Consider the integral \[I=\int_0^\infty\frac{\sin(x)}{e^x-1}dx\]
We can approximate it with a Riemann sum in J:
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Using Euler’s formula, \(\sin(x)=(e^{ix}-e^{-ix})/2i\), and the geometric series \[\frac1{1/r-1}=r+r^2+r^3+\ldots\] for \(r=e^{-x}\), we get \[I=\sum_{n=1}^\infty\int_0^\infty\frac1{2i}\left(e^{(i-n)x}-e^{(-i-n)x}\right)dx=\frac1{2i}\sum_{n=1}^\infty\left(\frac1{-i+n}+\frac1{-i-n}\right)\]
We can also combine both fractions to get \(1/(n^2+1)\).
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Using the series \[\pi\cot(\pi z)=\sum_{n=-\infty}^\infty\frac1{z+n}\] we then get \(I=(-\pi i\cot(-\pi i)-1)/2\), or \((\pi\coth(\pi)-1)/2\), or \[I=\frac\pi2\frac{e^\pi+e^{-\pi}}{e^\pi-e^{-\pi}}-\frac12\]
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A more constructive way to view the integral is to start with \(\int_0^\infty e^{-x}dx=1\) and scaling the \(x\)-axis by a factor \(n\) to get \[\int_0^\infty e^{-nx}dx=\frac1n\] We can then differentiate both sides \(k\) times with respect to \(n\) and rearrange: \[\int_0^\infty\frac1{k!}x^ke^{-nx}dx=\frac1{n^{k+1}}\]
Because of the Taylor series of \(\sin\) and the geometric series we used, \(I\) is then a double sum over positive \(n\) and positive odd \(k\), with alternating signs. Summing over \(k\) first, we recover the values \(1/(n^2+1)\) we found earlier. If we instead sum over all \(n\) first and then sum over \(k\), we get an alternating sum of values of the Riemann zeta function \[I=\zeta(2)-\zeta(4)+\zeta(6)-\zeta(8)+\ldots\]
The sum doesn’t converge, because we weren’t justified in exchanging the limits.
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The Cesàro sum is defined as the average of the partial sums and converges slowly to \(I\). Because \(\zeta(2k)-1\downarrow 0\), its alternating sum does converge quickly, but we have to add \(\frac12\).
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Finally, summing over all \(k,n\) in no particular order:
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