Evaluate the following integral

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  • #1
quasar987
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Any idea on how to evaluate

[tex]\int_{0}^{\infty} x^n e^{-ax^2} dx[/tex]

where n is even?
 
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  • #2
quantumdude
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quasar987 said:
where n is pair ?

What's that mean?
 
  • #3
quasar987
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oh sorry, pair = french for even.

Edited.
 
  • #4
quantumdude
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I would integrate by parts.

Hint:

[tex]
u=x^{n-1}
[/tex]

[tex]
dv=xe^{-ax^2}dx
[/tex]

Try that and see what you come up with.
 
  • #5
quasar987
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How could have I missed that?! :grumpy:

Thx Matt!

Btw - are you still learning diff forms?
 
  • #6
quantumdude
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quasar987 said:
How could have I missed that?! :grumpy:

Can happen to anyone.

Thx Matt!

The name's Tom, not Matt. How would you like it if I called you "nebula"? :biggrin:

Btw - are you still learning diff forms?

Yeah, my next set of notes is in preparation. I stopped posting to the thread because I was asked to teach a couple of summer courses. They're condensed into 6 weeks, so I didn't have time for anything else.

Keep watching...
 
  • #7
quasar987
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Sorry Tom. :biggrin:

I think I'm still missing an important point in your idea though.. in tegrating by parts like you advised I get

[tex]\int_{0}^{\infty} x^n e^{-ax^2} dx = \left[-\frac{x^{n-1}}{2a} e^{-ax^2} \right]_0^{\infty} + \frac{n-1}{2a} \int_0^{\infty} x^{n-2} e^{-ax^2} dx[/tex]

So we can perform the same technique of integration on the last integral, and so on, such that after n/2 times of performing this trick, the integral we have to evaluate is

[tex]\int_0^{\infty} e^{-ax^2} dx [/tex]

But what is that?! the integrator says its a constant times [itex]\mbox{Erf}[\sqrt{5}x][/itex] The heck is Erf ?
 
  • #8
quantumdude
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quasar987 said:
[tex]\int_0^{\infty} e^{-ax^2} dx [/tex]

But what is that?!

That is perhaps the most famous integral ever taught to students of multivariable calculus, at least in America. It is a standard way to introduce polar coordinates in double integration.

[tex]I=\int_0^{\infty} e^{-ax^2} dx [/tex]

[tex]I=\int_0^{\infty} e^{-ay^2} dy [/tex]

[tex]I^2=\int_0^{\infty}\int_0^{\infty} e^{-a(x^2+y^2)} dx [/tex]

Now convert to polar coordinates and evaluate.

the integrator says its a constant times [itex]\mbox{Erf}[\sqrt{5}x][/itex] The heck is Erf ?

Erf(x) is the error function.
 
  • #9
quasar987
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RIIIIGHT! I remember browsing my calculus book last year, and for some random reason, I began reading this random problem. And I found it so cute I had to do it, and of course it was proving that this integral is worth [itex]\sqrt{\pi}/2[/itex] by the method you outlined!
 
  • #10
TOKAMAK
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If you want to skip the integration by parts, you can choose to use the differentiation under the integral sign:

[tex] \int_{-\infty}^{\infty} x^{n} dx e^{-ax^{2}} = \int_{-\infty}^{\infty} dx \frac{d^{\frac{n}{2}}}{da^{\frac{n}{2}}}(-1)^{\frac{n}{2}}e^{-ax^{2}} [/tex]

So now you can take the derivative with respect to a out in front:

[tex] (-1)^{\frac{n}{2}} \frac{d^{\frac{n}{2}}}{da^{\frac{n}{2}}}\int_{-\infty}^{\infty} dx e^{-ax^{2}} [/tex]

So, once you've figured out what [tex] \int_{-\infty}^{\infty} dx e^{-ax^{2}} [/tex] is, you could also do it in this way.
Notice of course, that this is only useful if n is even, which I think you stipulated, and if the integral is a definite integral.
 
  • #11
quasar987
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I'm not familiar with the use of this method TOKAMAK (I'm not Feynman).

btw, if anyone carried out the problem til the end and want to compare, I arrive at

[tex]\sqrt{\frac{\pi}{2^n a^{n+1}}}\prod_{j=1,3,5,...}^{n-1} (n-j)[/tex]
 

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