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How to evaluate $\int_{0}^{\infty} x^4 e^{-0.5x^2} dx$?

I am trying to evaluate this integral, but to no avail. I am completely stuck and have tried all the classic methods (u-substitution, integration by parts, etc). Could someone point me in the right direction? It has been ages since I took calculus.

Gary
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graphtheory123
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  • Have you learned about the normal distribution in probability theory? This is similar to $ E\left[X^{4}\right],X\sim\mathcal{N}\left(0,1\right) $ which you can find using the characteristic function or using known results. – Danny Blozrov May 29 '23 at 23:25
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    I know about the normal distribution, but I am unclear on how to leverage that to solve this problem. – graphtheory123 May 29 '23 at 23:29
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    First perform the change of integration variables from $x$ to $t$ via $x=\sqrt{2 t}$. After that, use the gamma function. – Gary May 29 '23 at 23:31
  • Do you remember the shell method? Compute the volume of that function rotated about the $y$ axis using the shell method. The answer will be the square root of that volume. – Ninad Munshi May 29 '23 at 23:39
  • https://math.stackexchange.com/questions/1945448/methods-for-finding-raw-moments-of-the-normal-distribution This might help, all you have to do is change your integral because it is an even function $ \int\limits _{0}^{\infty}x^{4}e^{\frac{-x^{2}}{2}}dx=\frac{1}{2}\int\limits _{-\infty}^{\infty}x^{4}e^{\frac{-x^{2}}{2}}dx $ and also multiply the $ \frac{1}{\sqrt{2\pi}} $ factor – Danny Blozrov May 29 '23 at 23:39
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    use integration by parts: $u=x^3$, $dv= xe^{-0.5x^2}$. – user51547 May 29 '23 at 23:42
  • Another method (the same one used when the $x^4$ isn't there): multiply the integral by itself to turn it into a two-variable integral over the first quadrant, then switch to polar coordinates. – Greg Martin May 30 '23 at 00:23

1 Answers1

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Hint Using the symmetry of the integrand and applying $u$-substitution to the Gaussian integral gives: $$\int_0^\infty e^{-a^2 x^2} \,dx = \frac{\sqrt\pi}{2 a} .$$

Differentiating with respect to $a$ gives $$-2 a\int_0^\infty x^2 e^{-a^2 x^2} \,dx = -\frac{\sqrt\pi}{2 a^2} ,$$ and then rearranging gives $$\int_0^\infty x^2 e^{-a^2 x^2} \,dx = \frac{\sqrt\pi}{4 a^3} .$$

Now, differentiate again.

We get $$-2 a\int_0^\infty x^4 e^{-a^2 x^2} \,dx = -\frac{3\sqrt\pi}{4 a^4},$$ so $$\boxed{\int_0^\infty x^4 e^{-a^2 x^2} \,dx = \frac{3\sqrt\pi}{8 a^5}} .$$ Taking $a = \frac{1}{\sqrt{2}}$ gives $$\boxed{\int_0^\infty x^4 e^{-\frac{1}{2} x^2} \,dx = 3 \sqrt\frac{\pi}{2}} .$$

More generally we have

$$\int_0^\infty x^k e^{-a^2 x^2} \,dx = \frac{1}{2 a^{k + 1}} \int_0^\infty t^\frac{k - 1}{2} e^{-t} = \frac{1}{2 a^{k + 1}} \Gamma\left(\frac{k + 1}{2}\right) ,$$ where $\Gamma$ denotes the gamma function

Travis Willse
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