Prove that:$$\int_0^{\infty}\frac{\ln x}{x^2+bx+c^2}\,dx=\frac{2\ln c}{\sqrt{4c^2-b^2}}\cot^{-1}\left(\frac{b}{\sqrt{4c^2-b^2}}\right),$$ where $4c^2-b^2>0, c>0.$
We have:
\begin{align}
4(x^2+bx+c^2)&=(2x+b)^2-\left(\underbrace{\sqrt{4c^2-b^2}}_{=k\text{ (say)}}\right)^2\\\\
&=(2x+b+k)(2x+b-k).
\end{align}
Thus,
\begin{align}
I&=4\int_0^{\infty}\frac{\ln x}{(2x+b+k)(2x+b-k)}\,dx\\\\
&=\frac4{2k}\int_0^{\infty}\left(\frac1{2x+b-k}-\frac1{2x+b+k}\right)\ln x\, dx\\\\
&=\frac2k\left[\underbrace{\int_0^{\infty}\frac{\ln x}{2x+b-k}\,dx}_{=I_1}-\underbrace{\int_0^{\infty}\frac{\ln x}{2x+b+k}\,dx}_{=I_2}\right]
\end{align}
For $I_1:$ Letting $2x+b-k=t$ yields
\begin{align}
I_1&=\int_{b-k}^{\infty} \frac{\ln(t-b+k)-\ln 2}t\,dt\\\\
&=\int_{b-k}^{\infty}\frac{\ln(t-b+k)}t\, dt-\ln 2\int_{b-k}^{\infty}\frac{dt}t
\end{align}
Now $\ln(t-b+k)=\ln(k-b)+\ln\left(\frac{t}{k-b}+1\right).$ So,$$I_1=\ln(k-b)\int_{b-k}^{\infty}\frac{dt}t+\int_{b-k}^{\infty} \frac{\ln\left(\frac t{k-b}+1\right)}t\,dt-\ln 2\int_{b-k}^{\infty}\frac{dt}t.$$
If we let: $\frac t{k-b}=-u.$ Then $$I_1=\ln\left(\frac{k-b}2\right)\int_{b-k}^{\infty} \frac{dt}t+\int_1^{\infty}\frac{\ln(1-u)}u\,du.$$
It's getting messier and messier. Also it seems that $I_1$ and hence $I$ diverges. Please tell me, am I heading towards something useful? It has become difficult for me to handle this integral. Please show me a proper way.
Best Answer
Hint : Substitute $xy=c^2\to x=\frac{c^2}{y}$ then the integral becomes $$ I= \int_0^{\infty}\frac{\ln(c^2)-\ln y}{y^2+by+c^2}dy\\2I= \int_0^{\infty}\frac{\ln(c^2)}{y^2+by+c^2}dy$$ and since $$ y^2+by+c^2= \left(y+\frac{b}{2}\right)^2+\left(c^2-\frac{b^2}{4}\right)$$ and using elementary integral formula of $$\int_{0}^{\infty}\frac{dx}{x^2+a^2}=\frac{1}{a}\tan^{-1}\left(\frac{x}{a}\right)$$ we have our desired closed form.