \frac{ \sqrt{4e^{ \pi}}}{2} \int^{1}_{0} \sqrt{1+ \left( \frac{3x}{ \sqrt{4e^{ \pi}}}

\right)^{2}} dx [/spoiler]

Use the series formula <span class="math"> \displaystyle \sqrt{1+w} = \sum^{ \infty}_{n=0} \frac{(-1)^{n}

(2n)!}{(1-2n)(n!)^{2} (4^{n})}w^{n} [/spoiler]. It is important to note that this formula only

works for <span class="math"> |w| \leq 1 [/spoiler], fortunately this is true for all values <span class="math"> 0 \leq x

\leq 1 [/spoiler] on <span class="math"> w = \left( \frac{3x}{ \sqrt{4e^{ \pi}}} \right)^{2} [/spoiler]

<span class="math"> \displaystyle \frac{3}{4}+ \frac{ \sqrt{4e^{ \pi}}}{2} \int^{1}_{0} \sqrt{1+ \left(

\frac{3x}{ \sqrt{4e^{ \pi}}} \right)^{2}} dx = \frac{3}{4}+ \frac{ \sqrt{4e^{ \pi}}}{2} \int^

{1}_{0} \sum^{ \infty}_{n=0} \frac{(-1)^{n}(2n)!}{(1-2n)(n!)^{2} (4^{n})}x^{n} dx [/spoiler]
feel the need to post it.

>>6560267
<span class="math"> \displaystyle \frac{3}{4}+ \frac{ \sqrt{4e^{ \pi}}}{2} \int^{1}_{0} \sum^{ \infty}_{n=0}

\frac{(-1)^{n}(2n)!}{(1-2n)(n!)^{2} (4^{n})} \left( \frac{3x}{ \sqrt{4e^{ \pi}}}\right)^{n} dx

= \frac{3}{4}+ \frac{9 \sqrt{4e^{ \pi}}}{8e^{ \pi}} \sum^{ \infty}_{n=0} \frac{(-1)^{n}(2n)!}

{(1-2n)(n+1)(n! )^{2}(4^{n})} [/spoiler]

I'll post a closed form later, maybe. Typing <span class="math"> \int \sqrt{1+t^{2}} dt [/spoiler] into Wolfram

Alpha will tell you what I'd be posting though (with all steps included), so I don't really feel the need to post it.

I think I need to take a break before I go insane with all my stupid errors.

>>6560267
<span class="math"> \displaystyle \frac{3}{4}+ \frac{ \sqrt{4e^{ \pi}}}{2} \int^{1}_{0} \sqrt{1+ \left( \frac{3x}{ \sqrt{4e^{ \pi}}} \right)^{2}} dx = \frac{3}{4}+ \frac{ \sqrt{4e^{ \pi}}}{2} \int^ {1}_{0} \sum^{ \infty}_{n=0} \frac{(-1)^{n}(2n)!}{(1-2n)(n!)^{2} (4^{n})}x^{n} dx [/spoiler]

Should be:

<span class="math"> \displaystyle \frac{3}{4}+ \frac{ \sqrt{4e^{ \pi}}}{2} \int^{1}_{0} \sqrt{1+ \left( \frac{3x}{ \sqrt{4e^{ \pi}}} \right)^{2}} dx = \frac{3}{4}+ \frac{ \sqrt{4e^{ \pi}}}{2} \int^{1}_{0} \sum^{ \infty}_{n=0} \frac{(-1)^{n}(2n)!}{(1-2n)(n!)^{2} (4^{n})} \left( \frac{3x}{ \sqrt{4e^{ \pi}}}\right)^{n} dx [/spoiler]

>>6560267
And that's still wrong... I can't get anything done right today.

A thread where I can contriboot! Please halp!

