How to evaluate $$\int_0^{\infty } \left(\frac{1}{(x+1)^2 \log (x+1)}-\frac{\log (x+1) \tan ^{-1}(x)}{x^3}\right) \, dx = G – \gamma + \frac{1}{4} \pi \log 2 – \frac{3}{2}.$$
I made some progress.
Integrating the second term by parts shows that negative of the integral equals
$$
-I = 1/2 + 1/2 \int_0^{\infty} \,dx \frac{1}{x^2} \left[ \frac{ \ln(1+x)}{1+x^2} + \frac{ \tan^{-1} x }{1+x} \right] – \frac{1}{(1+x)^2 \ln(1+x)}.
$$
Integrating the arctan term by parts and using the partial fraction expansion of $\frac{1}{x^2(1+x)}$ yields
$$
-I = 1 – G/2 – \frac{\pi}{8} \ln 2 + \int_0^{\infty} \,dx \frac{1}{2(1+x^2)} \left[\frac{\ln(1+x)}{x^2} + \frac{1}{x} \right] – \frac{1}{(1+x)^2 \ln (1+x)}.
$$
Here I used the integrals $$ \int_0^{\infty} \,dx \frac{\ln x}{1+x^2} = 0$$ and $$ \int_0^{\infty} \,dx \frac{\ln (1+x)}{1+x^2} = G + \frac{\pi}{4} \ln 2.$$ The latter still requires proof.
The value of the integral follows immediately if we can show that
$$
\int_0^{\infty} \,dx \frac{1}{x} \left[ \frac{1}{1+x^2} – e^{-x} \right] = \gamma
$$
[Note that I made a substitution in the second term] and
$$
\int_0^{\infty} \,dx\left[ \frac{\ln(1+x)}{x^2(1+x^2)} – \frac{1}{(1+x)^2 \ln (1+x)} \right] = 1 + \gamma – G – \frac{\pi}{4} \ln 2,
$$
which are beautiful independent results. Can someone continue from here?
Best Answer
To show that $$ \int_0^{\infty} \frac{1}{x}\left(\frac{1}{1+x^2}-e^{-x}\right)=\gamma, $$ we can use integration by parts on $\frac{1}{x}$ that $$ \begin{aligned} \int_0^{\infty} \frac{1}{x}\left(\frac{1}{1+x^2}-e^{-x}\right) d x = & \int_0^{\infty}\left(\frac{1}{1+x^2}-e^{-x}\right) d(\ln x) \\ = & {\left[\ln x\left(\frac{1}{1+x^2}-e^{-x}\right)\right]_0^{\infty} } - \int_0^{\infty} \ln x\left[\frac{2 x}{\left(1+x^2\right)^2}+e^{-x}\right] d x\\=& -2 \int_0^{\infty} \frac{x \ln x}{\left(1+x^2\right)^2} d x+\gamma\\=& \gamma \end{aligned} $$ where $$ \int_0^{\infty} \frac{x \ln x}{\left(1+x^2\right)^2} d x \stackrel{x\mapsto\frac{1}{x}}{=} -\int_0^{\infty} \frac{x \ln x}{\left(x^2+1\right)^2} d x \Rightarrow \int_0^{\infty} \frac{x \ln x}{\left(1+x^2\right)^2} d x =0 $$