 Logarithmic parameter integral with Cauchy weight

For real $\alpha$, the displayed improper integral converges and is an even function of $\alpha$. For $\alpha >0$, differentiation under the integral sign and partial fraction decomposition giveFor $\alpha \ne 1$, use $x^2/[(1+{\alpha }^2{x}^2)(1+x^2)]=[1/(1+x^2)-1/(1+{\alpha }^2{x}^2)]/({\alpha }^2-1)$. Integrate the resulting rational functions and extend to $\alpha =1$ by continuity. Integrating in $\alpha$ and using $E(0)=0$ gives $E(\alpha )=\pi \ln (1+\alpha )$ for $\alpha \ge 0$; evenness gives the formula for negative parameters. Thus $E^{\prime }(\alpha )=\pi \mathrm{sgn}(\alpha )/(1+|\alpha |)$ away from zero. At zero the one-sided derivatives are opposite, so $E$ is continuous but not differentiable. Substituting $\alpha =0$ into the pointwise differentiated integrand would therefore be an invalid interchange.