5
$\begingroup$

The task is to calculate sum $\sum_{k=1}^{\infty}\frac{1}{a^2+k^2}$ using Fourier coefficients of $f(x)=e^{ax}$. First of all I calculated Fourier coefficients of the sum:

$$\hat{f(k)}=\frac{1}{2\pi}\int_{0}^{2\pi}e^{ax}e^{-ikx}dx=\frac{e^{a2\pi}-1}{2\pi(a-ik)}$$

Using Parseval identity I calculated the sum: $$\frac{e^{4a\pi}-1}{2a}=\int_{0}^{2\pi}e^{2ax}dx=\sum_{k=-\infty}^{\infty}|\hat{f(k)}|^2=\frac{(e^{2a\pi}-1)^2}{4\pi^2}\cdot \sum_{k=-\infty}^{\infty}\frac{1}{a^2+k^2}=\frac{(e^{2a\pi}-1)^2}{4\pi^2}\cdot (2\cdot\sum_{k=1}^{\infty}\frac{1}{a^2+k^2}-\frac{1}{a^2})$$

and thus:

$$\sum_{k=1}^{\infty}\frac{1}{a^2+k^2}=\frac{a\pi(e^{2a\pi}+1)}{2a^2(e^{2a\pi}-1)}-\frac{1}{2a^2}$$

And in this task there is a subtask to calculate $\sum_{k=1}^{\infty}\frac{1}{(a^2+k^2)^2}$, so it seems it has to be calculated using previous sum, but I found no way to connect these sums, so what could be a the idea to deal with this?

Thanks in advance!

$\endgroup$
1

1 Answer 1

5
$\begingroup$

HINT

Note that $$\dfrac{d}{da} \left(\dfrac1{a^2+k^2}\right) = - \dfrac{2a}{(a^2+k^2)^2}$$

$\endgroup$
2
  • $\begingroup$ Thank you, now I can just derive sum: $$\frac{d}{da}\sum\frac{1}{a^2+k^2}=-2a\sum\frac{1}{(a^2+k^2)^2}$$ This helped a lot. $\endgroup$
    – nakajuice
    Apr 17, 2013 at 20:27
  • $\begingroup$ @starovoitovs: You may need to look at this problem and see the theorem that justify term by term differentiation of a series. $\endgroup$ Apr 17, 2013 at 20:27

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .