Hot Integral - 20

The necessary integrals can be found in tables of standard integrals, but let us derive the answer properly.

For $0 < u < 1$ we have $I(u) \; = \; \int_{\mathbb{R}} \frac{e^{ux}}{\sinh x - i}\,dx \; = \; \int_{\mathbb{R}} \frac{2e^{ux}}{e^x - 2i - e^{-x}}\,dx \; = \; \int_{\mathbb{R}} \frac{2 e^{(u+1)x}}{(e^x-i)^2}\,dx$ Using the substitution $t = e^x$ we obtain $I(u) \; = \; \int_0^\infty \frac{2t^u}{(t-i)^2}\,dt \qquad \qquad 0 < u < 1 \;.$ The function $F(z) \; = \; \frac{2z^u}{(z-i)^2}$ can be defined, and is analytic, on the domain obtained by removing the positive real axis (so that $0 < \mathrm{Arg}z < 2\pi$ throughout) and the point $i$ . Integrating this function around the "keyhole" contour which runs:

• from $\epsilon$ to $R$ just above the positive real axis,
• from $R$ to $Re^{2\pi i}$ in a circle of radius $R$ and centre $0$ ,
• from $Re^{2\pi i}$ to $\epsilon e^{2\pi i}$ just below the positive real axis,
• from $\epsilon e^{2\pi i}$ to $\epsilon$ in a circle of radius $\epsilon$ and centre $0$

where $0 < \epsilon < 1 < R$ equals $2\pi i$ times the residue of $F(z)$ at $z=i$ , and so, letting $R \to \infty$ and $\epsilon \to 0$ , we see that $\begin{array}{rcl} I(u) - e^{2\pi u i}I(u) & = & 2\pi i \mathrm{Res}_{z=i} F(z) \; = \; 2\pi i \Big(\dfrac{d}{dz}(z-i)^2F(z)\Big)\Big\vert_{z=i} \\ & = & 2\pi i \big(2u z^{u-1}\big)\Big\vert_{z=i} \; = \; 4\pi u e^{\frac12\pi u i} \end{array}$ since $F(z)$ has a double pole at $z=i$ . Thus $\begin{array}{rcl} I(u) & = &\displaystyle -\frac{4\pi u e^{\frac12\pi u i}}{e^{2\pi u i} - 1} \; = \; 2i e^{\frac12\pi u i} \frac{2\pi u i}{e^{2\pi u i} - 1} \; = \; 2ie^{\frac12\pi u i} \sum_{m=0}^\infty B_m \dfrac{(2 \pi u i)^m}{m!} \\ & = & \displaystyle 2i \left( \sum_{m=0}^\infty \dfrac{(2\pi u i)^m}{4^m m!}\right)\left(\sum_{m=0}^\infty B_m \dfrac{(2\pi u i)^m}{m!}\right) \; = \; 2i \sum_{m=0}^\infty \left(\sum_{k=0}^m {m \choose k} \dfrac{B_k}{4^{k-m}}\right) \dfrac{(2\pi ui)^m}{m!} \end{array}$ for all $0 < u < 1$ , where $B_0,B_1,B_2,\ldots$ are the Bernouilli numbers, so that $J_m \; = \; \int_{\mathbb{R}} \frac{x^m}{\sinh x - i}\,dx \; = \; 2i(2\pi i)^m \sum_{k=0}^m {m \choose k}\dfrac{B_k}{4^{m-k}} \;, \qquad \qquad m \ge 0 \;.$ But then $\begin{array}{rcl} J_1 & = & -4\pi\left(\tfrac14B_0 + B_1\right) \; = \; \pi \\ J_2 & = & -8\pi^2i\left(\tfrac{1}{16}B_0 + \tfrac12B_1 + B_2\right) \; = \; \tfrac16\pi^2 i \\ J_4 & = & 32\pi^4 i \left( \tfrac{1}{256}B_0 + \tfrac{1}{16}B_1 + \tfrac38B_2 + B_3 + B_4\right) \; = \; \tfrac{7}{120}\pi^4i \end{array}$ and so the desired integral is $J_1 + J_2 + J_4 \; = \; \pi + i\left( \tfrac16\pi^2 + \tfrac{7}{120}\pi^4\right)$ making the answer equal to $1 + 2 + 6 + 7 + 4 + 120 \,=\, 140$ .