= Logarithmic parameter integral with Cauchy weight
{title2=$\int_0^\infty\ln(1+\alpha^2x^2)/(1+x^2)\,dx=\pi\ln(1+|\alpha|)$}
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> give
$$
E'(\alpha)=\int_0^\infty\frac{2\alpha x^2}{(1+\alpha^2x^2)(1+x^2)}\,dx=\frac\pi{1+\alpha}.
$$
For $\alpha\ne1$, use $x^2/[(1+\alpha^2x^2)(1+x^2)]=[1/(1+x^2)-1/(1+\alpha^2x^2)]/(\alpha^2-1)$. 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 $\alpha\ge0$; evenness gives the formula for negative parameters. Thus $E'(\alpha)=\pi\operatorname{sgn}(\alpha)/(1+|\alpha|)$ 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.
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