Question: How do you show that$$\sum\limits_{m\geq3}\left\{4\binom {1/2}m+\binom {1/2}{m-1}\right\}\frac 1{\left(m-\tfrac 52\right)!}\int\limits_{0}^{\infty}\frac {x^{m-3/2}}{e^x-1}\, dx=\int\limits_{0}^{\infty}\frac {3e^{-x}\left(2-2e^x+x+xe^x\right)}{2\sqrt{\pi}x^{3/2}\left(e^x-1\right)}\, dx$$
I’m not even sure where to begin. I’ve thought about expanding the sum, but that just seems to complicate things. Is there an easier way to transform the left-hand side to the right-hand side?
And if possible, is there a way to decompose the fraction on the right-hand side into the sum of different fractions? I still need a way to integrate the right-hand side. Integration via partial fractions is the first thing that comes to mind, but the integrand looks really ugly to decompose.
Note: On the left-hand side, I started off with$$\sum\limits_{m\geq3}\left\{4\binom {1/2}m+\binom{1/2}{m-1}\right\}\zeta\left(m-\tfrac 32\right)$$and used the integration representation of the zeta function to get the above equation.
It is straightforward to show that $$\begin{aligned}\frac {\color{red}{4\binom {1/2}m}+\color{green}{\binom {1/2}{m-1}}}{\color{blue}{\left(m-\tfrac 52\right)!}} &= \left[ {\color{red}{4\frac{{{{( - 1)}^{m - 1}}(2m - 2)!}}{{{2^{2m - 1}}m!(m - 1)!}}} + \color{green}{\frac{{{{( - 1)}^m}(2m - 4)!}}{{{2^{2m - 3}}(m - 1)!(m - 2)!}}}} \right]\color{blue}{\frac{{{2^{2m - 4}}(m - 2)!}}{{\sqrt{\pi}(2m - 4)!}}} \\ &= \frac{{3{{( - 1)}^{m - 1}}(m - 2)}}{{2\sqrt{\pi}m!}}\end{aligned} $$
Hence your sum $S$ equals to $$S = \frac{3}{{2\sqrt \pi }}\int_0^{ + \infty } {\frac{{\sum\limits_{m = 3}^\infty {\frac{{{{( - 1)}^{m - 1}}(m - 2)}}{{m!}}{x^m}} }}{{{x^{3/2}}({e^x} - 1)}}dx} $$
Using the series (directly obtainable from the series of $e^{-x}$) $$\sum\limits_{m = 3}^\infty {\frac{{{{( - 1)}^{m - 1}}(m - 2)}}{{m!}}{x^m}} = {e^{ - x}}(2 - 2{e^x} + x + x{e^x})$$ concludes the proof.