# 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.

• Come on, do you REALLY think that series does not converge? Don't treat computer algebra software as if it were the truth. It is there to help with computations, not to replace our brains. – Giuseppe Negro Dec 9 '17 at 15:30
• @GiuseppeNegro Exactly. People really should stop believe that blindly in technologies. Our brains are ways more powerful than any software, now and ever. ^_^ We just need to reason, study and do. – Von Neumann Dec 9 '17 at 15:31
• @GiuseppeNegro that's what I was thinking! – Matthew Dec 9 '17 at 15:39
• WolframAlpha says that last sum converges – Hurkyl Dec 10 '17 at 23:23

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.

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}

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.

• I'm pretty sure its user error, not WA error. – Hurkyl Dec 10 '17 at 23:23
• @Hurkyl I can't say, maybe you're right. I never use WA because it's full of bugs and mistakes. – Von Neumann Dec 11 '17 at 1:02