Integrals

[Anonymous No.4229278]

Let's see some integrals, /sci/.

<span class="math">\int_0^{2\pi} e^{\cos\theta}\cos(n \theta - \sin\theta) = \frac{2\pi}{n!}[/spoiler]

[Anonymous No.4229299]

>this thread makes me moist
<div class="math">\int_{0}^{1} \frac{dx}{x^x} = \sum_{n=1}^{\infty} \frac{1}{n^n}[/moot]

[Josef crossed out the Atlantic !!nUf2NflSAyw No.4229322]

Borwein integrals!
<div class="math"> \int_0^\infty \frac{\sin(x)}{x} \, dx=\pi/2 [/moot]<div class="math"> \int_0^\infty \frac{\sin(x)}{x}\frac{\sin(x/3)}{x/3} \, dx = \pi/2 [/moot]<div class="math"> \int_0^\infty \frac{\sin(x)}{x}\frac{\sin(x/3)}{x/3}\frac{\sin(x/5)}{x/5} \, dx = \pi/2 [/moot]<div class="math">\quad\vdots[/moot]<div class="math"> \int_0^\infty \frac{\sin(x)}{x}\frac{\sin(x/3)}{x/3}\cdots\frac{\sin(x/13)}{x/13} \, dx = \pi/2 [/moot]<div class="math">
\int_0^\infty \frac{\sin(x)}{x}\frac{\sin(x/3)}{x/3}\cdots\frac{\sin(x/15)}{x/15} \, dx = \frac{467807924713440738696537864469}{935615849440640907310521750000}\pi [/moot]
http://en.wikipedia.org/wiki/Borwein_integral

[Anonymous No.4229331]

solve it without mathematica

<div class="math">\int_{0}^{1} \frac{1}{x}-floor(\frac{1}{x})[/moot]

..go

[Colonel Coffee Mug !phJ7yIcs.Q No.4229334]

>>4229322

It's funny because the slashes almost line up.

[Anonymous No.4229341]

>>4229331
<div class="math">\int_{0}^{1} \frac{1}{x}-floor(\frac{1}{x}) \, dt= \frac{1}{x}-floor(\frac{1}{x})[/moot]

[Anonymous No.4229347]

>>4229322
otherwise known as troll integrals

[Anonymous No.4229351]

>>4229341
0/10

[Josef crossed out the Atlantic !!nUf2NflSAyw No.4229363]

>>4229331
Is there some nice solution to this? Also \lfloor, \rfloor:
<div class="math">\int_0^1\mathrm dx\left(\frac1x-\left\lfloor\frac1x\right\rfloor\right)[/moot]

[Anonymous No.4229371]

<div class="math">\int_{1}^{e} \frac{1}{x} dx = 1[/moot]

Babby-tier, inbound.

[Anonymous No.4229379]

>>4229347
this one always gets me hard
<div class="math">\int_{0}^{\pi}\int_{0}^{\pi}\int_{0}^{\pi} \frac{du\, dv\, dw}{3-\cos (u)-\cos (v)-\cos (w)}=\frac{\sqrt{6}}{96} \Gamma\left ( \frac{1}{24}\right ) \Gamma\left ( \frac{5}{24}\right ) \Gamma\left ( \frac{7}{24}\right ) \Gamma\left ( \frac{11}{24}\right ) [/moot]

[Anonymous No.4229383]

>>4229379
http://mathworld.wolfram.com/WatsonsTripleIntegrals.html

[Anonymous No.4229385]

>>4229322
Interesting fact, those terms are even so the integral from negative infinity to positive infinity is pi.

