# Use of Mellin transform for evaluation of a series

Show that $$\sum_{n=1}^\infty \frac{\sin an}{n}=\frac{\pi-a}{2} \ , \ 0

I was asked to use Mellin transform to prove this result. So I used a formula related to the general series as follows $$\sum_{n=1}^\infty f(nx)=\mathcal{M}^{-1}\{F(s)\zeta(s);x\}$$ where $$F(s)=\mathcal{M}\{f(x);s\}$$ and $$\zeta(s)$$ is Riemann zeta function. Now taking $$f(n)=\displaystyle\frac{\sin an}{n}$$ we have $$\displaystyle\mathcal{M}\bigg(\frac{\sin ax}{x}\bigg)=-\frac{\Gamma(s-1)\cos \frac{s\pi}{2}}{a^{s-1}}$$. Combining everything and using the formula that $$\pi^s\zeta(1-s)=2^{1-s}\Gamma(s)\zeta(s)\cos \frac{s\pi}{2}$$ we have $$\sum_{n=1}^\infty \frac{\sin an}{n}=-\frac{a}{2}\mathcal{M}^{-1}\bigg\{\bigg(\frac{2\pi}{a}\bigg)^s\frac{\zeta(1-s)}{s-1};x=1\bigg\}$$ Now if we use calculus of residues to evaluate the expression at RHS, we see that we have singularity at $$s=1$$. Using $$\zeta(0)=-\frac{1}{2}$$ we can easily calculate the residue at $$s=1$$, which is $$\displaystyle-\frac{\pi}{a}$$. So I get the answer $$\sum_{n=1}^\infty \frac{\sin an}{n}=-\frac{a}{2}.\bigg(-\frac{\pi}{a}\bigg)=\frac{\pi}{2}$$ which is wrong of course as there should be an extra term $$-\displaystyle\frac{a}{2}$$ on RHS. My doubt is I am missing a residue on the RHS while calculating the Mellin inverse, but I am unable to get that. I need a help in this regard. Any help is appreciated.

To use the Mellin summation formula, we must take care on the domain of validity to evaluate the inverse transform. The $$\zeta(s)$$ term is obtained for $$\Re(s)>1$$, while $$$$\displaystyle\mathcal{M}\left(\frac{\sin ax}{x}\right)=-\frac{\Gamma(s-1)\cos \frac{s\pi}{2}}{a^{s-1}}$$$$ is valid for $$0<\Re(s)<2$$. Then \begin{align} \sum_{n=1}^\infty \frac{\sin an}{n}&=-\frac{a}{2}\mathcal{M}^{-1}\left\lbrace\left(\frac{2\pi}{a}\right)^s\frac{\zeta(1-s)}{s-1};x=1\right\rbrace\\ &=-\frac{a}{2}\frac{1}{2i\pi}\int_{c-i\infty}^{c+i\infty}\left(\frac{2\pi}{a}\right)^s\frac{\zeta(1-s)}{s-1}\,ds \end{align} where $$1<\Re(c)<2$$. Closing the the contour with the left large half-circle needs to take into account both poles at $$s=1$$ (residue $$-\pi/a$$, as already noted) and $$s=0$$ (residue $$1$$, as $$\zeta(z)$$ has a simple pole at $$z=1$$, with residue $$1$$). The latter gives the missing term.
• Here $\lim_{s\to 0} s\zeta(1-s)=-1$? Dec 28 '19 at 19:04
• Yes, $\zeta(1-s)\sim -s^{-1}+O\left(s^0\right)$ Dec 28 '19 at 19:26
• See DLMF, for example, which gives the expansion near $1-s=0$. Dec 28 '19 at 19:39