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






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.

This entry was posted in Analysis, Compact, Math, Riemann Integrable, Series of Functions, Uniform Convergence. Bookmark the permalink.

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