Evaluation of Gaussian integrals. In Bender and Orszag's section on applying Laplace's method they state the equality
$$\int_{0}^{\infty}e^{-xt^2}\frac{1}{3}xt^4 dt = \frac{1}{2}\sqrt{\frac{\pi}{x}}\left(\frac{1}{4x}\right)$$
How does one calculate this integral ?
 A: Integration by parts does work: 
$$I=\int_{0}^{\infty}e^{-xt^{2}}\frac{1}{3}xt^{4}dt\\=\left[\frac{-1}{2x}e^{-xt^{2}}\frac{1}{3}xt^{3}\right]_{0}^{\infty}+\int_{0}^{\infty}\frac{1}{2}t^{2}e^{-xt^{2}}dt \\=\left[\frac{-t}{4x}e^{-xt^{2}}\right]_{0}^{\infty}+\frac{1}{4x}\int_{0}^{\infty}e^{-xt^{2}}dt \\=\frac{1}{4x}\int_{0}^{\infty}\frac{1}{\sqrt{x}}e^{-u^{2}}du=\frac{1}{4x}\frac{\sqrt{\pi}}{2\sqrt{x}}$$
A: Notice, the Laplace transform $$L[t^n]=\int_{0}^{\infty}e^{-st}t^ndt=\frac{\Gamma(n+1)}{s^{n+1}}$$ & $\Gamma\left(\frac{1}{2}\right)=\sqrt {\pi}$
Now, we have $$\int_{0}^{\infty}e^{-xt^2}\frac{1}{3}xt^4dt$$
$$=\frac{1}{3}x\int_{0}^{\infty}e^{-xt^2}t^4dt$$
Let, $t^2=u\implies 2tdt=du\iff dt=\frac{du}{2\sqrt u}$, we get 
$$\frac{1}{3}x\int_{0}^{\infty}e^{-xu}u^2\frac{du}{2\sqrt u}$$
$$=\frac{1}{6}x\int_{0}^{\infty}e^{-xu}u^{3/2}du$$
$$=\frac{1}{6}xL[u^{3/2}]_{s=x}$$
$$=\frac{1}{6}x\left[\frac{\Gamma\left(\frac{3}{2}+1\right)}{s^{\frac{3}{2}+1}}\right]_{s=x}$$
$$=\frac{1}{6}x\left[\frac{\frac{3}{2}\frac{1}{2}\Gamma\left(\frac{1}{2}\right)}{s^{5/2}}\right]_{s=x}$$
$$=\frac{x}{8}\left[\frac{\sqrt{\pi}}{x^{5/2}}\right]$$
$$=\frac{1}{2}\sqrt{\frac{\pi}{x}}\frac{1}{(4x)}$$ $$=\frac{1}{8x}\sqrt{\frac{\pi}{x}}$$
