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Hello fellow Stack heads,

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.

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You have that $$ \int_0^{+\infty}{\delta(\cos x)e^{-x}dx}=\sum_{n=0}^{+\infty}{e^{-\left(\frac{\pi}{2}+n\pi\right)}}=\frac{e^{-\frac{\pi}{2}}}{1-e^{-\pi}}=\frac{1}{2\mathrm{sh}\left(\frac{\pi}{2}\right)} $$

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  • $\begingroup$ Oh man that was quick! A bit more than I wanted but I suppose the solution was a one-liner anyway! Thank you so much! $\endgroup$ Commented Sep 14, 2019 at 20:13
  • $\begingroup$ I suppose you use property $\delta(f(x))=...$ given in math.stackexchange.com/q/2481114 for example. $\endgroup$
    – Jean Marie
    Commented Sep 14, 2019 at 20:42
  • $\begingroup$ I used the property $\int_{0}^{+\infty}{\delta(x)f(x)dx}=f(0)$. $\endgroup$
    – Tuvasbien
    Commented Sep 14, 2019 at 20:46

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