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

gaussian-integralimproper-integralsintegrationpolar coordinates

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.

Best Answer

  • 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 $t=1$.
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