How do I solve the integral of exp(-x^2) between -a and a?

[Swiss Army]

Could someone show me how the integral of f(v) equals 1, where f(v) is

(4/(Sqrt Pi)) x (m/2kT)^(3/2) x (v^2) x (e^(-mv^2/2kT))

I got through the integral and it didn't equal 1, but it's supposed to, so could someone help me out.

 

[Hurkyl]

Could you show us what you have done so far?

And what is the region of integration?

 

[Swiss Army]

Ooops, sorry. The integration is from zero to infinity. I'll type in what I've done so far if someone can tell me how to get the integral and various other mathematical characters in the post, it's much easier that way. But so far my methods have included substitution, integration by parts, and taking the limit of the function.

 

[Zurtex]

Ooops, sorry. The integration is from zero to infinity. I'll type in what I've done so far if someone can tell me how to get the integral and various other mathematical characters in the post, it's much easier that way. But so far my methods have included substitution, integration by parts, and taking the limit of the function.

Look here: https://www.physicsforums.com/misc/howtolatex.pdf

Or go to this thread and click on various examples and the code will pop up in a small window so you see how it is done: https://www.physicsforums.com/showthread.php?t=8997

 

[Parth Dave]

you should be able to get the indefinate integral using integration by parts. use v^2 = du and e^(-mv^2/2kt) = v. Than in the new integral you create, you can substitute e^u and it should work out fine (in other words, do integration by parts and than use substitution). Whenever you have a parametre for an integral as an infinite, you have to take the limit of the integral as it approaches that infinite. For example, in this case you would have to than show that (lim x--> infinite of F(x)) - F(0) = 1 (thats what you do when you have infinite parameters for integration). Try it out and see what you get.

 

[Swiss Army]

hmmm...

Yeah, those were the methods that I used to integrate it, integration by parts, substitution, and taking the limit. I talked to my Calculus 3 teacher yesterday about it, and she said it was something like a complex integral where you can't just solve it using those methods...or something like that.

 

[Parth Dave]

when you do integration by parts you will obtain a new integral that you have to solve. This [unless I am mistaken] can be solved using substitution. Altho you may have to do integration by parts again (i suspect you would)

 

[NSX]

Yeah .. it seems that way.

I tried to use IBP & substitution.

I assume k, m, T, & [tex]\pi[/tex] are constants.

Then I get:

[tex]\frac{4m^\frac{3}{2}}{\sqrt{\pi}(2kT)^\frac{3}{2}e^\frac{m}{2kT}}\int_{0}^{\infty}\frac{v^2}{e^v^2}dv[/tex]

I'm not sure how to do in an integral the likes of [itex]\int\frac{v^2}{e^v^2}[/itex], or [itex]\intv^2e^-v^2[/itex] for that matter.

With substitution of [itex]u=v^2[/itex], I get:

[tex]\frac{1}{2}\int\frac{u^\frac{1}{2}}{e^u}du[/tex]

 

[franznietzsche]

Yeah .. it seems that way.

I tried to use IBP & substitution.

I assume k, m, T, & [tex]\pi[/tex] are constants.

Then I get:

[tex]\frac{4m^\frac{3}{2}}{\sqrt{\pi}(2kT)^\frac{3}{2}e^\frac{m}{2kT}}\int_{0}^{\infty}\frac{v^2}{e^v^2}[/tex]

I'm not sure how to do in an integral the likes of [itex]\int\frac{v^2}{e^v^2}[/itex], or [itex]\intv^2e^-v^2[/itex] for that matter.

With substitution of [itex]u=v^2[/itex], I get:

[tex]\frac{1}{2}\int\frac{u^\frac{1}{2}}{e^u}du[/tex]

Maybe I'm just more tired than i tihnk, but with that substitution i believe you should get:

[tex]

\frac{1}{2}\int\frac{u}{e^u}du

[/tex]

 

[cookiemonster]

You are tired.

cookiemonster

 

[Ebolamonk3y]

Huh? Makes no sense..

 

[franznietzsche]

You are tired.

cookiemonster

oh...yeah...i didn't change the varible of integration, i left it [tex] dv [/tex] even though i typed [tex] du [/tex]. So yeah, disregard that post then.

edit: actually he never put the [tex] dv [/tex] up there, so I'm assuming that was my error.

 

[Ebolamonk3y]

yay! problem solved..

 

[franznietzsche]

yay! problem solved..

Just the roblem of my mistakes when sleep-deprived. Oh well...i suppose i could go actually try to solve it now...here i go

 

[Ebolamonk3y]

Dun try now... get sleep and then try. :)

 

[franznietzsche]

actually i think it can't be solved analytically...integration by parts fails, given that [tex] \frac{1}{e^x^2} [/tex] cannot be integrated. So...i can't seem to come up with any standard single variable calculus techniques that would help...so yeah, sleep first, boggle later.

 

[Swiss Army]

The thing is, the integral of the whole expression is supposed to divide out to 1. The integral is supposed to end up being the inverse of the constant you pull out when you integrate, giving you a 1/1 or 1 answer. This suggests that standard integration techniques won't work, because the when you integrate, you have to end up with the Sqrt( Pi ) / 4 in the answer to cancel out the 4/Sqrt ( Pi ) in the constant.

 

[NSX]

Mathematica?
Matlab?

haha

But like you said in your post, if you require concepts from Calc III, it's beyond what I can do.

 

[quantumdude]

This integral can be done the same way that the integral of exp(-x²) can be done. First, write the integral of x²exp(-x²) from zero to infinity. Then write the integral of y²exp(-y²) from zero to infinity (they're both exactly the same as your integral). Now multiply the integrands together double integrate over x and y. When you convert to polar coordinates, you will get an integral that can be done by parts. Just don't forget to take the square root at the end.

Note:

x²y²=r⁴sin²(θ)cos²(θ)
x²+y²=r²
dx dy=rdr dθ

 

[Morto]

Hello!

Sorry to hash up this very old thread, but I need help with this exact problem and I don't understand the solution given.

I'm trying to find the integral of x²exp(-x² / a²) from -infinity to infinity.

I've done what's recommended here, but I end up with a double integral of 1/4 sin²(2θ) r⁵ exp[-r²] drdθ from zero to infinity and zero to infinity. How do I solve this?

 

[Count Iblis]

Hello!

Sorry to hash up this very old thread, but I need help with this exact problem and I don't understand the solution given.

I'm trying to find the integral of x²exp(-x² / a²) from -infinity to infinity.

I've done what's recommended here, but I end up with a double integral of 1/4 sin²(2θ) r⁵ exp[-r²] drdθ from zero to infinity and zero to infinity. How do I solve this?

Well, what is recommended here is not really what I would recommend

This problem is actually so simple that you can do it all in your head.

First step: Recall that [tex]\int_{-\infty}^{\infty}\exp\left(-x^{2}\right)dx=\sqrt{\pi}[/tex]

Second step: The integral of [tex]\int_{-\infty}^{\infty}\exp\left(-a x^{2}\right)dx=\sqrt{\frac{\pi}{a}}[/tex] . Substitute [tex]x=\frac{y}{\sqrt{a}}[/tex] in this integral to see this.

Third step: Differentiate both sides w.r.t. a. This will bring down a factor [tex]x^{2}[/tex] in the integrand.

Fourth step: replace [tex]a\longrightarrow \frac{1}{a^{2}}[/tex]

 

[Namratha]

hey.. i really need some immediate help to solve the integral exp(-x^2) between the limits -a and a.. is it an underterminable integral?