2

How to evaluate the integral $\int_{0}^{\infty}t^{n}e^{-at^{2}}dt$ where n is a positive even number and $\int_{0}^{\infty}e^{-at^{2}}dt=\frac{1}{2}\sqrt{\pi /a}$

esege
  • 3,621

2 Answers2

3

Integrate by parts:

$$I_n = \int_{0}^{\infty}t^{n}e^{-at^{2}}dt = \frac 1a \int_{0}^{\infty} t^{n-1} (-at e^{-at^{2}}) dt \\ = \frac 1a \left[-t^{n-1}e^{-at^{2}}\right]_{0}^{\infty} + \frac{n-1}a \int_{0}^{\infty} t^{n-2}e^{-at^{2}}dt =\frac{n-1}a I_{n-2} $$

You can explicitely compute $I_1$ and you know $I_0$.

$$I_1 = \left[-\frac 1a e^{-at^{2}}\right]_{0}^{\infty}=\frac 1a \\ I_{2n+1} = \frac{2n(2n-2)\dots 2}{a^n}\frac 1a = \frac{2^nn!}{a^{n+1}} \\ I_{2n} = \frac{(2n-1)(2n-3)\dots 1}{a^n} \frac{\sqrt{\pi}}{2\sqrt{a}} = \frac{(2n)! \sqrt{\pi}}{2^{n+1} n!a^{n+1/2}} $$

mookid
  • 28,236
1

Let $t=\sqrt{x}$; then the integral is

$$\frac12 \int_0^{\infty} dx \, x^{(n-1)/2} \, e^{-a x} = \frac{\Gamma \left ( \frac{n+1}{2}\right )}{2 a^{(n+1)/2}}$$

Ron Gordon
  • 138,521
  • Thank you. Can we answer this question with just using Leibniz integral rule? – esege Apr 05 '14 at 14:06
  • 1
    @esege: please elaborate. – Ron Gordon Apr 05 '14 at 14:07
  • Well our teacher told us if we constantly take derivative of the first integral with respect to a, we can find some sort of pattern. He didn't mentioned gamma function when he asked this. – esege Apr 05 '14 at 14:09
  • 1
    @esege: You can do that, of course, but you will get higher and higher powers of $t$ inside the integral. In any case, for odd $n$, you get a factorial. – Ron Gordon Apr 05 '14 at 14:12