Evaluate the following integral

  • Thread starter quasar987
  • Start date
  • #1
quasar987
Science Advisor
Homework Helper
Gold Member
4,783
15
Any idea on how to evaluate

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

where n is even???
 
Last edited:

Answers and Replies

  • #2
Tom Mattson
Staff Emeritus
Science Advisor
Gold Member
5,500
8
quasar987 said:
where n is pair ???
What's that mean?
 
  • #3
quasar987
Science Advisor
Homework Helper
Gold Member
4,783
15
oh sorry, pair = french for even.

Edited.
 
  • #4
Tom Mattson
Staff Emeritus
Science Advisor
Gold Member
5,500
8
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
Science Advisor
Homework Helper
Gold Member
4,783
15
How could have I missed that?! :grumpy:

Thx Matt!

Btw - are you still learning diff forms?
 
  • #6
Tom Mattson
Staff Emeritus
Science Advisor
Gold Member
5,500
8
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
Science Advisor
Homework Helper
Gold Member
4,783
15
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
Tom Mattson
Staff Emeritus
Science Advisor
Gold Member
5,500
8
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
Science Advisor
Homework Helper
Gold Member
4,783
15
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
45
0
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
Science Advisor
Homework Helper
Gold Member
4,783
15
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]
 

Related Threads on Evaluate the following integral

  • Last Post
Replies
4
Views
1K
Replies
2
Views
894
  • Last Post
Replies
4
Views
1K
Replies
3
Views
1K
  • Last Post
Replies
6
Views
5K
  • Last Post
Replies
4
Views
4K
  • Last Post
Replies
6
Views
4K
  • Last Post
Replies
7
Views
1K
  • Last Post
Replies
3
Views
1K
Replies
16
Views
2K
Top