Would you please help me solve Exercise 18 of Chapter V in B. A. Fuchs & B. V. Shabat (1964) Functions of a complex variable and some of their applications: Vol. 1, Oxford: Pergamon Press.
Show that Schwarz's integral formula for the strip $-\pi/2 < y < \pi/2$ is $$f(z) = \frac1{2\pi} \int_{-\infty}^\infty \frac{u_+(\xi) + u_-(\xi)}{\cosh(\xi - z)} d\xi - \frac{i \sinh z}{2\pi} \int_{-\infty}^\infty \frac{u_+(\xi) - u_-(\xi)}{\cosh(\xi - z) \cosh \xi} d\xi$$ where $u_+(\xi)$ and $u_-(\xi)$ are the values of the real parts of $f(z)$ on the lines $y = +\pi/2$ and $y = -\pi/2$, respectively.
In other words, $u_+(\xi) =$ Re$f(\xi + \pi i/2),$ and $u_-(\xi) =$ Re$f(\xi - \pi i/2)$ where $\xi$ is real.
One approach is to start with Schwarz's formula for the upper half-plane, $$f(z') = \frac1{\pi i} \int_{-\infty}^\infty \frac{u(\xi')}{\xi' - z'} d\xi', $$ where $u(\xi')$ is the value of the real part of $f(z')$ on the $x'$-axis, i.e., $u(\xi') =$ Re$f(\xi' + 0i)$ where $\xi'$ is real, and then use $z' = ie^z,$ which is the conformal map from the strip to the upper half-plane. That map sends the line $y = \pi/2$, i.e., the top boundary of the strip, from $y = \infty + \pi i/2$ to $y = -\infty + \pi i/2$ to the negative $x'$-axis in the half plane from $-\infty + 0i$ to $0 + 0i$, and $y = -\pi/2$ from $y = -\infty - \pi i/2$ to $y = \infty - \pi i/2$ to the positive $x'$-axis from $0 + 0i$ to $\infty + 0i$. As a result, we have to separate the integral into two parts, $$f(z') = \frac1{\pi i} \int_{-\infty}^0 \frac{u(\xi')}{\xi' - z'} d\xi' + \frac1{\pi i} \int_0^\infty \frac{u(\xi')}{\xi' - z'} d\xi',$$ before we substitute for $z'$ and $\xi'.$ After substitution, $u$ becomes $u_+$ in the first integral and $u_-$ in the second, and the limits of integration change to $-\infty$ and $\infty$ with proper attention, or course, to the direction of integration. In the first integral, $\xi' = ie^{\xi + i\pi/2} = -e^\xi;$ in the second, $\xi' = ie^{\xi - i\pi/2} = e^\xi.$
After substitution and rearrangement, I get something other than the requested formula: $$f(z) = \frac1{2\pi} \int_{-\infty}^\infty \frac{u_+(\xi) + u_-(\xi)}{\cosh(\xi - z)} d\xi + \frac i{2\pi} \int_{-\infty}^\infty \frac{u_+(\xi) - u_-(\xi)}{\cosh(\xi - z)} e^{\xi - z} d\xi.$$
A second approach is to derive Poisson's integral formula for the strip, and then show that that formula is the real part of Schwarz's integral formula for the strip. I have proofs of Poisson's integral formula for the disc and upper half-plane, but I do not know how to start the derivation for the strip.