[Anonymous No.4229386]

>>4229331
is this some kind of troll math? it froze my mathcad

[Anonymous No.4229387]

<div class="math">s=\int \sqrt{g_{\mu \nu}dx^\mu dx^\nu }[/moot]

[Anonymous No.4229397]

<div class="math">\int_0^1\mathrm dx\left(\frac1x-\left\lfloor\frac1x\right\rfloor\right)=\sum_{k=1}^{\infty} \int_{1 \over k+1}^{1 \over k}\mathrm dx\left(\frac1x-\left\lfloor\frac1x\right\rfloor\right) =\sum_{k=1}^{\infty}\left( \ln \left( {k+1 \over k} \right) +1 \right)= \infty[/moot]

[Anonymous No.4229409]

>>4229397
<div class="math">\int_0^1\mathrm dx\left(\frac1x-\left\lfloor\frac1x\right\rfloor\right)=\sum_{k=1}^{\infty} \int_{1 \over k+1}^{1 \over k}\mathrm dx\left(\frac1x-\left\lfloor\frac1x\right\rfloor\right) =\sum_{k=1}^{\infty} \ln \left( {k+1 \over k} \right) = \lim_{k \to \infty} \ln(k+1)=\infty[/moot]

>fixed

[Anonymous No.4229415]

>>4229409
Sounds like what I'd do, indeed.

[Josef crossed out the Atlantic !!nUf2NflSAyw No.4229425]

>>4229397
Since <span class="math">\frac1x\geq\left\lfloor\frac1x\right\rfloor[/spoiler], <span class="math">0\leq\frac1x-\left\lfloor\frac1x\right\rfloor[/spoiler]; because in general <span class="math">x-\lfloor x\rfloor\leq1[/spoiler] we get the estimate <span class="math">0\leq\frac1x-\left\lfloor\frac1x\right\rfloor<1[/spoiler], so the integral is finite.
Not sure where the error in the calculation above is though.

[Anonymous No.4229427]

<div class="math">
\int_0^1\mathrm dx\left(\frac1x-\left\lfloor\frac1x\right\rfloor\right)=\sum_{k=1}^{\infty} \int_{1 \over k+1}^{1 \over k}\mathrm dx\left(\frac1x-\left\lfloor\frac1x\right\rfloor\right) =\sum_{k=1}^{\infty}\left( \ln \left( {k+1 \over k} \right) +{1 \over k} \right)= \gamma[/moot]

[Anonymous No.4229429]

>>4229427
-1/k instead of +1/k obviously.

[Josef crossed out the Atlantic !!nUf2NflSAyw No.4229434]

>>4229429
Dammit. I knew it looked like the integral definition of <span class="math">\gamma[/spoiler], but couldn't find out how to get there.

[Anonymous No.4229438]

Fuckin' nerds.

[Josef crossed out the Atlantic !!nUf2NflSAyw No.4229447]

>>4229427
On second thought, why's that <span class="math">\gamma[/spoiler]? The argument of the log is kind of weird. Plus my numerical results say there's something like <span class="math">0.42[/spoiler] coming out, whereas <span class="math">\gamma[/spoiler] is <span class="math">0.58[/spoiler].

[Anonymous No.4229467]

>>4229447
<div class="math">
\sum_{k=1}^{\infty}\left( \ln \left( {k+1 \over k} \right) -{1 \over k} \right)=\lim_{K \to \infty}\sum_{k=1}^{K}\left( \ln \left( {k+1 \over k} \right) -{1 \over k} \right)=\lim_{K \to \infty} \left( \sum_{k=1}^{K} \ln \left( {k+1 \over k} \right) - \sum_{k=1}^{K} {1 \over k} \right)=\lim_{K \to \infty} \left( ln(K+1) - \sum_{k=1}^{K} {1 \over k} \right)=\lim_{K \to \infty} \left( ln(K) - \sum_{k=1}^{K} {1 \over k} \right)=\gamma
[/moot]

[Josef crossed out the Atlantic !!nUf2NflSAyw No.4229477]

>>4229467
Alright, all doubts gone, congratulations <span class="math">\ddot\smile[/spoiler]

[Anonymous No.4229494]

>>4229438
<div class="math">\frac{2}{\pi}\int_{0}^{\pi /2}\frac{1}{\left ( \theta ^2+\ln \left ( \cos\left ( \theta\right ) \right ) ^2\right ) ^{2n+2}}\, \sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\cdots +\frac{1}{2}\sqrt{\frac{\ln\left ( \cos\left ( \theta\right ) \right ) ^2}{\theta ^2 + \ln\left ( \cos\left ( \theta\right ) \right ) ^2}}}}\, d\theta = \ln (2)^{\frac{1}{2^n}}[/moot]

[Anonymous No.4229526]

>>4229427
how did you go from the floor of 1/x to 1/k?

