# Prove that $\int_0^\infty \frac{1+2\cos x+x\sin x}{1+2x\sin x +x^2}dx=\frac{\pi}{1+\Omega}$ where $\Omega e^\Omega=1$

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 integrand function is even, hence the result follows from the residue theorem - have a look at the techniques used here: math.stackexchange.com/questions/45745/… Commented Jun 21, 2017 at 20:02
• sigh... Unfortunately, I have not yet learned the residue theorem. I suppose I should take some time to acquaint myself with it... Commented Jun 21, 2017 at 20:03
• Unfortunately I am not aware of more elementary methods for proving such identity. Are you confident with the Lagrange inversion formula (sometimes known as Lagrange-Buhrmann inversion formula)? Commented Jun 21, 2017 at 20:05
• Yes. You think that I can use that to solve this problem? Commented Jun 21, 2017 at 20:07
• It is not straightforward but I think you can use Lagrange's inversion formula together with the Laplace transform to prove such identity. Something similar is done here: math.stackexchange.com/a/2113495/44121 Commented Jun 21, 2017 at 20:11

A quite simpler solution.

Firstly notice that $$\frac{1+2\cos x+x\sin x}{1+2x\sin x+x^2}=\Re\left[\frac{1+2e^{ix}}{1-ixe^{ix}}\right]$$ Set $$\displaystyle f(z)=\dfrac{1+2e^z}{1-ze^z}$$ and integrate along the righter large semicircle, i.e. the imaginary axis and the arc with radius $$R\to\infty$$ and $$\theta\in(-\pi/2,\pi/2)$$.

The former integral reduces to the desired one. The imaginary part of $$f(ix)$$ is odd and vanishes as we are integrating on a even region. $$\int_lf(z)dz=\lim_{R\to\infty}\int_{-R}^R f(ix)i~dx=i\int_{-\infty}^\infty\frac{1+2\cos x+x\sin x}{1+2x\sin x+x^2}dx$$

Meanwhile, the latter is simple $$\int_{C}f(z)dz=\lim_{|z|\to\infty}\int_C\dfrac{1+2e^z}{1-ze^z}dz=-2\lim_{|z|\to\infty}\int_C\frac{dz}z=-2\int_{\pi/2}^{-\pi/2}id\theta=2\pi i$$ The sum of the two integrals equals to $$-2\pi i$$ times the sum of residues in the contour since the contour is clockwise. The poles of the integrand are the solutions to $$ze^z=1$$, namely $$W_n(1)~~ n\in \mathbb Z$$ where $$n$$ denotes the different branches.

It is well known that when $$x>0$$, among $$W_n(x)$$ s there is only one with positive real part, which is precisely $$W_0$$ and is real. One can verify the statement by just referring to the graph of the function, or after some rigorous yet elementary calculations.

After all, the only pole locates at $$z=\Omega$$ with order $$1$$, and the residue is easy $$\text{Res}[f(z),z=\Omega]=\frac{1+2e^z}{(1-ze^z)'}\Big|_{z=\Omega}=-\frac{1+2e^\Omega}{(1+\Omega)e^\Omega}=-1-\frac1{1+\Omega}$$ Apply the residue theorem $$i\int_{-\infty}^\infty\frac{1+2\cos x+x\sin x}{1+2x\sin x+x^2}dx+2\pi i=-2\pi i\left(-1-\frac1{1+\Omega}\right)$$ Finally, note that the integrand is even and the result follows

$$\int_0^\infty\frac{1+2\cos x+x\sin x}{1+2x\sin x+x^2}dx=\frac{\pi}{1+\Omega}$$

where $$\Omega$$ is the unique real solution to $$xe^x=1$$.

• This $\Omega$ constant is very similar to Dottie number. I wonder whether they can be expressed via each other. Commented May 21, 2023 at 6:55

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 $$\left(\int_{-R}^R+\int_{\gamma_R}\right)f(z)=2\pi i\sum_{f(z^*)=0\\\Im z^*>0\\|z^*| where $$\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)|. 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 $$2I+\lim_{R\to+\infty}\int_0^\pi f(Re^{i\theta})iRe^{i\theta}\,d\theta=\pi+\frac\pi{1+W_0(1)}-\pi\lim_{R\to+\infty}\sum_{0<|k|<\max K\\|W_K(1)| where the divergent limits are interpreted as $$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}$$ so \begin{align}\lim_{R\to+\infty}\int_0^\pi f(Re^{i\theta})iRe^{i\theta}\,d\theta&=\lim_{R\to+\infty}\int_0^\pi\frac{Re^{i\theta}(i-Re^{i\theta}/2)e^{-iRe^{i\theta}}}{iRe^{i\theta}e^{-iRe^{i\theta}}+R^2e^{2i\theta}}\,d\theta\\&=\lim_{R\to+\infty}\int_0^\pi\frac{i-Re^{i\theta}/2}{i+Re^{iRe^{i\theta}}e^{i\theta}}\,d\theta\\&\stackrel{(*)}=\lim_{R\to+\infty}\int_{-\pi}^\pi\frac{i-Re^{i\theta}/2}{i+Re^{iRe^{i\theta}}e^{i\theta}}\,d\theta\\&=\lim_{R\to+\infty}\oint_{|z|=R}\frac{i-z/2}{i+ze^{iz}}\,\frac{dz}{iz}\\&=\lim_{R\to+\infty}\frac12\oint_{|z|=R}\frac{2i-z}{z(ize^{iz}-1)}\,dz\\&=\small\pi i\left(-2i+i\left(1+\frac1{1+W_0(1)}\right)+\lim_{R\to+\infty}\sum_{0<|k|<\max K\\|W_K(1)| where $$(*)$$ follows since \begin{align}\lim_{R\to+\infty}\int_{-\pi}^0\frac{i-Re^{i\theta}/2}{i+Re^{iRe^{i\theta}}e^{i\theta}}\,d\theta&=\lim_{R\to+\infty}\int_0^\pi\frac{ie^{i\theta}-R/2}{ie^{i\theta}+Re^{iRe^{-i\theta}}}\,d\theta\to0\end{align} as the denominator exponentially increases on the order of $$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.