How to prove $\int^{\infty}_0 e^{-c^2/a^2}c^4\,dc=\frac{3}{8}a^5\sqrt\pi$?

Question

How do i prove this integral can someone give me proof i've been trying this too long i tried direct integration by parts, differentiation all to no avail.Please support us :)

$$\int^{\infty}_0 e^{\frac{-c^2}{a^2}}c^4\,dc=\frac{3}{8}a^5\sqrt\pi$$

Answer 1

Note that, with $I(t)=\int_0^\infty e^{-t x^2}dx= \frac {\sqrt{\pi}}{2\sqrt t }$ $$ I''(t) = \int_0^\infty x^4 e^{-t x^2}dx= \frac{3\sqrt{\pi}}{8 \ t^{5/2}}$$

Thus

$$\int^{\infty}_0 e^{-c^2/a^2}c^4\,dc=I''(1/a^2)=\frac{3\sqrt\pi}{8}a^5$$

Answer 2

Two times by parts:
$\int\limits_0^\infty e^\frac{-x^2}{a^2}dx= \left.xe^\frac{-x^2}{a^2}\right|_0^\infty+ \int\limits_0^\infty \frac{2x^2}{a^2}e^\frac{-x^2}{a^2}dx= \frac{2}{a^2}\int\limits_0^\infty e^\frac{-x^2}{a^2}d\left(\frac{x^3}{3}\right)=$
$ \frac{2}{a^2}\left(\left.\frac{x^3}{3}e^\frac{-x^2}{a^2}\right|_0^\infty+\int\limits_0^\infty \frac{x^3}{3}\cdot\frac{2x}{a^2}e^\frac{-x^2}{a^2}dx\right)= \frac{4}{3a^4}\int\limits_0^\infty x^4e^\frac{-x^2}{a^2}dx$
Let alone $\int\limits_0^\infty e^{-x^2}dx=\frac{\sqrt{\pi}}{2}$ or here be a known result, we obtain
$$\int\limits_0^\infty x^4e^\frac{-x^2}{a^2}dx=\frac{3a^4}{4}\int\limits_0^\infty e^\frac{-x^2}{a^2}dx= \frac{3a^4}{4}\cdot a\int\limits_0^\infty e^{-x^2}dx=\frac{3a^5\sqrt{\pi}}{8}\hbox{, QED}$$