$\newcommand{\pv}{\mathrm{P.V.}}\newcommand{\Log}{\mathrm{Log}}\newcommand{\d}{\,\mathrm{d}}$This is exercise 4 from a paper - pages 53, 54 - about the contour integration of Malmsten's integrals and related logarithmic forms.
I successfully completed exercise $6$ $a$ through $c$, but am stuck on the applications of these results. I shown that, where $a,t$ are positive reals:
$$\begin{align}\pv\int_{-\infty}^\infty\frac{\ln(x^2+a^2)}{\sinh x\pm\sinh t}\d x &= \pm\frac{1}{\cosh t} \left\{4t\ln2\pi+\pi^2-2\pi\arctan\frac{a}{t} \vphantom{\frac12}\right.\\ &\left. {} + 4\pi\cdot\Im\left[\Log\Gamma\left(\frac{a+it}{2\pi}\right)-\Log\Gamma\left(\frac{1}{2}+\frac{a-it}{2\pi}\right)\right]\right\} \end{align}$$
With the corollaries (for $0\lt t\le1$):
$$\begin{align} &\int_{-\infty}^\infty\frac{\ln(x^2-t^2+1)}{\sinh x\pm\sinh t}\d x \\ & \hspace{30pt} = \pm\frac{1}{\cosh t}\left\{4t\ln2\pi+\pi^2-2\pi\arctan\frac{\sqrt{1-t^2}}{t}\right. \\ & \hspace{90pt}\left.{} + 4\pi\cdot\Im\biggl[\Log\Gamma\Bigl(\frac{\sqrt{1-t^2}+it}{2\pi}\ \Bigr)-\Log\Gamma\Bigl(\frac{1}{2}+\frac{\sqrt{1-t^2}-it}{2\pi}\Bigr)\biggr]\right\} \\ &\int_0^\infty\frac{\ln(x^2-t^2+1)}{\sinh^2t-\sinh^2x}\d x \\ & \hspace{30pt} =\frac{1}{\sinh2t} \left\{4t\ln2\pi+\pi^2-2\pi\arctan\frac{\sqrt{1-t^2}}{t} \right.\\ & \hspace{90pt}\left.{} + 4\pi\cdot\Im\biggl[\Log\Gamma\Bigl(\frac{\sqrt{1-t^2}+it}{2\pi}\Bigr)-\Log\Gamma\Bigl(\frac{1}{2}+\frac{\sqrt{1-t^2}-it}{2\pi}\Bigr)\biggr]\right\} \end{align}$$
The author, without only the hint that these may be found "with the help of the last formula", leaves the following integrals to be evaluated:
$$\begin{align}\tag{1}\int_{-\infty}^\infty\frac{x+t}{\sinh x+\sinh t}\d x&=2\int_0^\infty\frac{x\sinh x-t\sinh t}{\sinh^2x-\sinh^2t}\d x=\frac{\pi^2+4t^2}{2\cosh t}\\\tag{2}\int_{-\infty}^\infty\frac{x^2-t^2}{\sinh x+\sinh t}\d x&=-\frac{t(\pi^2+4t^2)}{3\cosh t}\end{align}$$
Etc.
I have spent hours staring at these now. I was able to show the above integral formulae but I don't have a clue how to apply them to these integrals. It is strongly unclear to me how to convert the numerator of the form $x\sinh x-t\sinh t$ into one of the form $\ln(x^2-t^2+1)$. Common themes in this paper have been taking limits as the constant $a$ goes to zero, and substitutions of the form $x\mapsto e^x$ to convert hyperbolics into polynomials, and indeed even taking principal values and then applying them to convergent integrals; I've thought about all of this, and just cannot see the link. I considered contour integration; the problem is, which contour, and what integrand? If these could be evaluated by plain and simple contour integration, they wouldn't be in this paper! Those integrals are supposed to be related somehow to the three formulae at the top, especially the third one, and the provided answers are similar in spirit to the formulae, but many of the log terms, in particular the log gamma term, are missing. This suggests I find integrals $(1)$ and $(2)$ by taking the difference of two integrals with numerator of the form $\ln(x^2-t^2+1)$, but they are different enough that I cannot see the connections.
I've been happy with my ability to do these exercises so far, so I'd really just like a constructive hint to get the ball rolling on this one.
Many thanks in advance.
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 $$\int_{-\infty}^{\infty} \frac{x^{2}-t^{2}}{\sinh x + \sinh t} \, \mathrm dx = \frac{2t^{3} - \pi^{2}t}{3\cosh t} - \frac{2t^{3}}{\cosh t} = -\frac{t \left(\pi^{2}+ 4 t^{2} \right)}{3\cosh t}.$$