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The stated derivative formula needs a domain qualification. The original PDF calls $\alpha$ real, but the integral is even in $\alpha$; its derivative therefore cannot equal the printed positive expression on both sides of zero.

First take $\alpha >0$. The improper integral converges, with an $O(x^2)$ integrand near zero and $O(\ln x/x^2)$ decay at infinity. On compact positive parameter intervals, an integrable bound also justifies differentiation under the integral sign:For $\alpha \ne 1$, the partial fraction decomposition isIntegrating the rational functions givesAt $\alpha =1$, the same result follows by continuity, or directly from $2{\int }_0^{\infty }{x}^2/(1+x^2{)}^{2}dx=\pi /2$.

Integrating gives $E(\alpha )=\pi \ln (1+\alpha )+C$ on the positive half-line. To fix $C$, take $\alpha \to 0^{+}$: for $0\le \alpha \le 1$ the integrand is bounded by $\ln (1+x^2)/(1+x^2)$, an integrable function. The dominated convergence theorem gives $E(\alpha )\to E(0)=0$, so $C=0$. Evenness then extends the logarithmic parameter integral with Cauchy weight to its full stated domain:For negative $\alpha$, the derivative is $-\pi /(1-\alpha )$. At zero the right and left derivatives are $\pi$ and $-\pi$, so $E$ is not differentiable there. The pointwise derivative of the integrand at zero is zero, but interchanging it with the infinite integral at that parameter is invalid. Finally, substituting $\alpha =1$ gives the requested special value

Use the principal branch of the complex logarithm for $\mathrm{Log}(z+i)$. Its branch cut is the ray $z=-i-s$, $s\ge 0$, lying below the real axis. Thus it is analytic in the upper half-plane, where the only pole of the integrand is $z=i$.

Close the real interval by a large upper semicircle. The estimation lemma bounds the arc integral by $O(\ln R/R)$, so it vanishes. The simple-pole residue isThe residue theorem then givesFor real $x$, $\mathrm{Re}\mathrm{Log}(x+i)=\frac{1}{2}\ln (x^2+1)$. Taking real parts and then using evenness of the real integrand yieldsBoth endpoints converge: the integrand is $O(x^2)$ near zero and $O(\ln x/x^2)$ at infinity. This also agrees with the logarithmic parameter integral with Cauchy weight at parameter one.