How to evaluate integral: $ \int_{-\infty}^{\infty} x^{n} e^{-\alpha x^{2}} dx $ only using methods of multidimensional integration

Question

I need to solve this integral $$ \int_{-\infty}^{\infty} x^{n} e^{-\alpha x^{2}} dx. $$ where n is natural number and $\alpha > 0$.

I know how to solve it using derivation according parameter $\alpha$ or with integration by parts but my task is to solve it using only methods of multidimensional integration.I am not sure how to do so.I have tried to use polar coordinates but without success.

It would be great if anyone can show me how to do so.Thanks in advance.

Answer 1

I will assume that $n=2k$ since for odd $n$ the integral is obviously zero.

Hint: Let $$ I(\alpha,k)=\frac12\int_{-\infty}^\infty x^{2k}e^{-\alpha x^2}dx=\int_0^\infty x^{2k}e^{-\alpha x^2}dx. $$

Then $$ I^2(\alpha,k)=\int_0^\infty x^{2k}e^{-\alpha x^2}dx\int_0^\infty y^{2k}e^{-\alpha y^2}dy\\ =\int_0^{\pi/2}\cos^{2k}\theta\sin^{2k}\theta\,d\theta\int_0^\infty r^{2k+1}e^{-\alpha r^2}dr. $$

Can you take it from here?

Answer 2

You can try with partial integration with $u=x^{n-1}$ and $dv=xe^{-\alpha x^2}dx$, then use induction.

Answer 3

$$I(n)=\int_{-\infty}^\infty x^ne^{-ax^2}dx$$ $$I'(n)=n\int_{-\infty}^\infty x^{n-1}e^{-ax^2}dx=nI(n-1)$$ $$I''(n)=n(n-1)I(n-2)$$ and we know that: $$I(n)=\sum_{i=0}^\infty\frac{I^{(i)}(a)}{i!}(n-a)^i$$ choosing $a=0$ we get: $$I(n)=\sum_{i=0}^n\frac{n!}{(n-i)!i!}n^iI(n-i)$$

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Using the total derviative definition we can also show that: $$I(N,A)=N\int_0^N\int_{-\infty}^\infty x^{n-1}e^{-ax^2}dxda-2\int_0^A\int_{-\infty}^\infty x^{n+1}e^{-ax^2}dxda$$