Determine $I_n=\int_{-\infty}^\infty t^ne^{-\frac{t^2}{a}}dt$

Question

if $\Large \int_0^\infty t^ne^{-t}dt =n!$

What's about $\Large I_n=\int_{-\infty}^\infty t^ne^{-\frac{t^2}{a}}dt$ ; $a\in \mathbb{R}$

Answer 1

Let $t=\frac{x^2}a$ in $\int_0^\infty t^m e^{-t}dt =m!$ to get

$$\frac2{a^{m+1}} \int_0^\infty x^{2m+1}e^{-\frac{x^2}a}= m! $$

Then, for even $n\ge2$, substitute $n=2m+1$ in above result

\begin{align} I_n &= \int_{-\infty}^\infty t^ne^{-\frac{t^2}{a}}dt =a^{\frac{n+1}2}\left(\frac{n-1}2\right)! =a^{\frac{n+1}2}\frac{(n-1)(n-3)...1}{2^{n-2}}\frac{\sqrt\pi}2 \end{align}

where $(\frac12)! = \frac{\sqrt\pi}2$.

Answer 2

Hint: Substitute $t^2=ap$, then $\mathbf{I_n = \frac{1}{2} \, a^{\frac{n+1}{2}} \int_{0}^\infty dp \exp(-p) \, p^{\frac{n-1}{2}}}$ and you may be interested in gamma functions.

Answer 3

If $t=0$, then $\frac{t^2}{a} = 0$ too, so the bounds don't change. Now set $\frac{t^2}{a} = v \implies$$ 2tdt = adv, t= \sqrt{av}$. The expression is equal to $$ \frac{1}{2}\int_{0}^{\infty} t^{n-1}e^{-\frac{v^2}{a}}2tdt = \frac{a}{2}\int_{0}^{\infty}(av)^{\frac{n-1}{2}} e^{-v}dv $$ Can you handle from here?

Answer 4

Hint:

WLOG, $a=2$ (your job to perform the rescaling of the variable). Then by parts,

$$I_n=\int_{-\infty}^\infty t^ne^{-t^2/2}dt=\int_{-\infty}^\infty t^{n-1}\,t\,e^{-t^2/2}dt=-\left.t^{n-1}e^{-t^2/2}\right|_{-\infty}^\infty+(n-1)\int_{-\infty}^\infty t^{n-2}e^{-t^2/2}dt \\=(n-1)I_{n-2}.$$

This gives you a recurrence relation, to be completed with the well-known Gaussial integral

$$I_0=\int_{-\infty}^\infty e^{-t^2/2}dt=\sqrt{2\pi}.$$