Can anyone offer any advice on computing - analytically if possible - the following related, Fourier-like integrals: $$I_{1}(p, B, b) \equiv \int_{0}^{\infty}dx\, e^{ipx - B\frac{\tanh(bx)}{bx}}; \qquad I_{2}(p, c, D) \equiv \int_{-\infty}^{\infty} dx\, \frac{x}{\sinh(c x)}e^{i p x - D x \coth(cx)},$$ where $p$, $B$, $b$, $c$ and $D$ are arbitrary real constants.
For $I_{2}$ I have the following attempt at a solution: When $D = 0$, $I_{2}$ can can be computed by continuing $x$ to the complex plane and deforming the contour of integration (along the real line) so that it closes just above the first non-zero pole of $\sinh(cx)$ on the imaginary axis -- see here, for example.
Unfortunately the same trick does not work when $D \neq 0$. The x multiplying $\coth(cx)$ means that the integral along the contour just above the pole is not simply the negative of $I_{2}$. Furthermore, at the poles of $\sinh(cx)$ the D-dependent part of the exponent either blows up due to the same factor in the denominator of $\coth(cx) = \frac{\cosh(cx)}{\sinh(cx)}$ leaving either $e^{ipx}$ or a divergent quantity.
I also tried writing $$I_{2} = \int_{0}^{\infty} dx \,- \frac{\partial_{D}}{\cosh(cx)} e^{ipx - Dx \coth(cx)}$$ which transfers the poles to those of $\cosh(cx)$, on the imaginary axis half way between each of the poles of $\sinh(cx)$. Then I tried exchanging the integral and differential: unfortunately at the new poles, the residue is independent of $D$, due to the $\cosh(cx)$ in the numerator of the $\coth(cx)$, so that the resulting integral is independent of $D$....I guess it's one of those cases where one cannot exchange the limiting process of calculating the residue at the poles with the limiting process of differentiating with respect to $D$.
The integral $I_{1}$ does not have a pole structure to take advantage of so it looks as if it must be expressed in terms of special functions. When $D \neq 0$ perhaps the same is true of $I_{2}$. I'd appreciate some advice on making progress with these calculations.
