Show that $\int_0^\pi \frac{x\cos x \sinh x}{\sin^2x+\sinh^2x}dx=\frac{\pi}{2}\ln\left(\frac{e^\pi+1}{e^\pi-1}\right)$

Question

I am trying to show that $$I=\int_0^\pi \frac{x\cos x \sinh x}{\sin^2x+\sinh^2x}dx=\frac{\pi}{2}\ln\left(\frac{e^\pi+1}{e^\pi-1}\right)$$ I know there is an antiderivative in terms of dilogarithms and logarithms, but this nice closed form makes me think there is a clever way to get this result for these specific bounds of integration. So please avoid posting proofs with the antiderivative.

What I managed to do for now is to notice that $$I=-\Im\int_0^\pi\frac{x}{\sin(x(1+i))}dx$$ which you get by simple computation. From here how to proceed? I'm not sure if the substitution $t=x(1+i)$ is allowed here, since I think this would become a problem of contour integration, which I usually avoid.

Here is my question:

Is there a way to obtain this result without using the antiderivative and contour integration?

Answer 1

Note the KEY trick: $$\boxed{\sin^2x+\sinh^2x=\cosh^2x-\cos^2x}\tag{0}$$

Hence, we can use $\color{red}{\text{partial fraction}}$ to split the integral:

$$\int_0^\pi \frac{x\cos x \sinh x}{\sin^2x+\sinh^2x}dx=\frac12\int_0^\pi \frac{x \sinh x}{\cosh x-\cos x}dx-\frac12\int_0^\pi \frac{x \sinh x}{\cosh x+\cos x}dx$$

From the series representation $$ \frac1{1-e^{-t+ix}}=\sum_{k=0}^\infty e^{-kt}e^{ikx},~~~~\frac1{1-e^{-t-ix}}=\sum_{k=0}^\infty e^{-kt}e^{-ikx} $$ Add them to get the real part: $$ \boxed{\frac{\sinh t}{\cosh t-\cos x}=1+2\sum_{k=1}^\infty e^{-kt}\cos(kx)}\tag{1} $$

From the series representation $$ \frac1{1+e^{-t+ix}}=\sum_{k=0}^\infty (-1)^ke^{-kt}e^{ikx},~~~~\frac1{1+e^{-t-ix}}=\sum_{k=0}^\infty (-1)^ke^{-kt}e^{-ikx} $$ Add them to get the real part: $$ \boxed{\frac{\sinh t}{\cosh t+\cos x}=1+2\sum_{k=1}^\infty(-1)^k e^{-kt}\cos(kx)}\tag{2} $$ The rest are trivial computations,

$$\begin{align}\int_0^\pi \frac{x \sinh x}{\cosh x-\cos x}dx&=\int_0^\pi xdx+2\sum_{k=1}^\infty\int_0^\pi xe^{-kx}\cos(kx)dx\\ \\ &=\frac{\pi^2}2-\pi\sum_{k=1}^\infty\frac{(-e^{-\pi})^k}{k}\\ \\ &=\frac{\pi^2}2+\pi\ln(1+e^{-\pi})\end{align}$$

Similarly

$$\begin{align}\int_0^\pi \frac{x \sinh x}{\cosh x+\cos x}dx&=\int_0^\pi xdx+2\sum_{k=1}^\infty(-1)^k\int_0^\pi xe^{-kx}\cos(kx)dx\\ \\ &=\frac{\pi^2}2-\pi\sum_{k=1}^\infty\frac{(e^{-\pi})^k}{k}\\ \\ &=\frac{\pi^2}2+\pi\ln(1-e^{-\pi})\end{align}$$

Finally,

$$\boxed{\int_0^\pi \frac{x\cos x \sinh x}{\sin^2x+\sinh^2x}dx=\frac\pi2\ln\left( \frac{1+e^{-\pi}}{1-e^{-\pi}}\right)=\frac\pi2\ln\left( \coth\left(\frac\pi2\right)\right)~}$$

Answer 2

PREP $$\frac{1}{\sin(ax)}=\frac{2ie^{-aix}}{1-e^{-2aix}}=2ie^{-aix}\sum_{n=0}^\infty e^{-2anix}=2i\sum_{n=0}^\infty e^{(2n+1)aix}$$ $$\int_0^\pi xe^{mx}dx=\frac{\pi e^{\pi m}}{m}-\frac{e^{\pi m}}{m^2}+\frac{1}{m^2}$$

$$\sum_{n=0}^\infty\frac{x^{2n+1}}{2n+1}=\operatorname{arctanh} x, \space |a|^2<1$$ $$\sum_{n=0}^\infty\frac{1}{(2n+1)^2}=\frac{\pi^2} 8$$ Let $a=1+i$

$$I=\int_0^\pi\frac{x\cos(x)\sinh(x)}{\sin^2(x)+\sinh^2(x)}dx=-\Im \int_0^\pi\frac{x}{\sin(ax)}dx$$ $$I=-\Im \int_0^\pi\frac{x}{\sin(ax)}dx=-\Im \int_0^\pi 2ix\sum_{n=0}^\infty e^{(2n+1)aix}dx$$

$$=-\Im \sum_{n=0}^\infty 2i\int_0^\pi xe^{(2n+1)aix}dx=-\Im \sum_{n=0}^\infty 2i\left(\frac{\pi e^{(2n+1)ai\pi }}{(2n+1)ai}+\frac{e^{(2n+1)ai\pi }}{(2n+1)^2a^2}-\frac{1}{(2n+1)^2a^2}\right)$$

