/sci/ - Science & Math

sorry for being lazy about the pic...

Let's make it more interesting. Do this instead:

<span class="math">\int_{-\infty }^{\infty } e^{-ax^{2}}= \sqrt{\pi }[/spoiler]

>>1414874
<span class="math">\int_{-\infty }^{\infty } e^{-ax^{2}}[/spoiler]

oops, copy-pasta without fixing the eq...

I tried it and after 5 second, I realized I need to do separation of variable so no.

>>1414913
Bah... I am not being very careful with this. You shouldn't need to do separation of variables because there is only ONE variable, <span class="math">x[/spoiler].

<span class="math">\int_{-\infty }^{\infty } e^{-a\cdot x^{2}}dx[/spoiler]

where <span class="math">a>0[/math and <span class="math">e[/spoiler] is the natural logarithm base.[/spoiler]

>>1414838
you let c equal o to infinity of e^-x^2, and then you do it again with c equal to e^-y^2, then multiple them together to get c^2= (0 to infinity) integral of e^-(x^2+y^2)dxdy.

then change to polar to get c^2= (0 to infinity)(0 to pi/2) integral of re^-(r^2) dr d(theta).

that looks like shit, but anyway you will get sqrt(pi)/2, so just multiple by 2 to get the other side of infinity.

>>1414940
where <span class="math">a>0[/spoiler] and <span class="math">e[/spoiler] is the natural logarithm base.

>>1414940
>>1414913
wait, now I realized this is one of those function without integral durp.

>>1414913
"Separation of variables"? I thought such a term existed only for DEs.

Whatever, here's the proof, I hope I didn't introduce too many typos.

First note that, because <span class="math">e^x>0[/spoiler], <span class="math">\int_{-\infty}^\infty e^{-x^2} \mathrm dx > 0[/spoiler].
Now look at
<div class="math">\left(\int_{-\infty}^\infty e^{-x^2} \mathrm dx\right)^2</div>
i.e. the square of the integral. Tear it apart ...
<div class="math">= \left(\int_{-\infty}^\infty e^{-x^2} \mathrm dx\right)\left( \int_{-\infty}^\infty e^{-x^2} \mathrm dx\right)</div>
... rename x to y in the second integral ...
<div class="math">= \int_{-\infty}^\infty e^{-x^2} \mathrm dx \int_{-\infty}^\infty e^{-y^2} \mathrm dy</div>
... bring the independant integrals together ...
<div class="math">= \int_{-\infty}^\infty\int_{-\infty}^\infty e^{-(x^2+y^2)} \mathrm dx \mathrm dy</div>
... transform (x,y) to polar coordinates <span class="math">(r,\varphi)[/spoiler]; Jacobi determinant = r ...
<div class="math">= \int_0^\infty\int_0^\{2\pi} e^{-r^2} r \mathrm d\varphi \mathrm dr</div>
... calculate the <span class="math">\varphi[/spoiler] integral ...
<div class="math">= 2\pi \int_0^\infty e^{-r^2} r \mathrm dr</div>
... calculate the r integral, the JacobiDet-factor makes it possible to integrate the Gaussian ...
<div class="math">= 2\pi \left[-\frac{1}{2}e^{-r^2}\right]_0^\infty</div>
... and there you go,
<div class="math">= 2\pi (0+\frac{1}{2})
= \pi</div>

Now, as in the beginning I noticed that the integral is larger than zero, the positive root of this is the value of the integral, thus
<div class="math">\int_{-\infty}^\infty e^{-x^2} \mathrm dx = \sqrt \pi</div>

>>1414913
"Separation of variables"? I thought such a term existed only for DEs.

Whatever, here's the proof, I hope I didn't introduce too many typos.

First note that, because <span class="math">e^x>0[/spoiler], <span class="math">\int_{-\infty}^\infty e^{-x^2} \mathrm dx > 0[/spoiler].
Now look at
<div class="math">\left(\int_{-\infty}^\infty e^{-x^2} \mathrm dx\right)^2</div>
i.e. the square of the integral. Tear it apart ...
<div class="math">= \left(\int_{-\infty}^\infty e^{-x^2} \mathrm dx\right)\left( \int_{-\infty}^\infty e^{-x^2} \mathrm dx\right)</div>
... rename x to y in the second integral ...
<div class="math">= \int_{-\infty}^\infty e^{-x^2} \mathrm dx \int_{-\infty}^\infty e^{-y^2} \mathrm dy</div>
... bring the independant integrals together ...
<div class="math">= \int_{-\infty}^\infty\int_{-\infty}^\infty e^{-(x^2+y^2)} \mathrm dx \mathrm dy</div>
... transform (x,y) to polar coordinates <span class="math">(r,\varphi)[/spoiler]; Jacobi determinant = r ...
<div class="math">= \int_0^\infty\int_0^{2\pi} e^{-r^2} r \mathrm d\varphi \mathrm dr</div>
... calculate the <span class="math">\varphi[/spoiler] integral ...
<div class="math">= 2\pi \int_0^\infty e^{-r^2} r \mathrm dr</div>
... calculate the r integral, the JacobiDet-factor makes it possible to integrate the Gaussian ...
<div class="math">= 2\pi \left[-\frac{1}{2}e^{-r^2}\right]_0^\infty</div>
... and there you go,
<div class="math">= 2\pi (0+\frac{1}{2})
= \pi</div>

