Dirac-$\delta$ integral evaluation. Infinite series value?

Question

Evaluate $$\int_0^\infty \delta(cos(x))e^{-x}dx$$

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This is a homework problem I'm dealing with. Here's my solution:

$\delta$ function of a function is given by $$\delta (g(t))=\sum_n \frac{\delta (t-t_n)}{|g'(t_n)|}$$ where $t_n$ are the roots of $g(t)$. Let $t=cos(x)$ $$\delta (cos(x))=\sum_n \frac{\delta (x-(2n+1)\frac{\pi} 2)}{|-sin((2n+1)\frac{\pi} 2)|}$$ $$\int_0^\infty \delta(cos(x))e^{-x}dx=\int_0^\infty e^{-x}\sum_n {\delta (x-(2n+1)\frac{\pi} 2)}dx$$ $$=\sum_{n=0}^\infty e^{-(2n+1)\pi/2}=e^{-\pi/2}+e^{-3\pi/2}+e^{-5\pi/2}+...$$ I believe everything is correct so far. My question is, how can I determine the value this series converges to, or, is there a simpler representation of it?

Answer 1

You've done the vast majority of the work. Now you have

\begin{align}\sum_{n=0}^\infty e^{-(2n+1)\pi/2} &= e^{-\pi/2} \sum_{n=0}^\infty e^{-n \pi} \\ & = e^{-\pi/2} \sum_{n=0}^\infty \left ( e^{-\pi} \right )^n \\ & = \frac{e^{-\pi/2}}{1-e^{-\pi}} \end{align}

using the geometric series.