1
$\begingroup$

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?

$\endgroup$
  • 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
0
$\begingroup$

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

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.