Finding the closed form of $\int _0^{\infty }\frac{\ln \left(1+ax\right)}{1+x^2}\:\mathrm{d}x$

Question

I solved a similar case which is also a very well known integral $$\int _0^{\infty }\frac{\ln \left(1+x\right)}{1+x^2}\:\mathrm{d}x=\frac{\pi }{4}\ln \left(2\right)+G$$ My teacher gave me a hint which was splitting the integral at the point $1$, $$\int _0^1\frac{\ln \left(1+x\right)}{1+x^2}\:\mathrm{d}x+\int _1^{\infty }\frac{\ln \left(1+x\right)}{1+x^2}\:\mathrm{d}x=\int _0^1\frac{\ln \left(1+x\right)}{1+x^2}\:\mathrm{d}x+\int _0^1\frac{\ln \left(\frac{1+x}{x}\right)}{1+x^2}\:\mathrm{d}x$$ $$2\int _0^1\frac{\ln \left(1+x\right)}{1+x^2}\:\mathrm{d}x-\int _0^1\frac{\ln \left(x\right)}{1+x^2}\:\mathrm{d}x=\frac{\pi }{4}\ln \left(2\right)+G$$ I used the values for each integral since they are very well known.

My question is, can this integral be generalized for $a>0$?, in other words can similar tools help me calculate $$\int _0^{\infty }\frac{\ln \left(1+ax\right)}{1+x^2}\:\mathrm{d}x$$

Answer 1

You can evaluate this integral with Feynman's trick, $$I\left(a\right)=\int _0^{\infty }\frac{\ln \left(1+ax\right)}{1+x^2}\:dx$$ $$I'\left(a\right)=\int _0^{\infty }\frac{x}{\left(1+x^2\right)\left(1+ax\right)}\:dx=\frac{1}{1+a^2}\int _0^{\infty }\left(\frac{x+a}{1+x^2}-\frac{a}{1+ax}\right)\:dx$$ $$=\frac{1}{1+a^2}\:\left(\frac{1}{2}\ln \left(1+x^2\right)+a\arctan \left(x\right)-\ln \left(1+ax\right)\right)\Biggr|^{\infty }_0=\frac{1}{1+a^2}\:\left(\frac{a\pi \:}{2}-\ln \left(a\right)\right)$$

To find $I\left(a\right)$ we have to integrate again with convenient bounds, $$\int _0^aI'\left(a\right)\:da=\:\frac{\pi }{2}\int _0^a\frac{a}{1+a^2}\:da-\int _0^a\frac{\ln \left(a\right)}{1+a^2}\:da$$ $$I\left(a\right)=\:\frac{\pi }{4}\ln \left(1+a^2\right)-\int _0^a\frac{\ln \left(a\right)}{1+a^2}\:da$$

To solve $\displaystyle\int _0^a\frac{\ln \left(a\right)}{1+a^2}\:da$ first IBP. $$\int _0^a\frac{\ln \left(a\right)}{1+a^2}\:da=\ln \left(a\right)\arctan \left(a\right)-\int _0^a\frac{\arctan \left(a\right)}{a}\:da=\ln \left(a\right)\arctan \left(a\right)-\text{Ti}_2\left(a\right)$$ Plugging that back we conclude that $$\boxed{I\left(a\right)=\:\frac{\pi }{4}\ln \left(1+a^2\right)-\ln \left(a\right)\arctan \left(a\right)+\text{Ti}_2\left(a\right)}$$ Where $\text{Ti}_2\left(a\right)$ is the Inverse Tangent Integral.

The integral you evaluated can be proved with this, $$I\left(1\right)=\int _0^{\infty }\frac{\ln \left(1+x\right)}{1+x^2}\:dx=\frac{\pi }{4}\ln \left(2\right)-\ln \left(1\right)\arctan \left(1\right)+\text{Ti}_2\left(1\right)$$ $$=\frac{\pi }{4}\ln \left(2\right)+G$$ Here $G$ denotes the Catalan's constant.

Answer 2

As @Dennis Orton answered, Feynman trick is certainly the most elegant approach for the solution.

What you could also do is $$\frac 1 {1+x^2}=\frac i 2 \left( \frac 1 {x+i}-\frac 1 {x-i}\right)$$ and we face two integrals $$I_k=\int \frac {\log(1+ax)}{x+k i}\,dx=\text{Li}_2\left(\frac{1+a x}{1-i a k}\right)+\log (a x+1) \log \left(1-\frac{1+a x}{1-i a k}\right)$$ $$J_k=\int_0^p \frac {\log(1+ax)}{x+k i}\,dx=\text{Li}_2\left(\frac{i (1+a p)}{a k+i}\right)+\log (1+a p) \log \left(\frac{a (k-i p)}{a k+i}\right)-\text{Li}_2\left(\frac{i}{a k+i}\right)$$ Computing $\frac i 2(J_1-J_{-1})$ and making $p \to\infty$, assuming $a>0$ you should end with $$\int _0^{\infty }\frac{\log \left(1+ax\right)}{1+x^2}\,dx=\frac{1}{4} \pi \log \left(1+a^2\right)+\log (a) \cot ^{-1}(a)+\frac{1}{2} i \left(\text{Li}_2\left(-\frac{i}{a}\right)-\text{Li}_2\left(\frac{i}{a}\right) \right)$$