Hi! I need to show that
$\displaystyle \frac{8}{\pi^2} \sum_{k=0}^{\infty}\frac{\sin(\frac{1}{2}(2k+1)\pi t) \sin(\frac{1}{2}(2k+1)\pi s)}{(2k+1)^2}= \min(t,s)$ for $\displaystyle 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 $\displaystyle 0<T<\pi/2$, and define a function by $\displaystyle f(x) = \begin{cases}x&(0\leqslant x\leqslant T),\\ T&(T\leqslant x\leqslant\pi-T),\\ \pi-x&(\pi-T\leqslant x\leqslant\pi).\end{cases}$
Make f into an odd function on the interval $\displaystyle [-\pi,\pi]$ by defining $\displaystyle f(-x)=-f(x).$ Notice that $\displaystyle f(x) = \min\{x,T\}$ when $\displaystyle x\in [0,\pi/2].$
Since f is odd, its Fourier cosine coefficients will all be 0. The sine coefficients will be given by
$\displaystyle b_n = \displaystyle\frac2\pi\Bigl(\int_0^T\!\!x\sin nx\,dx + \int_T^{\pi-T}\!\!\!T\sin nx\,dx + \int_{\pi-T}^\pi(\pi-x)\!\sin nx\,dx\Bigr).$
Grind out thtose integrals, using integration by parts as necessary. You will find that $\displaystyle b_n = \tfrac{4\sin nT}{\pi n^2}$ if n is odd, and $\displaystyle b_n=0$ if n is even. The function f is continuous and piecewise smooth, so it is the sum of its Fourier series. Therefore $\displaystyle f(x) = \displaystyle\sum_{k=0}^\infty\frac{4\sin\bigl((2k +1)T\bigr)}{\pi(2k+1)^2}\sin\bigl((2k+1)x\bigr).$
Now put $\displaystyle x = \tfrac12\pi s$ and $\displaystyle T = \tfrac12\pi t$ to see that $\displaystyle \min\{\tfrac12\pi s,\tfrac12\pi t\} = \displaystyle\sum_{k=0}^\infty\frac{4\sin\bigl((2k +1)\tfrac12\pi t\bigr)}{\pi(2k+1)^2}\sin\bigl((2k+1)\tfrac12\pi s\bigr).$ That's close enough to the answer that I needn't continue.