Determine the value of the integral $I=\int_{0}^{\infty} \frac{\ln\left(a^2+x^2\right)}{b^2+x^2}dx$ Determine the value of the integral $$I(a)=\int_{0}^{\infty} \frac{\ln\left(a^2+x^2\right)}{b^2+x^2}dx$$
My try:
$\to I'(a)=\int_{0}^{\infty}\frac{2a}{(a^2+x^2)(b^2+x^2)}dx=\frac{\pi}{b(a+b)}$
Hence $I(a)=\frac{\pi}{b}\ln(a+b)+C$
Question: Find C???
Thank you!
 A: Since integrand is even function, so
 $\int_{0}^{\infty} \frac{\ln\left(a^2+x^2\right)}{b^2+x^2}dx$=$\frac{1}{2}\int_{-\infty}^{\infty} \frac{\ln\left(a^2+x^2\right)}{b^2+x^2}dx$
so solving the R.H.S using residues--
$\int_{C} f(z)dz=\int_{Cr} f(z)dz +\int_{-R}^{R}\frac{1}{2} \frac{\ln\left(a^2+x^2\right)}{b^2+x^2}dx$
where C is closed loop in upper half of plane and f(z)= $ \frac{1}{2}\frac{\ln\left(a^2+z^2\right)}{b^2+z^2}$  
and f(z) becomes $\frac{1}{2}\frac{\ln\left(a^2+z^2\right)}{(z+bi)(z-bi)}$ by factorizing denominator
there for f(z) has singularities at z=+bi,-bi, but only z=bi lie in upper half of plane  
so,$\int_{C} f(z)dz=2(pi)(i)$[residue  of f(z) at $bi$]
and as R goes to infinity $\int_{Cr} f(z)dz-->0$
we get 
$\int_{0}^{\infty} \frac{\ln\left(a^2+x^2\right)}{b^2+x^2}dx=\int_{C} f(z)dz=2(\pi)(i)$[residue  of f(z) at $bi$] 
=$\frac{\pi}{2b}\ln(a^2-b^2)$
A: Firstly, we introduce a lemma:
\begin{equation}
\int_0^{\frac{\pi}{2}}\ln(\sin(x))dx=\int_0^{\frac{\pi}{2}}\ln(\cos(x))dx\\
\mbox{Then}\\
\int_0^{\frac{\pi}{2}}\ln(\tan(x))dx=0
\end{equation}
The proof see here.
Set $a=0$ and use the substitution that $x=b\tan(u)$, we have:
\begin{equation}
I(0)=2\int_0^\infty\frac{\ln(x)}{b^2+x^2}dx=\frac{2}{b^2}\int_0^\frac{\pi}{2}\frac{\ln(b\tan(u))}{(1+\tan(u)^2)\cos(u)^2}du\\
=\frac{2}{b^2}\int_0^\frac{\pi}{2}\ln(b\tan(u))du\\
=\frac{\pi\ln(b)}{b}
\end{equation}
Then, you can solve the constant $C$.
