Evaluate the integral: $\int_0^{\infty}\frac{\tan^{-1}(tx)}{x\left(1+x^2\right)} \mathrm{d}x$ I've been trying to evaluate the integral for a while now, and I've been unable to find it anywhere...
I tried substituting $\tan^{-1}(tx)$ as $u$ but got nowhere... I have done dozens of other substitutions...
I've been told to use the properties of gaussian integrals and the gamma function, but can't seem to figure out a way to bring in $\mathrm{e}$...
A few hints that point in the right direction to think will really help!
It would be really really helpful if you could solve this using only the properties of gaussian and gamma functions!
 A: Let:
$$
I(t)=\int_0^\infty\frac{\arctan tx}{x(1+x^2)}dx.
$$
Then
$$
I'(t)=\int_0^\infty\frac{1}{(1+t^2x^2)(1+x^2)}dx=
\frac1{t^2-1}\int_0^\infty\left[\frac{t^2}{1+t^2x^2}-\frac{1}{1+x^2}\right]dx\\
=\frac1{t^2-1} \left[t\arctan(tx)-\arctan x\right]_0^\infty=\frac\pi2\frac{|t|-1}{t^2-1}=\frac\pi2\frac1{|t|+1}.
$$
Finally integrating the last expression over $t$ one obtains:
$$
I(t)=I(0)+\frac\pi2\text {sgn}(t)\log(|t|+1)=\text {sgn}(t)\frac\pi2\log(|t|+1).
$$
A: We can write $\arctan(tx)=\int_0^t \frac{x}{1+y^2x^2}\,dy$.  Proceeding, we find that for $t>0$
$$\begin{align}
\int_0^\infty \frac{\arctan(tx)}{x(1+x^2)}\,dx&=\int_0^\infty \frac{1}{1+x^2}\int_0^t \frac1{1+y^2x^2}\,dy\,dx\\\\
&\overbrace{=}^{\text{Fubini}}\int_0^t \int_0^\infty \frac{1}{(1+x^2)(1+y^2x^2)}\,dx\,dy\\\\
&=\int_0^t \frac{\pi/2}{y+1}\,dy\\\\
&=\frac\pi2 \log(1+t)
\end{align}$$
Inasmuch as the integral of interest is an odd function of $t$, we have
$$\int_0^\infty \frac{\arctan(tx)}{x(1+x^2)}\,dx=\begin{cases}\frac\pi 2\log(1+t)&, t>0\\\\0&,t=0\\\\-\frac\pi2 \log(1-t)&,t<0\end{cases}$$
