Show $\int_{0}^{\infty}\frac{x}{x^{2}+1}\log\left(\left|\frac{x^r+1}{x^r-1}\right|\right)dx = \frac{\pi^2}{4r}$

absolute valueintegrationlogarithms

After playing around with a few values of $r$, I have the following conjecture:
$$\int_{0}^{\infty}\frac{x}{x^{2}+1}\log\left(\left|\frac{x^r+1}{x^r-1}\right|\right)dx = \frac{\pi^2}{4r}$$ for $r\gt 0$. My question is whether or not this is true and how to show as such.

I was inspired to ask this question after reading Cody's about showing $$\displaystyle \int_{0}^{\infty}\frac{x}{x^{2}+a^{2}}\log\left(\left|\frac{x+1}{x-1}\right|\right)dx=\pi\tan^{-1}(1/a)$$
One can see that the $r=1$ case of mine and the $a=1$ case of Cody's coincide.

That integral with $r=1$ lends itself to contour integration in the upper half plane because $\log\left(\left|\frac{x+1}{x-1}\right|\right)$ is an odd function, making the integrand even. However, my integrand lacks that nice symmetry for arbitrary $r$. For odd integers $r$, we regain that symmetry but branch points in the upper half plane add complexity. I'm not sure whether the integral with irrational $r$ can be handled with complex techniques, so I feel unsure of how to proceed.

Best Answer

I'll offer another approach. You can easily show $f(x):=\tfrac{x}{x^2+1}\ln|\tfrac{x^r+1}{x^r-1}|$ satisfies $f(\tfrac{1}{x})=f(x)$ so $$\int_0^\infty f(x)dx=\int_0^1 (1+\tfrac{1}{x^2})f(x)dx=\int_0^1 \tfrac{1}{x}\ln\tfrac{1+x^r}{1-x^r}dx.$$With $y=x^r$ your integral becomes $\tfrac{1}{r}\int_0^1 \tfrac{1}{y}\ln\tfrac{1+y}{1-y}dy$, so the $r=1$ case you already know completes the proof.

Related Solutions

Compute $\int_0^\infty \frac{\log(2+x^2)}{4+x^2}\,\mathrm dx$

Using Feynman's trick works:

For $a\geq0$, let $$G(a)=\int_0^\infty \frac{\ln(2+ ax^2)}{4+x^2}\,\mathrm dx.$$

Then the Leibniz rule implies that for $a>0$, $$G'(a)=\int_0^\infty \frac{x^2}{\left(x^2+4\right) \left(a x^2+2\right)}\,\mathrm dx.$$ The last integrand is a rational function and thus has a closed form anti-derivative that can be computed algortihmically. Namely,

$$G'(a)=\left. \frac{\frac{\sqrt{2} \tan ^{-1}\left(\frac{\sqrt{a} x}{\sqrt{2}}\right)}{\sqrt{a}}+2 \tan ^{-1}\left(\frac{2}{x}\right)}{2-4 a}\right|_{x=0}^{x=\infty}=\dots=\frac{\pi}{2\sqrt2(1+\sqrt2\sqrt a)\sqrt a}.$$

We know that $G(a)=G(0)+\int_0^a G'(b)\,\mathrm db$. Indeed, $$\int_0^a G'(b)\,\mathrm db = \frac{\pi}{2\sqrt2} \int_0^a \frac1{(1+\sqrt{2}\sqrt b)\sqrt b}\,\mathrm db=\frac\pi2\ln(1+\sqrt2\sqrt a).$$

Also, $$G(0)=\ln(2)\int_0^\infty \frac{1}{4+x^2}\,\mathrm dx=\frac{\ln(2)\pi}4.$$

Hence,

$$\bbox[15px,border:1px groove navy]{ \int_0^\infty \frac{\ln(2+ x^2)}{4+x^2}\,\mathrm dx =G(1)=\frac{\pi}2\left(\frac{\ln(2)}2+\ln(1+\sqrt2)\right). }$$

Evaluate $\int_{0}^{\infty} \ln(1+\frac{2\cos x}{x^2} +\frac{1}{x^4}) \, dx$

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

f[x_] := x^2*Exp[-I*x] + 1; 1/(2 Pi*I)*NIntegrate[f'[z]/f[z], {z, -1, 1, 1 + I, -1 + I, -1}]

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.

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