Dirac delta distribution satisfies the following rule: $$ \delta[g(x)] = \sum_m \frac {1} {|g'(x_m)|}\delta(x-x_m)$$ Where $g(x_m)=0,g'(x_m) \neq 0$.

Using that rule, show that $$ \int_0^\infty e^{-x}\delta(cos(x))dx=\frac 1 {2sinh(\frac {\pi} 2)}$$

Hint: using the rule (1) and inserting it into the integral (2), one can easily perform the integration and you should end up with the geometric series that can be explicitly summed.

Okay, so using the hint I plug in the rule and obtain the following: $$ \int_0^\infty e^{-x}\sum_m \frac {1} {|sin(x_m)|}\delta(x-x_m)$$ Now using the fact that the delta function integral goes to 1, and using $e^{-x}$ as my test function, f(x), I obtain the following equation. $$e^{-x_m}\sum_m \frac {1} {|sin(x_m)|} $$

I'm not sure if we can bring the $e^{-x_m}$ in, but here we go... $$\sum_m \frac {e^{-x_m}} {|sin(x_m)|}$$

Now that, I've gotten here I really have no idea how to sum this, or even look at it as a geometric series which I could sum.

Please correct my work above if I made a mistake, and where should I go from here?

  • $\begingroup$ You cannot pull out $e^{-x_m}$ of the penultimate summation. $\endgroup$
    – Jean Marie
    Commented Sep 6, 2016 at 11:33

1 Answer 1


Hint: Recall that the $x_m$'s are determined by $\cos x_m=0$

  • $\begingroup$ Okay, so I did that and got $x_m = -\frac {\pi} 2, + \frac {\pi} 2$, I then evaluate for those values: $$ e^{-(-\frac {\pi} 2)} + e^{-\frac {\pi} 2}$$ which ends up simplifying to: $$ 2cosh(\frac {\pi} 2)$$, and not what it should be: $$ \frac {1} {2sinh(\frac {\pi} 2)}$$ $\endgroup$ Commented Sep 6, 2016 at 19:35
  • $\begingroup$ You should look for the zeros in $(0,+\infty)$. The zeros are: $x_m = \pi/2+m \pi$ where $m=0,1,2,3,...$ and you should see a geometric series popping out! $\endgroup$
    – H. H. Rugh
    Commented Sep 6, 2016 at 19:37

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