From Coth to Cot

If $a,b$ are positive integers, then $\begin{aligned} \int_0^\infty e^{-ax} \mathrm{coth}\,x\,\sin bx\,dx & = \int_0^\infty e^{-ax} \left(\frac{1 + e^{-2x}}{1 - e^{-2x}}\right)\,\sin bx\,dx \\ & = \int_0^\infty e^{-ax} \left\{ 1 + 2\sum_{n \ge 1} e^{-2nx}\right\} \sin bx\,dx \\ & = \frac{b}{a^2 + b^2} + 2\sum_{n \ge 1} \frac{b}{(a + 2n)^2 + b^2} \\ & = \sum_{n \ge 0} \frac{b}{(a+2n)^2 + b^2} + \sum_{n \ge 1} \frac{b}{(a + 2n)^2 + b^2} \end{aligned}$ If we define $S_{a,b} \; = \; \int_0^\infty e^{-ax} \frac{\sin 2bx}{\sin x}\,dx$ then we have $S_{a,1} \; = \; 2\int_0^\infty e^{-ax} \cos x\,dx \; = \; \frac{2a}{a^2 + 1}$ while $\begin{aligned} S_{a,m+1} - S_{a,m} & = \int_0^\infty e^{-ax} \frac{\sin 2(m+1)x - \sin 2mx}{\sin x}\,dx \; = \; 2\int_0^\infty e^{-ax} \cos(2m+1)x\,dx \\ & = \frac{2a}{a^2 + (2m+1)^2} \end{aligned}$ and hence $S_{a,b} \; = \; 2a\sum_{k=0}^{b-1} \frac{1}{a^2 + (2k+1)^2}$ Thus $\begin{aligned} \int_0^\infty e^{-ax} \cot x \sin(2b+1)x\,dx & = \tfrac12\int_0^\infty e^{-ax} \frac{\sin2(b+1)x + \sin 2bx}{\sin x}\,dx \\ & = \tfrac12\big[S_{a,b+1} + S_{a,b}\big] \\ & = a\sum_{k=0}^b \frac{1}{a^2 + (2k+1)^2} + a \sum_{k=0}^{b-1} \frac{1}{a^2 + (2k+1)^2} \end{aligned}$ Putting this together, we have $\begin{aligned} \int_0^\infty &e^{-(2b+1)x}\big(\mathrm{coth}\,x + \cot x\big) \sin(2b+1)x\,dx \\ & = \begin{array}{l} \displaystyle \sum_{n \ge 0} \frac{2b+1}{(2b+1+2n)^2 + (2b+1)^2} + \sum_{n \ge 1} \frac{2b+1}{(2b+1 + 2n)^2 + (2b+1)^2} \\ \displaystyle + (2b+1)\sum_{k=0}^b \frac{1}{(2b+1)^2 + (2k+1)^2} + (2b+1) \sum_{k=0}^{b-1} \frac{1}{(2b+1)^2 + (2k+1)^2} \end{array} \\ & = 2(2b+1)\sum_{k=0}^\infty \frac{1}{(2b+1)^2 + (2k+1)^2} \\ & = 2(2b+1) \times \frac{\pi}{4(2b+1)} \tanh\big(\tfrac{(2b+1)\pi}{2}\big) \; = \; \tfrac12\pi \tanh\big(\tfrac{(2b+1)\pi}{2}\big) \end{aligned}$ In particular, we have $\int_0^\infty e^{-2017x}\big(\mathrm{coth}\,x + \cot x\big) \sin(2017x)\,dx \; =\; \tfrac12\pi \tanh\big(\tfrac{2017 \pi}{2}\big)$ making the answer $2 + 2017 + 2 = \boxed{2021}$ .