How does one convert an integrand of the form $\frac{x\sinh x-t\sinh t}{\sinh^2x-\sinh^2t}$ into the form $\frac{\ln(x^2-t^2+1)}{\sinh^2x-\sinh^2t}$?

Solution 1:

UPDATE:

After looking at the paper more carefully, I think the author is referring to the formula for $$\operatorname{PV} \int_{-\infty}^{\infty} \frac{e^{-iax}}{\sinh x + \sinh t} \, \mathrm dx.$$ I think they want the reader to differentiate under the integral sign.

There's not much information available about about when DUTIS can be justified when the integral only exists as a Cauchy principal value integral.

The following approach is more direct and easier to justify.

Let's integrate the function $$f(z) = \frac{z^{\color{red}{2}}}{\sinh z+ \sinh t}, \quad t \in \mathbb{R}, $$ around an indented rectangular contour $C_{R}$ with vertices at $z= \pm R, \pm R + 2 \pi i. $

There are simple poles on the contour at $z= -t$ and $z= -t + 2 \pi i$. There is also a simple pole inside the contour at $z= t+ \pi i $.

Letting $R \to \infty$, the integral vanishes on the vertical sides of the contour because the magnitude of $\sinh z$ grows exponentially as $\Re(z) \to \pm \infty$.

Therefore, we have $$ \begin{align} \lim_{R \to \infty}\int_{C_{R}} f(z) \mathrm \, dz &= \operatorname{PV} \int_{-\infty}^{\infty} \frac{x^{2}}{\sinh x + \sinh t} \, \mathrm dx - \operatorname{PV}\int_{-\infty}^{\infty} \frac{(x+2 \pi i)^{2}}{\sinh x + \sinh t} \, \mathrm dx \\ &= -4 \pi i \operatorname{PV} \int_{-\infty}^{\infty}\frac{x}{\sinh x + \sinh t} \, \mathrm dx + 4 \pi^{2} \operatorname{PV}\int_{-\infty}^{\infty} \frac{1}{\sinh x + \sinh t} \, \mathrm dx \\ &= \pi i \operatorname{Res}[f(z), -t] + \pi i \operatorname{Res} [f(z), -t+ 2 \pi i] + 2 \pi i\operatorname{Res}[f(z), t + \pi i] \\ &= \frac{8 \pi^{2} t}{\cosh t} - i \, \frac{2 \pi^{3}}{\cosh t}. \end{align} $$

Equating the imaginary parts on both sides of the equation, we get $$\operatorname{PV} \int_{-\infty}^{\infty}\frac{x}{\sinh x + \sinh t} \, \mathrm dx = \frac{\pi^{2}}{2\cosh t}. $$

And equating the real parts on both sides of the equation, we get $$\operatorname{PV} \int_{-\infty}^{\infty}\frac{1}{\sinh x + \sinh t} \, \mathrm dx = \frac{2t}{\cosh t}. $$

Therefore, $$ \int_{-\infty}^{\infty} \frac{x+t}{\sinh x + \sinh t} \, \mathrm dx = \frac{\pi^{2} }{2\cosh t} + \frac{2t^{2}}{\cosh t} = \frac{\pi^{2} + 4 t^{2}}{2 \cosh t}.$$

The Cauchy principal value sign was dropped because the integral converges in the traditional sense.

If we integrate $$g(z) = \frac{z^{3}}{\sinh z + \sinh t} $$ around the same contour and then equate the imaginary parts on both sides of the equation, we find that $$-6 \pi \operatorname{PV} \int_{-\infty}^{\infty} \frac{x^{2}}{\sinh x + \sinh t} \, \mathrm dx + 8 \pi^{3} \operatorname{PV}\int_{-\infty}^{\infty} \frac{1}{\sinh x + \sinh t} \, \mathrm dx = \frac{18\pi^{3}t - 4 \pi t^{3}}{\cosh t}. $$

Therefore, $$\operatorname{PV} \int_{-\infty}^{\infty} \frac{x^{2}}{\sinh x + \sinh t} \, \mathrm dx = - \frac{1}{6 \pi} \left(\frac{18\pi^{3}t - 4 \pi t^{3}}{\cosh t}- 8 \pi^{3} \frac{2t}{\cosh t} \right)= \frac{2t^{3} - \pi^{2}t}{3\cosh t}, $$ and $$\int_{-\infty}^{\infty} \frac{x^{2}-t^{2}}{\sinh x + \sinh t} \, \mathrm dx = \frac{2t^{3} - \pi^{2}t}{3\cosh t} - \frac{2t^{3}}{\cosh t} = -\frac{t \left(\pi^{2}+ 4 t^{2} \right)}{3\cosh t}.$$