[Anonymous No.4229528]

>>4229438
>>4229438
<div class="math">\frac{2}{\pi}\, \int_{0}^{\pi /2}\frac{1}{\left ( \theta ^2+\ln \left ( \cos\left ( \theta\right ) \right ) ^2\right ) ^{2n+2}}\, \sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\cdots +\frac{1}{2}\sqrt{\frac{\ln\left ( \cos\left ( \theta\right ) \right ) ^2}{\theta ^2 + \ln\left ( \cos\left ( \theta\right ) \right ) ^2}}}}}\, d\theta = \ln (2)^{\frac{1}{2^n}}[/moot]

[Anonymous No.4229538]

>>4229526
<div class="math">
\int_{1 \over k+1}^{1 \over k} \left\lfloor\frac1x\right\rfloor \mathrm dx=\int_{1 \over k+1}^{1 \over k} (k+1) \mathrm dx=(k+1)\left( {1 \over k} - {1 \over k+1} \right)={1 \over k}
[/moot]

[Anonymous No.4229552]

>>4229278

That one doesn't look right. Not saying it's wrong, just very unintuitive. Can anyone point me to a derivation?

[Anonymous No.4229641]

>>4229552
I think this works.
<div class="math">\int_0^{2\pi}e^{\cos\theta}\cos(n\theta-\sin\theta)d\theta = \text{Re}\left [\int_0^{2\pi}e^{\cos\theta}e^{i(n\theta-\sin\theta)}d\theta\right] = \text{Re}\left[\int_0^{2\pi}e^{in\theta}e^{\cos\theta-i\sin\theta}d\theta\right]=\text{Re}\left[\int
_0^{2\pi}\left(e^{i\theta}\right)^ne^{e^{-i\theta}}d\theta\right]=\text{Re}\left[\oint_{|z|=1}z^ne^{
\frac{1}{z}}\frac{dz}{iz}\right]=\text{Re}\left[\frac{1}{i}\cdot2\pi i\ \text{Res}\left(z^{n-1}e^{\frac{1}{z}},0\right)\right]=\frac{2\pi}{n!}[/moot]

[Anonymous No.4229650]

>>4229641
welp, take two
<div class="math">\int_0^{2\pi}e^{\cos\theta}\cos(n\theta-\sin\theta)d\theta = Re\left [\int_0^{2\pi}e^{\cos\theta}e^{i(n\theta-\sin\theta)}d\theta\right] = Re\left[\int_0^{2\pi}e^{in\theta}e^{\cos\theta-i\sin\theta}d\theta\right]=Re\left[\int_0^{2\pi}\left
(e^{i\theta}\right)^ne^{e^{-i\theta}}d\theta\right]=Re\left[\oint_{|z|=1}z^ne^{\frac{1}{z}}\frac{dz}
{iz}\right]=Re\left[\frac{1}{i}\cdot2\pi i\ Res\left(z^{n-1}e^{\frac{1}{z}},0\right)\right]=\frac{2\pi}{n!}[/moot]

[Anonymous No.4229652]

>>4229641
\mathrm?

[Anonymous No.4229681]

Where do you guys learn to solve these? I'm studying maths at uni and I haven't the faintest idea of where to start. Any books you recommend?

[Anonymous No.4229702]

<div class="math">\int_{\pi /2}^{\pi /4} \ln (\ln (\tan (x)))\, dx =\frac{\pi}{4}\, \ln\left ( \frac{4\pi ^3}{\Gamma\left ( \frac{1}{4}\right ) ^4\right ) [/moot]

[Anonymous No.4229709]

<div class="math">\int_{\pi /2}^{\pi /4} \ln (\ln (\tan (x)))\, dx =\frac{\pi}{4}\, \ln\left ( \frac{4\pi ^3}{\Gamma\left ( \frac{1}{4}\right ) ^4}\right ) [/moot]

[Anonymous No.4229769]

>>4229650

Yeah I guess. That's pretty cool.

