Computing integral involving Dirac Delta Function 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.
 A: I'll comment on calculating the sum. First, the sum $\sum\limits_{n=0}^{\infty}n^2 e^{-\pi n}$ certainly does converge, just as $\int_0^{\infty} x^2 e^{-\pi x} \, dx$ converges. Here's how I would go about finding a closed expression which is equal that sum:
Note that $n^2 e^{-\pi n} = \frac{d^2}{d\lambda^2}\Big|_{\lambda = \pi}e^{-\lambda n}$, and since we will be evaluating $\lambda$ at $\pi$, we can always assume $\lambda > 1$. Now:
$$
\sum_{n=0}^{\infty} n^2 e^{-\pi n} = \sum\limits_{n=0}^{\infty}\frac{d^2}{d\lambda^2}\Big|_{\lambda = \pi}e^{-\lambda n} = \frac{d^2}{d\lambda^2}\Big|_{\lambda = \pi}\sum\limits_{n=0}^{\infty}e^{-\lambda n} \\
=\frac{d^2}{d\lambda^2}\Big|_{\lambda = \pi} \sum\limits_{n=0}^{\infty} \left(e^{-\lambda}\right)^n  = \frac{d^2}{d\lambda^2}\Big|_{\lambda = \pi} \frac{1}{1 - e^{-\lambda}} = \cdots
$$
where the last evaluated equality comes from the geometric series formula. From here you only need to evaluate the differentiation.
A: 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,
$$
\begin{align}
2\pi^2\sum_{n=0}^\infty n^2e^{-\pi n}
&=2\pi^2\frac{e^{-\pi}+e^{-2\pi}}{\left(1-e^{-\pi}\right)^3}\\
&=\frac{\pi^2}2\frac{\cosh\left(\frac\pi2\right)}{\sinh^3\left(\frac\pi2\right)}
\end{align}
$$
A: Actually Wolfram Alpha is wrong.
The series
$$\sum_{n = 0}^{+\infty} n^2 e^{-n\pi}$$
Does converge to 
$$\frac{e^{\pi } \left(1+e^{\pi }\right)}{\left(e^{\pi }-1\right)^3}$$
Which is easy provable by using differentiation under the summation sign together with the geometric series.
