[Math] Integrate: $ \int_0^\infty \frac{\log(x)}{(1+x^2)^2} \, dx $ without using complex analysis methods

Question

Can this integral be solved without using any complex analysis methods: $$ \int_0^\infty \frac{\log(x)}{(1+x^2)^2} \, dx $$

Thanks.

Answer 1

[Corrected and justified]

The change of variables $x = e^{-t}$ converts the integral to $-\int_{-\infty}^\infty te^{-t} dt \left/(1+e^{2t})^2 \right.$. Splitting into $\int_{-\infty}^0 + \int_0^\infty$ and combining $t$ with $-t$ yields $$ - \int_0^\infty \frac{t(e^{-t}-e^{-3t})dt}{(1+e^{-2t})^2} = -\int_0^\infty t(e^{-t} - 3e^{-3t} + 5e^{-5t} - 7e^{-7t} + - \cdots) \, dt. $$ Integrating termwise yields $-(1 - \frac13 + \frac15 - \frac17 + - \cdots) = -\pi/4$. Termwise integration requires some justification because the sum does not converge absolutely, but no complex analysis is needed.

One way to justify the termwise integration in this setting is to write the integral as the limit as $N \rightarrow \infty$ of $$ - \int_0^\infty \frac{t(e^{-t}-e^{-3t}-2e^{-(4N+1)t})dt}{(1+e^{2t})^2}. $$ Expanding $(e^{-t}-e^{-3t}-2e^{-(4N+1)t}) \left/ (1+e^{-2t})^2 \right.$ in powers of $e^{-t}$, and integrating termwise, yields $$ S_N := -\left( 1 - \frac 13 + \frac15 - \frac17 + - \cdots \right. \phantom{\infty\infty\infty\infty\infty\infty\infty\infty\infty\inftyinfty\infty\infty} $$ $$ \phantom{\infty\infty\infty\infty\infty} \left. - \frac1{4N-1} + \frac{4N-1}{(4N+1)^2} - \frac{4N-1}{(4N+3)^2} + \frac{4N-1}{(4N+5)^2} - \frac{4N-1}{(4N+7)^2} + - \cdots \right), $$ this time justified by absolute convergence. The series $S_N$, like $-(1 - \frac13 + \frac15 - \frac17 + - \cdots)$, is an alternating series whose terms decrease in absolute value. The two series agree through the $1/(4N-1)$ term, and both remainders are bounded in absolute value by $1/(4N+1)$. Therefore as $N \rightarrow \infty$ the new series $S_N$ approaches $-(1 - \frac13 + \frac15 - \frac17 + - \cdots) = -\pi/4$, and we're done.