Start with NOT $$ f(z^\prime) = \frac1{\pi i} \int_{-\infty}^\infty \frac{u(\xi^\prime)}{\xi^\prime - z^\prime} d\xi^\prime$$ BUT $$ f(z^\prime) = \frac1{\pi i} \int_{-\infty}^\infty u(\xi^\prime)\left(\frac{1}{\xi^\prime - z^\prime}-\frac{\xi^\prime}{{\xi^\prime}^2+1}\right) d\xi^\prime. $$ This is the formula for a function with $\operatorname{Im }f(i)=0.$
Of course we separate the integral into two parts, $$ f(z^\prime) = \frac1{\pi i} \int_{-\infty}^0 u(\xi^\prime)\frac{\xi^\prime z^\prime +1}{\xi^\prime - z^\prime}\frac{d\xi^\prime}{{\xi^\prime}^2+1} +\frac1{\pi i} \int_0^\infty u(\xi^\prime) \frac{\xi^\prime z^\prime +1}{\xi^\prime - z^\prime}\frac{d\xi^\prime}{{\xi^\prime}^2+1}. \tag{1} $$ After substitution, the first integral will be \begin{align} &\frac1{\pi i} \int_\infty^{-\infty} u_+(\xi)\frac{-ie^\xi e^z +1}{-e^\xi -i e^z}\frac{-e^\xi}{e^{2\xi}+1} d\xi\\ &=\frac1{\pi i} \int_{-\infty}^{\infty} u_+(\xi)\frac{(ie^\xi e^z -1)(e^\xi -i e^z)}{(e^\xi +i e^z)(e^\xi -i e^z)}\frac{1}{e^{\xi}+e^{-\xi }} d\xi\\ &=\frac1{\pi i} \int_{-\infty}^{\infty} u_+(\xi)\frac{i(e^\xi +e^{-\xi}) +(e^z -e^{-z})}{e^{\xi-z} + e^{z-\xi }}\frac{1}{e^{\xi}+e^{-\xi }} d\xi\\ &=\frac1{2\pi} \int_{-\infty}^{\infty} u_+(\xi)\frac{\cosh \xi -i\sinh z}{\cosh(\xi-z)}\frac{1}{\cosh\xi} d\xi, \end{align} since $\xi ^\prime=ie^{\xi +i\pi/2}=-e^\xi $ and $z^\prime=ie^z.$
Similarly the second integral in $(1)$ will be $$ \frac1{2\pi} \int_{-\infty}^{\infty} u_-(\xi)\frac{\cosh \xi +i\sinh z}{\cosh(\xi-z)}\frac{1}{\cosh\xi} d\xi. $$
EDIT.
Schwarz's formula for the unit disc is the following: $$ F(z)=\frac{1}{2\pi i}\int_{|\xi|=1}\frac{\xi +z}{\xi -z}U(\xi )\frac{d\xi }{\xi }+i\operatorname{Im }F(0). $$ If $\operatorname{Im }F(0)=0$, of course it will be $$ F(z)=\frac{1}{2\pi i}\int_{|\xi|=1}\frac{\xi +z}{\xi -z}U(\xi )\frac{d\xi }{\xi }. $$ We start with this expression. Using substitution $$ z=\varphi (z^\prime)=\frac{z^\prime -i}{z^\prime +i},$$ we have \begin{align} \frac{\xi +z}{\xi -z}&=\frac{\varphi (\xi ^\prime)+\varphi (z^\prime) }{\varphi (\xi ^\prime)-\varphi (z^\prime) }=\frac{\xi ^\prime z^\prime +1}{i(\xi ^\prime-z^\prime)},\\ \frac{d\xi }{i\xi }&=\frac{2}{{\xi ^\prime}^2+1}d\xi ^\prime, \\ F(\varphi (z^\prime))&=\frac{1}{2\pi}\int_{-\infty}^\infty \frac{\xi ^\prime z^\prime +1}{i(\xi ^\prime-z^\prime)}U(\varphi (\xi^\prime))\frac{2}{{\xi ^\prime}^2+1}d\xi\\ &=\frac1{\pi i} \int_{-\infty}^\infty U(\varphi (\xi'))\left(\frac{1}{\xi^\prime - z^\prime}-\frac{\xi^\prime}{{\xi^\prime}^2+1}\right) d\xi^\prime. \end{align} Writing $F(\varphi (z^\prime))=f(z^\prime),$ $ U(\varphi (\xi ^\prime))=u(\xi ^\prime)$, We get $$f(z^\prime)=\frac1{\pi i} \int_{-\infty}^\infty u(\xi^\prime)\left(\frac{1}{\xi^\prime - z^\prime}-\frac{\xi^\prime}{{\xi^\prime}^2+1}\right) d\xi^\prime ,$$ which is the formula we have used above.