Now, as in the beginning I noticed that the integral is larger than zero, the positive root of this is the value of the integral, thus
<div class="math">\int_{-\infty}^\infty e^{-x^2} \mathrm dx = \sqrt \pi</div>

>>1415008
Excellent job.

>>1415008
oh wow, that's amazing!

>"Separation of variables"? I thought such a term existed only for DEs.
I'm taking DE during summer and been going to classes while sick and my head's still not working right.

>>1415008
thats clever, though i think they meant "integration by parts" i.e. u*dv = u*v- v*du

>>1415038
wow, how you read my mind bro, this thread is filled with amazing.

>>1415038
you cannot, that is impossible.

>>1415038
This is the standard way of solving the integral (which is basically a change of coordinates). I don't think you can make integration by parts work for it.

>>1415063
i think the erf might work.

>>1415075
Renaming the equation won't really help finding its solution, huh?

>>1415111
never mind the error function is just a transformation that is basically defined by ops equation.

>>1415008
Hi guys... I just took a Business Calc test today on logarithmic functions and such... I'll just be on my way then....

To take this even further, let's calculate
<div class="math">\int_0^\infty x^n e^{-ax^2} \mathrm dx</div>
... substitute <span class="math">x(t) = \sqrt{t/a}[/spoiler] to get
<div class="math">= \int_0^\infty \sqrt{\frac{t}{a}}^n e^{-t} \mathrm dt \frac{1}{2\sqrt{\frac{t}{a}}}</div>
... take constants out of the integral, get
<div class="math">= \frac{1}{2} a^{-\frac{n-1}{2}} \int_0^\infty t^{\frac{n-1}{2}} e^{-t} \mathrm dt</div>
Now compare the integral with the Gamma function, <span class="math">\Gamma(x) = \int_0^\infty e^{-t} t^{x-1}\mathrm dt[/spoiler]; you thus see that
<div class="math">= \frac{1}{2} a^{-\frac{n-1}{2}} \Gamma(\frac{n+1}{2})</div>
It's now useful to know some basic properties of the Gamma function: <span class="math">\Gamma(1) = 0[/spoiler], <span class="math">Gamma(1/2) = \sqrt\pi[/spoiler] to get a number out of that. :)

>>1415407
you math major?

>>1415407
<div class="math">\int_0^\infty \frac{x^s}{e^x-1} dx</div>

You should be able to solve this. /allusion /brofist

>>1415418
It's just tricky integration, not hardcore math. So no, I'm not a math major. What I've posted is about 2nd semester math. (Okay, the Gamma itself isn't, but at least its special values are)

>>1415428
Do you know how to calculate it? Is it nice or just annoying? If nice I'll give it a shot.

>>1415430
well it only requires 2nd semester calc to understand the proof but I don't think coming up with a proof is 2nd semester calc unless the student's brilliant.

>>1415455
Polylogarthimic function in Calc II? o.O

>>1415452
Yes, I was the one who solved it when it was first posted on /sci/. You were in the thread so I thought you'd remember it... Ah well, forget it.

Yes it has a nice answer. Hint: <span class="math">\zeta[/spoiler].

No residue theorem in this entire thread? I am disappointed.

>>1415461
It's not a polylog in the 2nd semester, it's an integral that will turn out to have been the polylog in the years to come.
For me, the Gamma integral was the first homework assignment in statistical physics (5th semester). I did it by hand over two pages and annoying as fuck, but then saw the similarity to Gamma. Raged, but happily.

>>1415464
Ah, Zeta. I thought it would be something similar to a Debye integral. Wikipedia says
<div class="math">\Gamma(s)\zeta(s) =\int_0^\infty\frac{x^{s-1}}{e^x-1}\,dx</div>
But hmm, no idea how to get there.

>>1415474
Split <span class="math">e^{-x} (1 - e^{-x})^{-1}[/spoiler] into a geometric sum, interchange the order of the summation and integration, do change of variables, separate the zeta from the gamma.

>>1415482
Aaah, now that you say it, I've seen (used?) the proof already somewhere.

Hmpf, can't do it. Ran into two dead ends now, I'll try again later. :(

>>1415474

fucked that one up, bro