Answer: The Gaussian-Polynomial Integral

Key Idea / Intuition

The Gaussian integral $\int_0^\infty e^{-x^2}dx = \frac{\sqrt{\pi}}{2}$ is the seed. Differentiating with respect to a parameter inserted into the exponent generates the $x^4$ factor "for free," turning a hard-looking integral into a consequence of the most famous integral in mathematics. Each differentiation brings down two extra powers of $x$, so two differentiations give $x^4$.

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Formal Proof / Solution

Step 1: Introduce a parameter.

Define $I(t) = \int_0^\infty e^{-t x^2} \, dx, \quad t > 0.$

By the standard substitution $u = \sqrt{t}\, x$, $I(t) = \frac{1}{\sqrt{t}} \int_0^\infty e^{-u^2} du = \frac{\sqrt{\pi}}{2} \cdot t^{-1/2}.$

Step 2: Differentiate under the integral sign (Feynman's trick).

$\frac{d}{dt} I(t) = \int_0^\infty \frac{\partial}{\partial t} e^{-tx^2} dx = -\int_0^\infty x^2 e^{-tx^2} dx.$

So $\int_0^\infty x^2 e^{-tx^2} dx = -I'(t) = \frac{\sqrt{\pi}}{4} t^{-3/2}.$

Step 3: Differentiate again.

$\frac{d^2}{dt^2} I(t) = \int_0^\infty x^4 e^{-tx^2} dx.$

Computing: $I''(t) = \frac{d}{dt}\left(-\frac{\sqrt{\pi}}{4} t^{-3/2}\right) = \frac{3\sqrt{\pi}}{8} t^{-5/2}.$

Step 4: Evaluate at $t = 1$.

$I = \int_0^\infty x^4 e^{-x^2} dx = I''(1) = \frac{3\sqrt{\pi}}{8}.$

Sanity check via Gamma function: Using $\int_0^\infty x^{2n} e^{-x^2}dx = \frac{(2n-1)!!}{2^{n+1}}\sqrt{\pi}$ with $n=2$: $\frac{3!!}{2^3}\sqrt{\pi} = \frac{3}{8}\sqrt{\pi}. \checkmark$

$\boxed{I = \dfrac{3\sqrt{\pi}}{8}}$