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.

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

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