TITLE: Computing integral involving Dirac Delta Function
QUESTION [7 upvotes]: Compute
$$
\int_{-\infty}^{\infty} t^2 \delta(\sin(t)) e^{-|t|} \mathrm dt
$$
In closed form, where $\delta(t)$ is the Dirac Delta function .
My attempt:    
$$
\int_{-\infty}^{\infty} t^2 \delta(\sin(t)) e^{-|t|} \mathrm dt = \int_{-\infty}^{\infty} \delta(\sin(t))t^2 e^{-|t|}\mathrm dt
$$
Then noting that $\sin(t)$ is zero whenever $t=n\pi$. By formula (2) and (7) in the above link,
\begin{align}
\int_{-\infty}^{\infty} \delta(\sin(t))t^2 e^{-|t|} \mathrm dt& = \sum_{n=-\infty}^{\infty} \frac{(n\pi)^2e^{-|n\pi|}}{|\cos(n\pi)|}
\\&
=2\pi^2\sum_{n=0}^{\infty} \frac{(n)^2e^{-n\pi}}{1}
\end{align}
However , I am stuck here, i do not know how to procede, Wolfram Alpha tells me that this sum doesn't converge so how can i compute it in closed form? I can only assume I have gone about this the wrong way or made a mistake. Any help would be great.

REPLY [5 votes]: Note that
$$
\begin{align}
\sum_{n=0}^\infty n^2x^n
&=x^2\sum_{n=0}^\infty n(n-1)x^{n-2}+x\sum_{n=0}^\infty nx^{n-1}\\
&=\frac{2x^2}{(1-x)^3}+\frac{x}{(1-x)^2}\\[6pt]
&=\frac{x+x^2}{(1-x)^3}
\end{align}
$$
Therefore,
$$
\begin{align}
2\pi^2\sum_{n=0}^\infty n^2e^{-\pi n}
&=2\pi^2\frac{e^{-\pi}+e^{-2\pi}}{\left(1-e^{-\pi}\right)^3}\\
&=\frac{\pi^2}2\frac{\cosh\left(\frac\pi2\right)}{\sinh^3\left(\frac\pi2\right)}
\end{align}
$$