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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.2012-05-26
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    @Mathemagician1234: No I didn't try this. How I can I go to polar coordinates? It seems to more difficult.2012-05-26
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    @passenger: A hint on the third one: symmetry.2012-05-26
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    @NateEldredge: Yes I have seen that. Since the integrated function is odd if the improper integral converges it must be equal to zero.2012-05-26
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    The third one, a simple substitution should work.2012-05-26
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    @MTurgeon:What substitution?2012-05-26
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    @passenger $u=-\frac{1}{2}(nx)^2$.2012-05-27

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