What is $\int_{0}^{\pi}\frac{x}{x^2+\ln^2(2\sin x)}\:\mathrm{d}x$?

Question

Could you prove that: \begin{align} \displaystyle 2\left(\int_{0}^{\pi}\frac{x}{x^2+\ln^2(2\sin x)}\:\mathrm{d}x\right)^{7} + & 53 \left(\int_{0}^{\pi}\frac{x}{x^2+\ln^2(2\sin x)}\:\mathrm{d}x\right)^{5} \\ + \left(\int_{0}^{\pi}\frac{x}{x^2+\ln^2(2\sin x)}\:\mathrm{d}x\right)^{4} + & \quad \left(\int_{0}^{\pi}\frac{x}{x^2+\ln^2(2\sin x)}\:\mathrm{d}x\right)^{3} \\ + & 19\int_{0}^{\pi} \frac{x}{x^2+\ln^2(2\sin x)}\:\mathrm{d}x = 2014 \end{align} Just for the fun of it.

Observe that $2\times53\times1\times1\times19=2014.$

EDIT: A different proof of $$ \int_{0}^{\pi} \frac{x}{x^2+\ln^2(2\sin x)}\:\mathrm{d}x = 2 $$ may be found here. Thanks.

Answer 1

Here is a proof, using complex analysis, that the integral is equal to $2$. Put $$ f(z) = {1\over\log{(i(1-e^{i2z}))}}, $$ the logarithm being the principal branch. As can be deduced from the comments, the original integral is equal to $$ \int_0^\pi \operatorname{Re}{f(x)}\,dx. $$ Next consider, for $R>\epsilon > 0$, the following contour:

It is straightforward to check that for each fixed $\epsilon$ and $R$, the function $f$ is analytic on an open set containing this contour (just consider where $i(1-e^{i2z})\leq0$; this can only happen when $\operatorname{Im}{z}<0$ or when $\operatorname{Im}{z} = 0$ and $\operatorname{Re}{z}$ is an integer multiple of $\pi$). It then follows from Cauchy's theorem that $f$ integrates to zero over it. First let's see that the integrals over the quarter-circular portions of the contour vanish in the limit $\epsilon \to 0$. I'll look at the quarter circle near $\pi$ (near the bottom right corner of the contour) but the one near zero is similar, if not easier. Writing $i(1-e^{i2z}) = i(1-e^{i2(z-\pi)})$, it is clear that $i(1-e^{i2z}) = O(z-\pi)$ as $z\to\pi$. It follows from this that $f(z) = O(1/\log{|z-\pi|})$ as $z\to0$, and therefore that the integral of $f$ over the bottom right quarter circle is $O(\epsilon/\log{\epsilon})$ as $\epsilon \to0$, hence it vanishes in the limit as claimed.

Thus for fixed $R$, we can let $\epsilon \to 0$ to see that $f$ integrates to zero over the rectangle with corners $0,\pi, \pi +iR,$ and $iR$. Now the vertical sides of this rectangle give the contribution \begin{align*} -\int_0^R f(iy)\,idy + \int_0^R f(iy+\pi)\,idy = 0, \end{align*} since $f(iy) = f(iy+\pi)$. It follows at once that for each $R$, we can set the contribution from the horizontal sides equal to zero, giving \begin{align*} \int_0^{\pi} f(x)\,dx - \int_0^\pi f(x+iR)\,dx=0, \qquad R>0. \tag{1} \end{align*} (Note that the above equation implies that the integral in $x$ of $f(x+iR)$ over the interval $[0,\pi]$ is constant as a function of $R$; another way to evaluate the integral is to prove this directly, which I'll add below in a moment.) Now $f(x+iR) \to 1/\log{i} = 2/\pi$ as $R\to\infty$, uniformly for $x\in[0,\pi]$. Thus \begin{align*} \lim_{R\to\infty} \int_0^\pi f(x+iR)\,dx = \pi\cdot {2\over \pi} = 2, \end{align*} and it follows from $(1)$ that the original integral is equal to $2$ as well.

(By the way, the above idea is basically a replication of the technique I used here and here, and which I originally got from Ahlfors' book on complex analysis.)

Answer 2

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\int_{0}^{\pi}{x \over x^{2} + \ln^{2}\pars{2\sin\pars{x}}}\,\dd x = 2: \ {\large ?}}$

\begin{align}&\color{#c00000}{% \int_{0}^{\pi}{x \over x^{2} + \ln^{2}\pars{2\sin\pars{x}}}\,\dd x} =-\,\Im\int_{0}^{\pi}{\dd x \over x\ic + \ln\pars{2\sin\pars{x}}} \\[3mm]&=-\,\Im \int_{\verts{z}\ =\ 1\atop{\vphantom{\Huge A}0\ <\ {\rm Arg}\pars{z}\ <\ \pi}} {\dd z/\pars{\ic z}\over \ln\pars{z} + \ln\pars{2\bracks{z^{2} - 1}/\bracks{2\ic z}}} \\[3mm]&=\Re \int_{\verts{z}\ =\ 1\atop{\vphantom{\Huge A}0\ <\ {\rm Arg}\pars{z}\ <\ \pi}} {1 \over \ln\pars{\bracks{1 - z^{2}}\ic}}\,{\dd z \over z} \end{align}

We close the contour with the 'line segment' $\ds{\braces{\pars{x,0}\ \mid\ x \in \pars{-1,1}}}$ with 'indented points' at $\ds{z = -1,0,1}$. I t turns out that the whole contribution to the integral arises from the 'indented point' at $\ds{z = 0}$: See my detailed calculation in one of my previous answers which is very close to the present one. Then,

\begin{align}&\color{#66f}{\large% \int_{0}^{\pi}{x \over x^{2} + \ln^{2}\pars{2\sin\pars{x}}}\,\dd x} =\left. -\lim_{\epsilon\ \to\ 0^{+}}\Re\int_{\pi}^{0} {1 \over \ln\pars{\bracks{1 - z^{2}}\ic}}\,{\dd z \over z} \right\vert_{\,z\ \equiv\ \epsilon\expo{\ic\theta}} \\[3mm]&=-\,\Re\int_{\pi}^{0}{1 \over \ln\pars{\ic}}\,\ic\,\dd\theta =-\,\Re\int_{\pi}^{0}{1 \over \pi\,\ic/2}\,\ic\,\dd\theta =-\,{2 \over \pi}\int_{\pi}^{0}\dd\theta =\color{#66f}{\Large 2} \end{align}