Whilst reading this Math SE post, I saw that the OP mentioned the integral $$\int_0^\infty \frac{1+2\cos x+x\sin x}{1+2x\sin x +x^2}dx=\frac{\pi}{1+\Omega}$$ where $\Omega$ is the unique solution to the equation $$xe^x=1$$ However, the question was about how to approximate the integral numerically. This is not a duplicate question; I would like to know how to exactly prove the above equality without approximation. Does anybody know how to do this?
Thanks!
The integral identity is true.
We will evaluate the integral $$I=\int_0^\infty\frac{1+2\cos x+x\sin x}{1+2x\sin x+x^2}\,dx$$ by taking a semicircular contour on the upper-half plane. As the integrand is even, it follows from the residue theorem that $$\left(\int_{-R}^R+\int_{\gamma_R}\right)f(z)=2\pi i\sum_{f(z^*)=0\\\Im z^*>0\\|z^*|<R}\operatorname{Res}(f(z),z^*)$$ where $\gamma_R$ is the anti-clockwise circular path from $\theta=0$ to $\pi$, and $$f(z)=\frac{1+2\cos z+z\sin z}{1+2z\sin z+z^2}=\frac{1+2\cos z+z\sin z}{(iz+e^{iz})(iz-e^{-iz})}.$$ The only values of $z^*$ meeting the three criteria under the summation are $iW_0(1)$, $-iW_k(1)$ and $-iW_{-k}(1)$ for all integers $k>0$ such that $|W_k(1)|<R$. It is useful to note that $W_k$ and $W_{-k}$ are complex conjugates.
To evaluate these residues, we take the limit of $(z-z^*)f(z)$ as $z\to z^*$, writing cosine and sine as complex exponentials, then making extensive use of the identity $e^{W_k(1)}=1/W_k(1)$. This yields \begin{align}\operatorname{Res}(f(z),iW_0(1))&=\frac1{2i}\left(1+\frac1{1+W_0(1)}\right)\\\operatorname{Res}(f(z),-iW_k(1))&=-\frac1{2i}\left(1+\frac1{1+W_k(1)}\right),\quad\forall|k|>0.\end{align} Therefore, the result from the residue theorem reads $$2I+\lim_{R\to+\infty}\int_0^\pi f(Re^{i\theta})iRe^{i\theta}\,d\theta=\pi+\frac\pi{1+W_0(1)}-\pi\lim_{R\to+\infty}\sum_{0<|k|<\max K\\|W_K(1)|<R}\left(1+\frac1{1+W_k(1)}\right),$$ where the divergent limits are interpreted as $R$ increasing at the same rate on both sides. Now we focus on the arc contribution \begin{align}f(Re^{i\theta})&=\frac{1+e^{iRe^{i\theta}}+e^{-iRe^{i\theta}}+Re^{i\theta}\frac{e^{iRe^{i\theta}}-e^{-iRe^{i\theta}}}{2i}}{1-iRe^{i\theta}(e^{iRe^{i\theta}}-e^{-iRe^{i\theta}})+R^2e^{2i\theta}}.\end{align} Since $e^{iRe^{i\theta}}\to0$, the only term that survives on the numerator is $(1-Re^{i\theta}/(2i))e^{-iRe^{i\theta}}$ as it dominates all the other terms as $R\to+\infty$. Likewise, the denominator only remains as $iRe^{i\theta}e^{-iRe^{i\theta}}+R^2e^{2i\theta}$ so \begin{align}\lim_{R\to+\infty}\int_0^\pi f(Re^{i\theta})iRe^{i\theta}\,d\theta&=\lim_{R\to+\infty}\int_0^\pi\frac{Re^{i\theta}(i-Re^{i\theta}/2)e^{-iRe^{i\theta}}}{iRe^{i\theta}e^{-iRe^{i\theta}}+R^2e^{2i\theta}}\,d\theta\\&=\lim_{R\to+\infty}\int_0^\pi\frac{i-Re^{i\theta}/2}{i+Re^{iRe^{i\theta}}e^{i\theta}}\,d\theta\\&\stackrel{(*)}=\lim_{R\to+\infty}\int_{-\pi}^\pi\frac{i-Re^{i\theta}/2}{i+Re^{iRe^{i\theta}}e^{i\theta}}\,d\theta\\&=\lim_{R\to+\infty}\oint_{|z|=R}\frac{i-z/2}{i+ze^{iz}}\,\frac{dz}{iz}\\&=\lim_{R\to+\infty}\frac12\oint_{|z|=R}\frac{2i-z}{z(ize^{iz}-1)}\,dz\\&=\small\pi i\left(-2i+i\left(1+\frac1{1+W_0(1)}\right)+\lim_{R\to+\infty}\sum_{0<|k|<\max K\\|W_K(1)|<R}i\left(1+\frac1{1+W_k(1)}\right)\right)\\&=\pi-\frac\pi{1+W_0(1)}-\pi\lim_{R\to+\infty}\sum_{0<|k|<\max K\\|W_K(1)|<R}\left(1+\frac1{1+W_k(1)}\right)\end{align} where $(*)$ follows since \begin{align}\lim_{R\to+\infty}\int_{-\pi}^0\frac{i-Re^{i\theta}/2}{i+Re^{iRe^{i\theta}}e^{i\theta}}\,d\theta&=\lim_{R\to+\infty}\int_0^\pi\frac{ie^{i\theta}-R/2}{ie^{i\theta}+Re^{iRe^{-i\theta}}}\,d\theta\to0\end{align} as the denominator exponentially increases on the order of $Re^R$. We finally obtain $$2I-\frac\pi{1+W_0(1)}=\frac\pi{1+W_0(1)}\implies I=\frac\pi{1+W_0(1)}$$ as desired.