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Calculate the following improper integrals

$ \displaystyle{ \int_{e}^{\infty} e^{-\frac{1}{2} (nx)^2} dx , \quad \int_{e}^{\infty}x^2 e^{-\frac{1}{2} (nx)^2} dx \quad ,\int_{-\infty}^{\infty} xe^{-\frac{1}{2} (nx)^2} dx }$.

I know that $ \displaystyle{ \int_{0}^{\infty} e^{-a x^2} dx =\frac{1}{2} \sqrt{\frac{\pi}{2}} }$ where $ a>0$

I have also read that $ \displaystyle{ \int_{0}^{\infty} e^{-a x^2} x^{2n}dx= \frac{1 \cdot 3 \cdot 5 \cdot \cdots \cdot (2n-1)}{2^{n+1} a^n} \sqrt{ \frac{\pi}{2}} }$ where $ n \in \mathbb N$.

So I think that for the first it is enough to compute $ \displaystyle{ \int_{0}^{e} e^{-\frac{1}{2} (nx)^2} dx } $ and for the second it is enough to compute $\displaystyle{\int_{0}^{e} x^{2} e^{-\frac{1}{2} (nx)^2} dx }$.

But I have no idea from here on.

Is there another approach which works better ?

I would really appreciate some help on any of the above integrals.

Thank's in advance!

edit: Sorry for putting the three integrals in one question but it seems to me that there is a connection between them ( on how I calculate them) since they are similar.

edit2: Can someone give me a proof or a link to see how can I get the series in the link http://people.math.sfu.ca/~cbm/aands/page_932.htm which was given in the answer below.

Thank's!

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Have you tried a simple integration by parts after converting to plane polar coordinates? This has the advantage of reducing the infinite domain to a finite region in the limit. I'll have to sit down and actually try it,but that's what I'd try first. –  Mathemagician1234 May 26 '12 at 21:09
    
@Mathemagician1234: No I didn't try this. How I can I go to polar coordinates? It seems to more difficult. –  passenger May 26 '12 at 21:13
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@passenger: A hint on the third one: symmetry. –  Nate Eldredge May 26 '12 at 21:23
    
@NateEldredge: Yes I have seen that. Since the integrated function is odd if the improper integral converges it must be equal to zero. –  passenger May 26 '12 at 21:26
    
The third one, a simple substitution should work. –  M Turgeon May 26 '12 at 21:31
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1 Answer

There is no solution in a finite number of terms involving the elementary functions for the first integral; its value is $a(1-\Phi(y))$ where the values of $a$ and $y$ can be determined via a change of variables that transform the integral into the integral formula for $aP\{X > y\}$ where $X$ is a standard normal random variable.

The second integral should be integrated by parts to simplify it a little. Try going the other way first: what is the derivative of $e^{-n^2x^2/2}$? Can you express the integral you need to compute in terms of $\int x\ \mathrm d(e^{-n^2x^2/2})$ and use integration by parts?

The third question you have solved already as your response to Nate Eldredge's hint seemed to indicate, but if not, look again to my suggestion of the derivative of $e^{-n^2x^2/2}$.

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If I do this integration by parts two the second integral I get again the first one, which you are telling that we can calculate. Are you sure that the first integral has no solution? –  passenger May 26 '12 at 22:01
    
Yes, in part 2, you do get the first integral again. That's why I said to integrate by parts to simplify it a little. And yes, I am sure that the first integral has no solution in terms of elementary functions (such as $\exp$, $\log$, trigonometric functions etc.) –  Dilip Sarwate May 26 '12 at 22:03
    
So no one of the first two can we calculate? –  passenger May 26 '12 at 22:07
    
$\Phi(\cdot)$ is well-tabulated, and various rational approximations to it have been developed. It can be expressed as an infinite series. Or you can leave the answer in terms of $\Phi$ or the related function $\text{erf}$ and see if that is acceptable. –  Dilip Sarwate May 26 '12 at 22:12
    
How can be exressed as series? Can you give a link or something to see it. Thank you! –  passenger May 26 '12 at 22:13
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