The problem, as originally stated, was to prove that

$$\int_0^{\infty} dx \frac{x}{x^2+a^2}\log{\left(\frac{x+1}{x-1}\right)} = \pi \arctan{\frac{1}{a}}$$

This is a challenging integral with unexpected twists and turns in its evaluation. Ultimately, though, it all works out, albeit not the way the OP expected: there is a nonzero imaginary part, which is why he put the argument of the log in absolute values later on. On the other hand, I will work with the original integral, so my result will be complex; the real part will agree with the originally posted result.

We begin by integrating by parts, something we normally have no business doing, but the infinities cancel:

$$\int_0^{\infty} dx \frac{x}{x^2+a^2}\log{\left(\frac{x+1}{x-1}\right)} = \\ \lim_{N \rightarrow \infty}\left[\frac{1}{2} \log{(x^2+a^2)}\log{\left(\frac{x+1}{x-1}\right)} \right]_0^N + \frac{1}{2} \int_0^{\infty} dx \log{(x^2+a^2)}\left(\frac{1}{x-1}-\frac{1}{x+1} \right)$$

Now, the limit in the first term on the RHS of the above equation converges to $-i \pi \, \log{a}$. The second term, the integral, has a singularity about which we may deform the interval of integration; the contribution of this deformation is

$$\lim_{\epsilon \rightarrow 0} i \frac{1}{2} \epsilon \int_0^1 d\phi e^{i \phi} \frac{\log{(1+a^2)}}{\epsilon e^{i \phi}} = i \frac{\pi}{2} \log{(1+a^2)}$$

Thus the integral is now equal to

$$PV \int_0^{\infty} dx \frac{\log{(x^2+a^2)}}{x^2-1} + i \frac{\pi}{2} \log{\left ( 1+ \frac{1}{a^2} \right )}$$

To evaluate this, let $I(a)$ be the real part of the above expression (the integral), and differentiate $I(a)$ with respect to $a$ (which I assume is a valid step, I will not prove it here) and get

$$\frac{\partial I}{\partial a} = 2 a \, PV \int_0^{\infty} \frac{dx}{x^2+a^2} \frac{1}{x^2-1} = a PV \int_{-\infty}^{\infty} \frac{dx}{x^2+a^2} \frac{1}{x^2-1}$$

Note that this integral is equal to

$$a \oint_C \frac{dz}{z^2+a^2} \frac{1}{z^2-1}$$

where $C$ is a semicircular arc in the upper half plane, just above the real axis (so we can ignore the poles at $z= \pm 1$, as the integral is a Cauchy principal value). This integral is equal to $i 2 \pi$ times the sum of the residues at the poles inside $C$. The only pole inside $C$ is at $z=ia$, so the integral is

$$a PV \int_{-\infty}^{\infty} \frac{dx}{x^2+a^2} \frac{1}{x^2-1} = a \, i 2 \pi \frac{1}{i 2 a} \frac{1}{(-a^2-1)} = -\frac{\pi}{a^2+1}$$

Integrating with respect to $a$, we find that

$$I(a) = K – \pi \arctan{a}$$

where $K$ is a constant of integration. We determine $K$ as being equal to $I(0)$. So the problem now reduces to evaluating

$$I(0) =2 \int_0^{\infty} dx \frac{\log{x}}{x^2-1}$$

Note that the singularity at $x=1$ is removable in this integral and therefore we do not need to use a Cauchy principal value. We evaluate this integral by once again appealing to the residue theorem, but this time, we consider

$$\oint_{C’} dz \frac{\log^2{z}}{z^2-1}$$

where $C’$ is a keyhole contour with respect to the positive real axis. By integrating around this contour and noting that the integrand vanishes sufficiently fast as the radius of the circular section of $C’$ increases without bound, we get

$$\oint_{C’} dz \frac{\log^2{z}}{z^2-1} = -i 4 \pi \int_0^{\infty} dx \frac{\log{x}}{x^2-1} + 4 \pi^2 \int_0^{\infty} dx \frac{1}{x^2-1}$$

This is equal to, by the residue theorem, $i 2 \pi$ times the sum of the residues of the poles of the integrand of the complex integral within $C’$. As the only pole is at $z=-1$, we see that

$$\begin{align}\oint_{C’} dz \frac{\log^2{z}}{z^2-1} &= i 2 \pi \frac{\log^2{(-1)}}{2 (-1)} \\ &= i 2 \pi \frac{\pi^2}{2}\end{align}$$

Now, the real part of the integral above is split into a Cauchy principal value and a piece indented about the singularity at $x=1$. The Cauchy principal value is zero:

$$\begin{align}PV \int_0^{\infty} dx \frac{1}{x^2-1} &= \lim_{\epsilon \rightarrow 0} \left [\int_0^{1-\epsilon} dx \frac{1}{x^2-1} + \int_{1+\epsilon}^{\infty} dx \frac{1}{x^2-1}\right]\\ &= \lim_{\epsilon \rightarrow 0} \left [\int_0^{1-\epsilon} dx \frac{1}{x^2-1} + \int_0^{1/(1+\epsilon)} \left (-\frac{dx}{x^2} \right ) \frac{1}{(1/x^2)-1} \right ]\\ &= \lim_{\epsilon \rightarrow 0} \left [\int_0^{1-\epsilon} dx \frac{1}{x^2-1} – \int_0^{1-\epsilon} \frac{dx}{x^2-1} \right ] \\ &= 0\end{align}$$

The indent in the contour, however, produces a contribution; let $x=1+\epsilon e^{i \phi}$ and $\phi \in [\pi,0]$:

$$4 \pi^2 i \epsilon \int_{-\pi}^0 d\phi \frac{e^{i \phi}}{2 \epsilon e^{i \phi}} = i \frac{\pi}{2} 4 \pi^2$$

so that

$$-i 4 \pi \int_0^{\infty} dx \frac{\log{x}}{x^2-1} = i 2 \pi \frac{\pi^2}{2} – i \frac{\pi}{2} 4 \pi^2 = -i 2 \pi \frac{\pi^2}{2}$$

Therefore

$$K = 2 \int_0^{\infty} dx \frac{\log{x}}{x^2-1} = \frac{\pi^2}{2}$$

and

$$I(a) = \frac{\pi^2}{2} – \pi \arctan{a} = \pi \left ( \frac{\pi}{2} – \arctan{a}\right ) = \pi \arctan{\frac{1}{a}}$$

and

$$\int_0^{\infty} dx \frac{x}{x^2+a^2}\log{\left(\frac{x+1}{x-1}\right)} = \pi \arctan{\frac{1}{a}}+ i \frac{\pi}{2} \log{\left ( 1+ \frac{1}{a^2} \right )}$$

Note the imaginary part, which differs from the original problem statement. By using the absolute values, this imaginary part goes away and the stated result is true.

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