Prove that$ \int_0^\infty x^{2n}e^{-ax^2}dx = \frac{(2n-1)(2n-3) \cdots1}{2(2a)^n}\sqrt{\frac\pi a} $

Question

I have the following question:

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Verify $ \int_0^\infty x^{2n}e^{-ax^2}dx = \frac{(2n-1)(2n-3) ... 1}{2(2a)^n}\sqrt{\frac{\pi}{a}} $ by differentiating $ \int_0^\infty e^{-ax^2}dx $ with respect to $a$, $n$ times.

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How would I go about doing this? I am unsure how differentiating the latter integral links to the former result.

Answer 1

You are looking for Leibniz's differentiation under the integral sign rule.

We know that

$$\int_0^{\infty} e^{-ax^2} \, dx = \frac{1}{2} \sqrt{\frac{\pi}{a}}.$$

Differentiating with respect to $a$ gives us

$$\int_0^{\infty} -x^2 e^{-ax^2} \, dx = \frac{1}{2} \left( - \frac{1}{2} \right) \sqrt{\frac{\pi}{a^3}},$$

or

$$\int_0^{\infty} x^2 e^{-ax^2} \, dx = \frac{1}{4a} \sqrt{\frac{\pi}{a}}.$$

By repeated application of the method you'll arrive at the answer.