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I'm having difficulty evaluating this integral:

$\int_{-\infty}^{\infty} {x^6 e^{-x^2}} dx$

All I've been able to do is separate them and evaluate them separately although I haven't been able to successfully do that:

$\int_{-\infty}^{0} {x^6 e^{-x^2}} dx + \int_{0}^{\infty} {x^6 e^{-x^2}} dx$

I was given this hint although I can't understand how they are equal (the integral on the RHS is the Gaussian integral):

$\int_{0}^{\infty} {x^6 e^{-x^2}} dx = -\frac{d^3}{da^3} \int_{0}^{\infty} { e^{-ax^2}} dx $

That's pretty much all I've got so far. Any help would really be appreciated!

Edit: This is different from the supposed duplicate because in the other one, I was asked to make one post per question and the answer was the hint but I didn't understand how it was equal.

Kath Kiyan
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  • the solution for the indefnite integral containes the error function ! – Dr. Sonnhard Graubner Aug 10 '17 at 12:55
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    It's https://math.stackexchange.com/q/941570 with $a=2$ and $n=3$ – PhyM Aug 10 '17 at 13:00
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    One method: integrate $I_k = \int_{-\infty}^{\infty} {x^k e^{-x^2}} dx$ by parts to get a reduction formula, in this case for $I_k$ in terms of $I_{k-2}$. Then all you need is the well-known $I_0$. – GEdgar Aug 10 '17 at 13:34

2 Answers2

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Hint:

First note that :

$$\int_{0}^{\infty} e^{-ax^2} = \frac{1}{2}\sqrt{\frac{\pi}{a}}$$

Secondly, convince yourself that $$\int_{-\infty}^{0} x^6e^{-x^2} = \int_{0}^{\infty} x^6e^{-x^2} $$

Dionel Jaime
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The hint you were given perhaps needs some justification. What they're doing is differentiating under the integral (Or Feynman Integraton). As it turns out, in some cases, you are free to differentiate under the integral, then evaluate the integral, and then re-integrate again later on. So to say, $$ -\frac{d^3}{da^3} \int_0^{\infty} e^{-ax^2} dx = -\int_0^{\infty} \frac{d^3}{da^3} e^{-ax^2} dx = \int_0^{\infty} x^6 e^{-ax^2} $$ They chose to generalize the equation for any $a$, and then later on put back $a$ as -$1$.

Other than this, integration by parts (with or without the tabular method) should be able to solve this. As Dionel hinted, $ \int_{0}^{\infty} e^{-ax^2}dx = \frac{\sqrt{\pi}}{2\sqrt{a}} $, and also showing that it is symmetric.

Max
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