# Difficult Integral Involving the Dirac Delta Function

I am stuck on a difficult integral that almost looks like it can be accomplished with a one-sided Laplace transform but more than likely can be solved using Dirac Delta simplification techniques. I must confirm that

$$I=\displaystyle{\int_{0}^{-\infty}}dx\,\delta(\cos(x))e^{-x}=\dfrac{1}{2\sinh\bigg(\dfrac{\pi}{2}\bigg)}.$$

Here is what I have so far using a previous question:

$$\delta(\cos(x))\Rightarrow g(x)=\cos(x)=0 \Rightarrow x=\dfrac{\pi}{2}(2n+1), \text{ where } n\in\mathbb{Z}_{+}\cup\{0\}.$$ Then $$g'(x)=-\sin(x).$$ So I have $$I=\displaystyle{\int_{0}^{-\infty}}dx\,\dfrac{\delta[x-\dfrac{\pi}{2}(2n+1)]}{\big\lvert\,-\sin(\dfrac{\pi}{2}(2n+1))\big\rvert}e^{-x}=\dfrac{1}{2\sinh\bigg(\dfrac{\pi}{2}\bigg)}.$$ Now I am SOL. Do I proceed with a Laplace Transform? Please help me by providing a hint and I will finish the problem and arrive at the solution with honor and dignity. Thank you all for your help and your time.

You have that $$\int_0^{+\infty}{\delta(\cos x)e^{-x}dx}=\sum_{n=0}^{+\infty}{e^{-\left(\frac{\pi}{2}+n\pi\right)}}=\frac{e^{-\frac{\pi}{2}}}{1-e^{-\pi}}=\frac{1}{2\mathrm{sh}\left(\frac{\pi}{2}\right)}$$
• I suppose you use property $\delta(f(x))=...$ given in math.stackexchange.com/q/2481114 for example. Commented Sep 14, 2019 at 20:42
• I used the property $\int_{0}^{+\infty}{\delta(x)f(x)dx}=f(0)$. Commented Sep 14, 2019 at 20:46