## Winter 2008 #5

Problem Statement: Let $f(x)=\sum\limits_{k=1}^{\infty}\dfrac{\cos(kx)}{k^3}$. Prove $f(x)$ is Riemann Integrable on $[0,\frac{\pi}{2}]$ and $\int\limits_{0}^{\frac{\pi}{2}}f(x)dx=\sum\limits_{k=0}^{\infty}\dfrac{(-1)^k}{(2k+1)^4}$

Proof: First note that each $f_k(x)=\dfrac{\cos(kx)}{k^3}$ is continuous on $[0,\frac{\pi}{2}]$ and since $f(x)$ is a sum of continuous functions it follows that $f(x)$ is continuous on $[0,\frac{\pi}{2}]$. Furthermore, since $[0,\frac{\pi}{2}]$ is a compact set it follows that this convergence is uniform. Since we have uniform convergence we get the following:

$\int\limits_{0}^{1}\sum\limits_{k=1}^{\infty}\dfrac{\cos(kx)}{k^3} dx=\sum\limits_{k=1}^{\infty}\int\limits_{0}^{\frac{\pi}{2}}\dfrac{\cos(kx)}{k^3} dx$

$=\sum\limits_{k=1}^{\infty}\frac{1}{k^3}\int\limits_{0}^{\frac{\pi}{2}}\cos(kx)dx$

$=\sum\limits_{k=1}^{\infty}\dfrac{\sin(\frac{k\pi}{2})}{k^4}$

$=\frac{1}{1}=\frac{1(0)}{2^4}+\frac{1(-1)}{3^4}+\frac{1(0)}{4^4}+\frac{1(1)}{5^4}+\dots$

$=\sum\limits_{k=0}^{\infty}\dfrac{(-1)^k}{(2k+1)^4}$

$\Box$

Reflection: At first glance this problem scared me. But, after being reminded by Suz that convergence of a sum to a function on a compact set gives us uniform convergence my life was made easier. This is another one of those handy facts to remember.