一类反正切的含参积分

青青子衿

\begin{align*}
\int^{+\infty}_{0}\,\frac{\arctan{x}}{x(a^2+x^2)}\mathrm{d}x&=\frac{\pi\ln(1+a)}{2a^2}\\
\int^{+\infty}_{0}\,\frac{\arctan{x}}{x(a^2+x^2)^2}\mathrm{d}x&=\frac{\pi\left[2(1+a)\ln(1+a)-a\right]}{4a^4(1+a)}\\
\end{align*}
Table of Integrals, Series, and Products (8th Edition)
Ch. 4.537  P. 607

hbghlyj

\[\frac{1}{2}\int^\infty_{-\infty}\frac{\arctan{x}}{x(a+x^2)}\ {\rm d}x=-\Re\pi \operatorname*{Res}_{z=i\sqrt{a}}\frac{\ln(1-iz)}{z(z^2+a)}=\frac{\pi\ln(1+\sqrt{a})}{2a}\]MSE