Let $b=e^{ai\pi}=e^{i(1+i)\pi}=-e^{-\pi}$. Note that $\frac{2i}{a^2}=1$

$$=-\Im \sum_{n=0}^\infty 2i\left(\frac{\pi b^{(2n+1) }}{(2n+1)ai}+\frac{b^{(2n+1) }}{(2n+1)^2a^2}-\frac{1}{(2n+1)^2a^2}\right)$$

$$=-\Im\left(-\frac{\pi^2}{8}+\frac{2\pi}{1+i}\operatorname{arctanh}(-e^{-\pi})+\sum_{n=0}^\infty\frac{b^{2n+1}}{(2n+1)^2}\right)$$

The remaining sum is real, so we can discard it $$I=\pi\operatorname{arctanh}(e^{-\pi})$$

Answer 3

(Not an answer to your question since this involves contour integration near the beginning. Just doing this for fun since you've already accepted an answer.)

Starting from where you left off, we let $x + ix \mapsto x$ to get

$$I := -\Im \int_0^\pi x \csc(x+ix)dx = -\Im \frac{1}{(1+i)^2}\int_0^{\pi + i\pi}x\csc{x} dx = \frac{1}{2}\Re\int_0^{\pi+i\pi}x\csc{x}dx.$$

Let $f(z) = z\csc{z}$. Centering at $z=0$, we can expand this as the series $\displaystyle 1 + \frac{z^2}{6} + \frac{7z^4}{360} + O\left(z^6\right).$ This converges as $z \to 0$, making $z=0$ a removable singularity.

With that in mind, we construct a contour shaped as a right triangle such that its vertices are on the points $(0,0), (0,\pi),$ and $(\pi, \pi)$. We also traverse around it clockwise. We use Cauchy's Residue Theorem to get

$$ \begin{align} 0 &= \int_{\pi+i\pi}^0 f(z)dz + \int_0^{i\pi}f(z)dz + \int_{i\pi}^{\pi+i\pi}f(z)dz \\ \int_{0}^{\pi+i\pi} f(z)dz &= \int_0^{i\pi}f(z)dz + \int_{i\pi}^{\pi+i\pi}f(z)dz. \end{align} $$

Since the contour integral over the hypotenuse is what we need to retrieve $I$, we apply $\frac{1}{2}\Re$ on both sides. Then we have this long calculation:

$$ \begin{align} I &= \frac{1}{2}\Re\int_0^{i\pi}f(z)dz + \frac{1}{2}\Re\int_{i\pi}^{\pi+i\pi}f(z)dz \\ \\ &= \frac{1}{2}\Re\int_0^\pi f(iy)d(iy) + \frac{1}{2}\Re\int_0^\pi f(x+i\pi)d(x+i\pi)\\ \\ &= \frac{1}{2}\Re i\int_0^\pi iy\csc(iy)dy + \Re\int_0^\pi \frac{x+i\pi}{2\sin(x+i\pi)}dx \\ \\ &= \frac{1}{2}\Re i \int_0^\pi y\operatorname{csch}ydy + \int_{0}^{\pi}\left(\frac{\cosh\left(\pi\right)x\sin x}{\cosh2\pi-\cos2x}+\frac{\pi\sinh\left(\pi\right)\cos x}{\cosh2\pi-\cos2x}\right)dx \\ \\ &= \frac{1}{2}\cdot 0+\cosh\left(\pi\right)\int_{0}^{\pi}\frac{\left(\pi-x\right)\sin\left(\pi-x\right)}{\cosh2\pi-\cos\left(2\left(\pi-x\right)\right)}dx\\ &~~~~~~~~~~~~~~+\pi\sinh\pi\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{\cos\left(x+\frac{\pi}{2}\right)}{\cosh2\pi-\cos\left(2\left(x+\frac{\pi}{2}\right)\right)}dx \\ \\ &= \pi\cosh\pi\int_{0}^{\pi}\frac{\sin x}{\cosh2\pi-\cos2x}dx-\cosh\pi\int_{0}^{\pi}\frac{x\sin x}{\cosh2\pi-\cos2x}dx + \pi \sinh{\pi} \cdot 0\\ \\ &= \frac{\pi\cosh\pi}{2}\int_{0}^{\pi}\frac{\sin x}{\cosh2\pi-\cos2x}dx \\ \\ &= -\frac{\pi\cosh\pi}{2}\int_{0}^{\pi}\frac{\sin x}{2\cos^{2}x-\cosh2\pi-1}dx \\ \\ &= -\frac{\pi\cosh\pi}{2}\int_{-1}^{1}\frac{1}{2x^{2}-\cosh2\pi-1}dx \\ \\ &= \frac{\pi\cosh\pi}{2+2\cosh2\pi}\int_{-1}^{1}\frac{1}{1-\frac{2x^{2}}{1+\cosh2\pi}}dx \\ \\ &= \frac{\pi\cosh^{2}\pi}{2+2\cosh2\pi}\int_{-\operatorname{sech}\pi}^{\operatorname{sech}\pi}\frac{1}{1-x^{2}}dx \\ \\ &= \frac{\pi\cosh^{2}\pi}{2+2\cosh2\pi}\left(\tanh^{-1}\left(\operatorname{sech}\pi\right)-\tanh^{-1}\left(-\operatorname{sech}\pi\right)\right) \\ \\ &= \frac{\pi}{2}\ln\left(\frac{e^{\pi}+1}{e^{\pi}-1}\right) \end{align} $$