# 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!

• Call this integral $f(t)$. Evaluate $f^\prime (t)$ with partial fractions, then use $f(0)=0$. – J.G. Mar 20 '19 at 22:06
• Thank you, I understand exactly what you are talking about... – Pratik Apshinge Mar 20 '19 at 22:08
• How will the derivative of the function give me information about the function itself? – Pratik Apshinge Mar 20 '19 at 22:17
• Maybe you can integrate the derivative of the function. – Minus One-Twelfth Mar 20 '19 at 22:17
• And, also, is there any possible way to do this with gaussian integrals or the gamma function? – Pratik Apshinge Mar 20 '19 at 22:18

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: $$I(t)=I(0)+\frac\pi2\text {sgn}(t)\log(|t|+1)=\text {sgn}(t)\frac\pi2\log(|t|+1).$$

• Wouldn't $\left[t\tan^{-1}\left(tx\right)-\tan^{-1}x\right]^{ }$ from 0 to infinity be 0? – Pratik Apshinge Mar 20 '19 at 22:51
• @PratikApshinge No. It will be $(t-1)\pi/2$. – user Mar 20 '19 at 22:54
• I'm so sorry, made a calculation mistake! – Pratik Apshinge Mar 20 '19 at 22:54
• No problem. You're welcome. – user Mar 20 '19 at 22:56

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

$$\int_0^\infty \frac{\arctan(tx)}{x(1+x^2)}\,dx=\begin{cases}\frac\pi 2\log(1+t)&, t>0\\\\0&,t=0\\\\-\frac\pi2 \log(1-t)&,t<0\end{cases}$$