# Summation to Integral

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.

• Some context about where you saw the question would be helpful, or at least some reason to believe that these are equal. – Erick Wong Oct 14 '17 at 5:48
• @Erick Wong I’m trying to prove one of Ramanujan’s identities $\sum\limits_{n\geq 0}(\sqrt{n+1}-\sqrt{n})^3=\frac {3}{2\pi}\zeta\left(\tfrac 32\right)$, and I got to that by the Taylor expansion of the left expression. (I’m following an answer by Jack D Aurizio) – Crescendo Oct 14 '17 at 14:21

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.
• Good answer, but I’m curious. How do you see that$$\sum\limits_{m\geq3}\frac {(-1)^{m-1}(m-2)}{m!}x^m=e^{-x}(2-2e^x+x+xe^x)$$ I’m not seeing an easy way... – Crescendo Oct 15 '17 at 2:55
• @Crescendo Use $$\sum_{m\geq 0} \frac{(-1)^{m}x^m}{m!} = e^{-x}$$ differentiate to get $$\sum_{m\geq 0} \frac{(-1)^{m}mx^{m-1}}{m!} = -e^{-x}$$ and the desired series follows by shifting terms – pisco Oct 15 '17 at 4:12