Evaluate $\int_{0}^{\infty} \frac{\ln x}{x^2+6x+10}dx$

Question

Evaluate $$\int_{0}^{\infty} \frac{\ln x}{x^2+6x+10}dx$$

The given answer is $0.370429$. Is there any method to solve this? Thanks in advance.

Answer 1

Actually Complex Analysis (or the dilogarithm machinery) is not strictly needed to solve the given integral, it is enough to exploit a substitution and a hidden symmetry.

$$\mathfrak{I}=\int_{0}^{+\infty}\frac{\log x}{x^2+6x+10}\,dx \stackrel{x\mapsto z\sqrt{10}}{=}\sqrt{10}\int_{0}^{+\infty}\frac{\tfrac{1}{2}\log(10)+\log z}{10z^2+6\sqrt{10}\, z+10}\,dz$$ but by enforcing the substitution $z\mapsto\frac{1}{z}$ we also have $$\mathfrak{I}=\sqrt{10}\int_{0}^{+\infty}\frac{\tfrac{1}{2}\log(10)\color{red}{-}\log z}{10z^2+6\sqrt{10}\,z+10}\,dz$$ from which it follows that $$ 2\mathfrak{I} = \sqrt{10}\log(10)\int_{0}^{+\infty}\frac{dz}{10z^2+6\sqrt{10}\,z+10} $$ and the problem boils down to the evaluation of an elementary integral.

Answer 2

Note $$ \int_0^\infty\frac{x^a}{x+1}dx=-\frac{\pi}{\sin(a\pi)}$$ and hence $$ \int_0^\infty\frac{x^a}{x+b}dx=-\frac{b^a\pi}{\sin(a\pi)}. $$ So \begin{eqnarray} &&\int_{0}^{\infty} \frac{\ln x}{x^2+6x+10}dx\\ &=&\lim_{a\to0}\frac{d}{da}\int_{0}^{\infty} \frac{x^a}{x^2+6x+10}dx\\ &=&\lim_{a\to0}\frac{d}{da}\int_{0}^{\infty} \frac{x^a}{(x+3)^2+1}dx\\ &=&\lim_{a\to0}\frac{i}{2}\frac{d}{da}\int_{0}^\infty\left(\frac{x^a}{x+3+i}-\frac{x^a}{x+3-i}\right)dx\\ &=&\lim_{a\to0}\frac{i}{2}\frac{d}{da}\bigg[-\frac{(3+i)^a}{\sin(a\pi)}+\frac{(3-i)^a}{\sin(a\pi)}\bigg]\\ &=&\lim_{a\to0}\frac{i}{2}\frac{d}{da}\frac{(3-i)^a-(3+i)^a}{\sin(a\pi)}\bigg]\\ &=&\lim_{a\to0}\frac{i}{2}\frac{[(3-i)^a \ln (3-i)-(3+i)^a \ln (3+i)]\sin(a\pi)-[(3-i)^a-(3+i)^a]\cos(a\pi)}{\sin^2(a\pi)}\\ &=&\frac{i}{4}[\ln^2(3-i)-\ln^2(3+i)]\\ &=&\frac{i}{4}\ln[(3-i)(3+i)]\ln[(3-i)/(3+i)]\\ &=&\frac{1}{2}\ln10\arctan(\frac13). \end{eqnarray}