I'm trying to solve the following integral:

$\int_{-\infty}^{\infty} \frac{\cos(\pi (x+1))}{x}[\sum_{n=1}^{\infty}\delta (x-n)]dx$

So this seems pretty terrible, and there is also a hint

Hint: "Don't be afraid". Nevertheless, I am afraid.

How do you start solving this? I know I'm supposed to show some effort, but I really have no idea where to start. Maybe integration by parts?

  • 1
    $\begingroup$ Do you know what $\cos(n \pi)$ evaluates to for an integer number $n$? $\endgroup$ – s.harp Oct 7 '15 at 17:35
  • $\begingroup$ What would the answer be if you replaced the sum of deltas with one delta? $\endgroup$ – fred Oct 7 '15 at 17:36
  • $\begingroup$ Ofcourse. It's $(-1)^{n}$ $\endgroup$ – Rick Joker Oct 7 '15 at 17:36
  • $\begingroup$ Do you know what the delta distribution does to functions? It may be better for you to consider $\sum_{n=0}^N \delta (x-n)$, do the integral and then take the limit $N \to \infty$. $\endgroup$ – s.harp Oct 7 '15 at 17:38
  • $\begingroup$ It should cancel out all the none integer values so the integrand is equivalent to $\sum_{n=1}^{\infty} (-1)^n$ correct? $\endgroup$ – Rick Joker Oct 7 '15 at 17:42

We simply use the "sifting" property of the Dirac Delta to obtain

$$\begin{align} \int_{-\infty}^{\infty}\frac{\cos(\pi(x+1))}{x}\sum_{n=1}^{\infty}\delta(x-n)\,dx&=\sum_{n=1}^{\infty}\frac{\cos(\pi(n+1))}{n}\\\\ &=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}\\\\ &=\log 2 \end{align}$$


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