Numerical evidence suggests the following
$$I=\int_{0}^{\infty} \ln\left(1+\frac{2\cos x}{x^2} +\frac{1}{x^4}\right) \, dx\stackrel{?}{=}4\pi W\left(\frac{1}{2}\right)=4.42\cdots$$
where $W(x)$ is the Lambert $W$ (product log) function.
Here’s the context for the problem:
I was playing around with other $\ln$ integrals, and evaluated the following:
$$\int_{0}^{\infty} \ln \left(1+2x^{-2} \cos(\phi)+x^{-4}\right)\,dx=2\pi\cos \left(\frac{\phi}{2}\right)$$ There are several questions on MSE on similar integrals to the above, such as this one.
The above integral can be evaluated in several ways, but I just chose to introduce a $\ln(z)$ integral representation. I was curious to know what the integral would instead be if I replaced $\cos \phi$ with $\cos x$. $I$ converges because it is bounded above by the integral $\int_{0}^{\infty} \ln (1+2x^{-2}+x^{-4})\,dx=2\pi$.
Here are my attempts:
Although I’m much more comfortable with real analysis, I currently have no method that I can apply real analytic techniques or representations on $I$. I tried instead contour integration:
Firstly notice that $$I \stackrel{\text{IBP}}{=}2\int_{0}^{\infty} \frac{2+2x^2 \cos x+x^3 \sin x}{1+x^4 +2x^2 \cos x} \, dx$$
I then converted these trig functions into exponential forms and factorised the denominator to determine where the poles are- they are the four solutions to $x^2+e^{i x} = 0$ and $x^2+e^{-i x}=0$ which are
$$x=\pm 2i W\left(\pm \frac{1}{2}\right)$$
I then took a semi-circular arc in the upper half plane to be my contour as the integrand is even, and computed the following residues:
$$\text{Res} \left[f(z),\,2i W\left(\frac{1}{2}\right)\right] =-2i W\left(\frac{1}{2}\right)$$
$$\text{Res} \left[f(z),\,-2i W\left(-\frac{1}{2}\right)\right] =2i W\left(-\frac{1}{2}\right)$$
where $f(x)$ is the integrand function above.
So we have then that
$$\oint_C f(z) \, dz=2I + \int_{\Gamma} f(z) \, dz = -4\pi \left(W\left(-\frac{1}{2}\right) -W\left(\frac{1}{2}\right)\right)$$
where $\int_{\Gamma}$ traces the semi-circular arc path. Here is where I’m stuck however because I’m not sure how to evaluate this arc integral.
Best Answer
For $a>0$, let $F(a) = \int_{-\infty}^\infty \log(1+\frac{e^{aix}}{x^2})\, dx$, your $I$ equals $\Re F(1)$. By parts gives $$F(a) = \int_{-\infty}^\infty \frac{2-i a x}{1+x^2 e^{-i a x}} \,dx$$ When $a=1$, I claim $f_a(z) = 1+z^2 e^{-a iz}$ has a unique root $z=2i W(1/2)$ on the upper half plane $\mathcal{H}$, then residue theorem (one also shows integral on big semicircle tends to $0$, this is not difficult) yields your claim.
For the claim: let $z_0$ be a root of $f_a$ in $\mathcal{H}$, then $|z_0|^2 = e^{-a \Im(z_0)} < 1$. Therefore all roots of $f_a$ in $\mathcal{H}$ lie in semicircle of radius $1$.
The number of roots can be found by evaluating (numerically) the contour integral (always an integer): $\frac{1}{2\pi i}\int_\gamma f_a'(z)/f_a(z) dz$. The Mathematica command
evaluates above integral over rectangle with indicated vertices, it outputs $1$, so there is only one root of $f_1$ in $\mathcal{H}$. One checks $z=2iW(1/2)$ is a root, so the only root.
For general $a>0$, $f_a$ might have many roots in $\mathcal{H}$, $F(a)$ is a finite sum of Lambert W over different branches.