INFINITE SUM cos(kx)/a^+k^2

In today´s post we will proof the following beautiful infinite sum

$\sum_{k=1}^\infty \frac{\cos (kx)}{a^2+k^2}=\frac{\pi \cosh(a\pi-ax)}{2a\sinh(a \pi)}-\frac{1}{2a^2 }$

To this end we will use Fourier series.

Recall the Fourier series representation

$f(x)=\frac{a_0}{2}+\sum_{k=1}^\infty a_k\cos (kx) + \sum_{k=1}^\infty b_k \sin(kx)$ (1)

where the coefficients are given by

$\begin{aligned} &a_{0}=\frac{1}{\pi} \int_{-\pi}^{\pi} f(x) d x \\ &a_{k}=\frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \cos (k x) d x \\ &b_{k}=\frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin (k x) d x \end{aligned}$

If we choose $f(x)=\cosh(ax)$

We obtain

$\cosh(ax)=\frac{\sinh(a \pi)}{a \pi}+\frac{2a\sinh(a\pi)}{\pi}\sum_{k=1}^\infty \frac{(-1)^k\cos (kx)}{a^2+k^2}$ (2)

Proof:

$\begin{aligned} a_0&=\frac1 \pi\int_{-\pi}^{\pi} \cosh(ax)\,dx\\ &=\frac{\sinh(ax)}{a\pi}\Bigg|_{-\pi}^{\pi}\\ &=\frac{\sinh(a\pi)-\sinh(-a\pi)}{a \pi}\\ &=\frac{\sinh(a\pi)+\sinh(a\pi)}{a \pi}\\ &=\frac{2\sinh(a\pi)}{a \pi} \qquad \blacksquare\\ \end{aligned}$

$\begin{aligned} a_k&=\frac1 \pi\int_{-\pi}^{\pi}\cos(kx)\cosh(ax)\,dx\\ &=\frac2 \pi\int_{0}^{\pi}\cos(kx)\cosh(ax)\,dx\\ &=\frac2 \pi\int_{0}^{\pi}\cos(kx)\left( \frac{e^{ax}+e^{-ax}}{2}\right)\,dx\\ &=\frac1 \pi\int_{0}^{\pi}e^{ax}\cos(kx)\,dx+\int_{0}^{\pi}e^{-ax}\cos(kx)\,dx\\ &=\frac1 \pi Re\left(\int_{0}^{\pi}e^{ax}e^{ikx}\,dx+\int_{0}^{\pi}e^{-ax}e^{ikx}\,dx\right)\\ &=\frac1 \pi Re\left(\int_{0}^{\pi}e^{(a+ik)x}\,dx+\int_{0}^{\pi}e^{-(a-ik)x}\,dx\right)\\ &=\frac1 \pi Re\left(\frac{e^{(a+ik)x}}{a+ik}\Bigg|_{0}^{\pi}-\frac{e^{-(a-ik)x}}{a-ik}\Bigg|_{0}^{\pi}\right)\\ &=\frac1 \pi Re\left(\frac{e^{a\pi}\cos(k \pi)-1}{a+ik}+\frac{1-e^{-a\pi}\cos(k \pi)}{a-ik}\right)\\ &=\frac1 \pi Re\left(\frac{e^{a\pi}(-1)^k-1}{a+ik}+\frac{1-e^{-a\pi}(-1)^k}{a-ik}\right)\\ &=\frac1 \pi Re\left(\frac{\left(a-ik\right)\left(e^{a\pi}(-1)^k-1\right)}{a^2+k^2}+\frac{\left(a+ik\right)\left(1-e^{-a\pi}(-1)^k\right)}{a^2+k^2}\right)\\ &= \frac{a(-1)^k\left(e^{a\pi}-e^{-a\pi} \right)}{\pi(a^2+k^2)}\\ &=\frac{2a(-1)^k\sinh(a\pi)}{\pi(a^2+k^2)} \qquad \blacksquare \end{aligned}$

$\begin{aligned} b_k&=\frac1 \pi\int_{-\pi}^{\pi}\sin(kx) \cosh(ax)\,dx\\ &=\frac1 \pi\int_{-\pi}^{0}\sin(kx) \cosh(ax)\,dx+\frac1 \pi\int_{0}^{\pi}\sin(kx) \cosh(ax)\,dx\\ &=\frac1 \pi\int_{0}^{\pi}\sin(-kx) \cosh(-ax)\,dx+\frac1 \pi\int_{0}^{\pi}\sin(kx) \cosh(ax)\,dx\\ &=-\frac1 \pi\int_{0}^{\pi}\sin(kx) \cosh(ax)\,dx+\frac1 \pi\int_{0}^{\pi}\sin(kx) \cosh(ax)\,dx\\ &=0 \qquad \blacksquare \end{aligned}$

Plugging the coefficients in (1) and rearranging we obtain

$\sum_{k=1}^\infty \frac{(-1)^k\cos (kx)}{a^2+k^2}=\frac{\pi\cosh(ax)}{2a\sinh(a \pi)}-\frac{1}{2a^2 }$ (3)

Letting $x \to \pi-x$

$\begin{aligned} \frac{\pi \cosh(a\pi-ax)}{2a\sinh(a \pi)}-\frac{1}{2a^2 }&=\sum_{k=1}^\infty \frac{(-1)^k\cos (k\pi-kx)}{a^2+k^2}\\ &=\sum_{k=1}^\infty \frac{(-1)^k\cos (k\pi)\cos (kx)}{a^2+k^2}+\sum_{k=1}^\infty \frac{(-1)^k\sin (k\pi)\sin (kx)}{a^2+k^2}\\ &=\sum_{k=1}^\infty \frac{(-1)^k(-1)^k\cos (kx)}{a^2+k^2}\\ &=\sum_{k=1}^\infty \frac{\cos (kx)}{a^2+k^2} \qquad \blacksquare\\ \end{aligned}$