>>6560282
Starting from this last step use u-substitution to yield the following conclusion:

<span class="math"> \displaystyle \int^{1}_{0} \sqrt{e^{ \pi}+3x \sqrt{e^{ \pi}+3x \sqrt{e^{ \pi}+3x \sqrt {e^{ \pi}+3x \cdots}}}} dx = \frac{3}{4}+ \frac{2e^{ \pi}}{3} \sum^{ \infty}_{n=0} \frac{(-1)^{n}(2n)!} {(1-2n)(n+1)(n! )^{2}(4^{n})} \left( \frac{3}{ \sqrt{4e^{ \pi}}}\right)^{n+1} [/spoiler]

It took way to long but I think that's it.

>>6560267

Simply use that method for recursion shown in the posts above by inserting an extra variable (e^pi +3xy = y) to obtain a closed expression for the function; then just integrate.

I'd demonstrate it, but I don't know LaTeX.

>>6560610
I already did that. It becomes:

<span class="math"> \int^{1}_{0} \frac{3x+ \sqrt{4e^{ \pi}+(3x)^2}}{2} dx [/spoiler]

Type <span class="math"> \int \sqrt{1+t^{2}} dt [/spoiler] into Wolfram Alpha to see the form of the closed expression. There are allot of steps and I don't feel like doing it right now.

>>6560616

Oh, my apologies! Didn't catch the first square root, so I thought it simplified to something much easier. I should've written sqrt(e^pi + 3xy) = y, then it becomes the mess you showed.

>>6560630
Haha, I bet you were looking for something elegant like what is shown in the other examples!

I understand what you were thinking though, I thought it would be much nicer too.

>>6560433
That one looks like quite a doozy.

>>6560642
The book said "It's easy to demonstrate that: [...]".
I suppose we're all pretty dumb.

>>6560779
What book are you reading?

>>6560779
I don't get why books do this. Purposely handwave and trivialize stuff that isn't trivial. It's as if the author thinks the reader is going to step back and say "wow this author must be really smart to say all this stuff is easy".

>>6559583
What book are you using to study real analysis?

>>6560098
People in this thread are finding the areas of complicated shapes

>>6560572
Still wrong... Fuck me with a rake.
<span class="math"> \displaystyle \int^{1}_{0} \sqrt{e^{ \pi}+3x \sqrt{e^{ \pi}+3x \sqrt{e^{ \pi}+3x \sqrt {e^{ \pi}+3x \cdots}}}} dx = \frac{3}{4}+ \frac{2e^{ \pi}}{3} \sum^{ \infty}_{n=0} \frac{(-1)^{n}(2n)!} {(1-2n)(2n+1)(n! )^{2}(4^{n})} \left( \frac{3}{ \sqrt{4e^{ \pi}}}\right)^{2n+1} [/spoiler]

>>6561927
>cant integrate sqrt(1 + x^2) without series

>>6561955
It's been a long time since calc 2, sorry for such a shamefur dispray sensei.

I got this integral on a test a couple of days ago:

<span class="math">
\int_0^{2\pi}\frac{1}{a^2-2a\sin(\theta)+1}d\theta
[/spoiler]

for <span class="math">a>1[/spoiler].

It's not really that difficult, but I wouldn't have been able to solve it last year.

>>6561966
<span class="math">
\int_0^{2\pi}\frac{1}{a^2-2a\sin(\t heta)+1}d\theta
[/spoiler]

>>6561966
<span class="math"> \int_0^{2 \pi} \frac{1}{a^2-2a \sin( \t heta)+1}d \theta [/spoiler]
Put spaces before your \ commands.

>>6561961
haha, i just thought it was funny: the series makes it look much harder

I got this integral on a test the other day:
<span class="math">
\int_0^{2\pi}\frac{1}{a^2-2a\sin(\t heta)+1}d\theta
[/spoiler]
for <span class="math">a>1[/spoiler].

It's not really that difficult, but I wouldn't have been able to solve it last year.

>>6561970
Well, I had the series for the square root memorized so it meant I would have to do less work. At least that was the plan. My math skills have been so rusty that what should have taken minutes took much longer.

>>6561972
<span class="math"> \int_0^{2 \pi} \frac{1}{a^2-2a \sin( \theta)+1}d \theta [/spoiler]
for a > 1

I got this integral on a test the other day:
<span class="math">
\int_0^{2\pi}\frac{1}{a^2-2a\sin(\theta)+1}d\theta
[/spoiler]
for <span class="math">a>1[/spoiler].

It's not really that difficult, but I wouldn't have been able to solve it last year.

I remember around 2 years ago when the last question on an exam was ridiculous. I pretty much spent an hour solving it (and most of the time was dedicated to the integration.) I remember lots of logarithms, and lots of "cheap tricks" (i.e. functions within functions within functions, situations where simplification actually complicates the problem, more terms added as you progress, and some physics babble) in an effort to make it more probable that you fuck up and get the whole thing wrong.

The answer was several pages long and remarkably, I somehow got it right (and was the only one in class who did so) The answer was pretty simple, with a pi fraction minus some e^gamma bullshit.

I still don't know how I did it. I wasn't having a good day and I just kind of "went with the flow" for that motherfucker.

>>6556896
(3e^x^2)/x^3

Havent done intergration in years just trying to reverse engineer it from what i remember doing for differentiating something like that basically it was a guess

>>6562142
<span class="math"> \int e^{x^2}dx [/spoiler] doesn't have a closed form solution.

>>6562150
After i gave it a go i typed it into wolfram come out with this and i just dont understand how root pi even comes into it, makes me feel so stupid sometimes

>>6562142
>>6562150

maybe it was supposed to be
<span class="math">\int e^{x^3} x^2 dx[/spoiler]

>>6562152
Don't feel stupid, the sqrt(pi) comes from the mostly arbitrary definition of erfi(x). It's a natural normalization when you compare with e^(-x^2/2), whose integral over R is finite and equals sqrt(2pi).

>>6562161
Oh right I see, I've just dug out my old maths text books and I think I can see why I was wrong, thanks

>>6559122
f(x) = 1 if x is irrational
=0 if x is rational
<div class="math"> \int f(x) \, \mathrm{d} x </div>?

>>6559630
Bumping for worked solution.

>>6562193
Doing this integral should be the same as doing <span class="math">\int 1 dx[/spoiler], right?

The rationals are countable, so the difference between <span class="math">\int f(x) dx[/spoiler] and <span class="math">\int 1 dx[/spoiler] over any closed interval should be 0.

<div class="math">\int _0^1 f(x) \mathrm {d}x</div>>pic related is <span class="math">f[/spoiler]

>>6556881
/r/ing the 13-integral for string theory

http://www.durofy.com/5-most-beautiful-questions-from-integral-calculus/

This link has some pretty neat stuff

>>6562974
> beautiful
lol no those integrals are disgusting

Beautiful integrals are stuff like the Dirichlet and Gaussian integrals, the gamma integral for n! etc

http://www.wolframalpha.com/input/?_=1398440565192&i=int+%28x^3+%2b+y^3+%2b+z^3%29%2fsqrt%28x^2+%2b+y^2+%2b+z^2%29+dx+dy+dz&fp=1&incTime=true

/thread

>triple integrals
>not even once

>>6562982
Agreed. Those are some of the ugliest, dumbest, and most worthless integrals I have ever seen outside of high school calculus.

>www.durofy.com
This site is now blocked on my computer

>>6562982
These aren't beautiful, they don't even seem insightful.

>>6562193
>>6562824
Why don't you read the threads before posting in it?

>>6556911 and its replies

>>6563427
seeing as how I wrote both of them
and some of the replies
and they are different
and didn't get an answer to>>6559122
so I relinked it with the new integral

maybe you should read the replies closer before accusing others

>>6563427
People answer the same questions numerous times in math even when knowing it's something that has already been properly answered with a proof. We do it not because we have to, but because it's fun.