[Anonymous No.4229810]

>>4229650
Residues are magic.

[Anonymous No.4229826]

>>4229538
thanks. (I accidentally set the floor to 1/(k+1) while do it)

[Anonymous No.4229830]

>>4229681
most of it is just calc 2 and freshman complex analysis

[Anonymous No.4229847]

<div class="math">\int_{0}^{1}\int_{0}^{1} \frac{x-1}{\ln (xy) -xy\ln (xy)}\, dx\, dy = \gamma[/moot]
<div class="math">\int_{0}^{1}\int_{0}^{1} \frac{x-1}{\ln (xy) +xy\ln (xy)}\, dx\, dy = \ln\left ( \frac{4}{\pi}\right )[/moot]

>i came

[Anonymous No.4229915]

<div class="math">\int_{0}^{1}\frac{\arctan\left ( \sqrt{x^2+1}\right ) }{(x^2+1)^{\frac{3}{2}}}\, \mathrm{d}x=\frac{\left ( 1-2\sqrt{2}\right ) \pi +6\sqrt{2}\arctan\left ( \sqrt{2}\right ) }{4}[/moot]
<div class="math">\int_{0}^{1}\frac{\arctan\left ( \sqrt{x^2+2}\right ) }{(x^2+1)\sqrt{x^2+2}}\, \mathrm{d}x=\frac{5}{96}\pi ^2[/moot]

[Anonymous No.4229928]

<div class="math">\int e^x =\frac{f(u)^n}{\mathrm{d}x}[/moot]

[Anonymous No.4229931]

<div class="math">\int_0^{\pi/2}\frac{dx}{1+\mathrm{tan}(x)^\alpha} = \frac{\pi}{4}[/moot]

[Anonymous No.4229972]

>>4229931
>implying that it is true for all <span class="math">\alpha[/spoiler]
>ok, it's true for <span class="math">\alpha \in \mathbb{R}[/spoiler]
>let <span class="math">\alpha = \infty[/spoiler]

[Anonymous No.4229983]

>>4229972

[Anonymous No.4229988]

>>4229972
You're retarded. Think about what the graph of 1/(1+tan(x)^infinity) would look like. It would be equal to 1 on the interval [0,pi/4), 1/2 at pi/4, and 0 on the interval (pi/4,pi/2]. If you integrate that you still get pi/4.

>also, it is true for all complex alpha

[Anonymous No.4229995]

>>4229983
>saging a math thread with many interesting identities to prove
>math and science board

[Anonymous No.4230006]

>>4229995
more like responding to some inane post of someone who is unable to understand simple postulates of the real number system without bumping the thread like an asinine idiot.

it's called a polite sage, as nothing was contributed.

[Anonymous No.4230044]

<div class="math">-\sqrt{3}\int_{0}^{1}\frac{\ln (x)}{1-x+x^2}\, \mathrm{d}x=2\sum_{k=1}^{\infty}\frac{\sin\left ( \frac{1}{3}k\pi\right ) }{k^3}[/moot]

[Anonymous No.4230079]

>>4229538
then at k = 1
<div class="math">\int_{\frac{1}{2}}^1 \floor{\frac{1}{x}} dx = 1[/moot]? shouldn't it be 1/2?

[Anonymous No.4230093]

>>4229538
then at k = 1
<div class="math">\int_{\frac{1}{2}}^1 \left\lfloor\frac{1}{x}\right\rfloor dx = 1[/moot]
? shouldn't it be 1/2?

[Anonymous No.4230144]

>>4230044
I would think that this is pretty easy to prove, as sin(k pi/3)=0 for k=0 mod 3, the sum simplifies to two alternating series and a factor of sqrt(3)/2, which cancels out the coefficients on both sides (and getting rid of that sine). Don't know where to go from there, thoguh