$I = \int_{0}^\infty t^2 e^{-t^2/2} dt$

Question

Q: Evaluate the integral $I = \int_\limits{0}^\infty t^2 e^{-t^2/2} dt$

Hint, write $I^2$ as the following iterated integral and convert to polar coordinates:

\begin{align*} I^2 &= \int_\limits{0}^\infty \int_\limits{0}^\infty x^2 e^{-x^2/2} \cdot y^2 e^{-y^2/2} \, dx \, dy \\ \end{align*}

I can see the final answer is $\frac{\pi}{2}$ but I don't see how to get this.

This problem is very similar to the Gaussian Integral: $I = \int_\limits{-\infty}^\infty e^{-x^2} dx = \sqrt{\pi}$. I can follow the derivation to this.

The Gaussian Integral technique of converting to polar coordinates doesn't seem to work as cleanly on this problem.

I can convert to polar coordinates:

\begin{align*} I^2 &= \int_\limits{0}^\infty \int_\limits{\pi/2}^\pi r^5 \sin^2 \theta \cos^2 \theta e^{-r^2/2} \, d\theta \, dr \\ \end{align*}

That doesn't look easy to evaluate.

Answer 1

• Why not integrate by parts from the initial integral? One has $$ I=-\frac12\int_0^\infty x \cdot \left(-2xe^{-x^2} \right)dx=-\frac12\int_0^\infty x \cdot \left(e^{-x^2} \right)'dx $$ then one may use the Gaussian integral to conclude.
• Another path is to differentiate the gaussian identity $$ \int_0^\infty e^{-tx^2}dx=\frac{\sqrt{\pi }}{2 \sqrt{t}} \qquad t>0, $$ with respect to $t$ and put $t=1$.

Answer 2

Let T a random variable normally distributed mean 0 , variance 1

$E(T^2)=E(T^21_{T>0})+E(T^21_{T<=0})$. By symmetry of T, we have

$var(T)=E(T^2)=2E(T^21_{T>0})=1$

Moreover, $E(T^21_{T>0})=\frac{1}{\sqrt{2\pi}}\int_\limits{0}^{\infty}t^2e^{-\frac{t^2}{2}}dt$. Then we can conclude

Answer 3

I thought it might be instructive to present a way forward that relies on "Feynman's Trick" for differentiating under the integral sign.

Proceeding, we let $I(a)$ be the integral given by

$$\begin{align} I(a)&=\int_0^\infty e^{-at^2}\,dt\\\\ &=\frac12\sqrt{\frac{\pi}{a}} \tag 1 \end{align}$$

Differentiating $(1)$ reveals

$$-\int_0^\infty t^2 e^{-at^2}\,dt=-\frac14\frac{\sqrt\pi}{a^{3/2}} \tag 2$$

Setting $a=1/2$ in $(2)$ and multiplying by $-1$ yields the coveted result

$$\bbox[5px,border:2px solid #C0A000]{\int_0^\infty t^2e^{-t^2/2}\,dt=\sqrt{\frac\pi{2}}}$$

And we are done!

Answer 4

I would use the definition of gamma function

$I=\int_{0}^{\infty}t^2e^{-t^2/2}dt $

Let $t^2/2 = s$

then, $ I = \sqrt{2}\int_{0}^{\infty}s^{1/2}e^{-s}ds $

$ = \sqrt2 \Gamma(3/2) $

$=\sqrt2 \frac{\sqrt{\pi}}{2} $

$=\sqrt\frac{\pi}{2}$

Answer 5

If you want to use your original hint, you have: $$I^2 = \int_\limits{0}^\infty \int_\limits{\pi/2}^\pi r^5 \sin^2 \theta \cos^2 \theta e^{-r^2/2} \, d\theta \, dr $$ First, we can switch the order of integration and start with $$\int_\limits{0}^\infty r^5 e^{-r^2/2}dr=\int_\limits{0}^\infty (r^4)(r e^{-r^2/2})dr $$ That can be integrted by parts twice, using the fact that $\int re^{-r^2/2}dr=-e^{-r^2/2}$, and one final integration yields the result $$\int_\limits{0}^\infty r^5 e^{-r^2/2}dr=8$$ And our integral becomes$I^2 =\int_\limits{\pi/2}^\pi 8\sin^2 \theta \cos^2 \theta d\theta$

Using the identity $\cos^2(x)\sin^2(x)=\frac{1}{4}\sin^2(2\theta)$ we have $I^2 =\int_\limits{\pi/2}^\pi 8 \frac{1}{4}\sin^2(2\theta) d\theta=\int_\limits{\pi/2}^\pi 2\sin^2(2\theta) d\theta$ And so $$I^2=\int_\limits{\pi/2}^\pi 2\sin^2(2\theta) d\theta=\frac{\pi}{2}$$ $$I=\sqrt\frac{\pi}{2}$$