Compute a sum

**DavidEriksson** · September 22nd 2010, 12:54 AM

Hi! I need to show that
$\frac{8}{\pi^2} \sum_{k=0}^{\infty}\frac{\sin(\frac{1}{2}(2k+1)\pi t) \sin(\frac{1}{2}(2k+1)\pi s)}{(2k+1)^2}= \min(t,s)$ for $t,s \in (0,1)$ .

**Opalg** · September 22nd 2010, 05:08 AM

Originally Posted by DavidEriksson

Hi! I need to show that
$\frac{8}{\pi^2} \sum_{k=0}^{\infty}\frac{\sin(\frac{1}{2}(2k+1)\pi t) \sin(\frac{1}{2}(2k+1)\pi s)}{(2k+1)^2}= \min(t,s)$ for $t,s \in (0,1)$ .

This looks like the sort of sum that arises in Fourier theory, so I looked for a function that might have this as its Fourier series.

Let $0<T<\pi/2$ , and define a function by $f(x) = \begin{cases}x&(0\leqslant x\leqslant T),\\ T&(T\leqslant x\leqslant\pi-T),\\ \pi-x&(\pi-T\leqslant x\leqslant\pi).\end{cases}$

Make f into an odd function on the interval $[-\pi,\pi]$ by defining $f(-x)=-f(x).$ Notice that $f(x) = \min\{x,T\}$ when $x\in [0,\pi/2].$

Since f is odd, its Fourier cosine coefficients will all be 0. The sine coefficients will be given by

$b_n = \displaystyle\frac2\pi\Bigl(\int_0^T\!\!x\sin nx\,dx + \int_T^{\pi-T}\!\!\!T\sin nx\,dx + \int_{\pi-T}^\pi(\pi-x)\!\sin nx\,dx\Bigr).$

Grind out thtose integrals, using integration by parts as necessary. You will find that $b_n = \tfrac{4\sin nT}{\pi n^2}$ if n is odd, and $b_n=0$ if n is even. The function f is continuous and piecewise smooth, so it is the sum of its Fourier series. Therefore $f(x) = \displaystyle\sum_{k=0}^\infty\frac{4\sin\bigl((2k +1)T\bigr)}{\pi(2k+1)^2}\sin\bigl((2k+1)x\bigr).$

Now put $x = \tfrac12\pi s$ and $T = \tfrac12\pi t$ to see that $\min\{\tfrac12\pi s,\tfrac12\pi t\} = \displaystyle\sum_{k=0}^\infty\frac{4\sin\bigl((2k +1)\tfrac12\pi t\bigr)}{\pi(2k+1)^2}\sin\bigl((2k+1)\tfrac12\pi s\bigr).$ That's close enough to the answer that I needn't continue.

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