Calculate the integral $\int_1^\infty \frac{x\ln x}{(x+1)(x^2+1)}dx$.

Question

Calculate the integral $\int_1^\infty \frac{x\ln x}{(x+1)(x^2+1)}dx$

I tried partial fraction decomposition on the denominator but that results in imaginary numbers. Some hints would be greatly appreciated.

Answer 1

Integrate as follows

\begin{align} &\int_1^\infty \frac{x\ln x}{(x+1)(x^2+1)}dx\overset{x\to \frac1x}=-\int_0^1 \frac{\ln x}{(x+1)(x^2+1)}dx\\=&-\frac12\int_0^1 \frac{\ln x}{1+x^2}dx+\frac12\int_0^1 \frac{x\ln x}{1+x^2}\overset{x^2\to x}{dx}-\frac12\int_0^1 \frac{\ln x}{1+x}dx\\ =& -\frac12(-G) -\frac38 \int_0^1 \frac{\ln x}{1+x}dx = \frac12G-\frac38(-\frac{\pi^2}{12})\\ =& \frac12G+\frac{\pi^2}{32} \end{align} where $\int_0^1 \frac{\ln x}{1+x^2}dx=-G$ and $\int_0^1 \frac{\ln x}{1+x}dx=-\frac{\pi^2}{12} $.

Answer 2

Too long for a comment.

It is not bad using partial fraction decomposition $$\frac x {(x+1)(x+i)(x-i)}=\frac{1-i}4 \frac 1{x-i}+\frac{1+i}4 \frac 1{x+i}-\frac 12 \frac 1{x+1}$$ So, for the antiderivative, you have three integrals (use integration by parts) $$I_a=\int \frac{\log(x)}{x+a}\,dx=\text{Li}_2\left(-\frac{x}{a}\right)+\log (x) \log \left(1+\frac{x}{a}\right)$$ $$J_a=\int_1^t \frac{\log(x)}{x+a}\,dx=\log (t) \log \left(1+\frac{t}{a}\right)+\text{Li}_2\left(-\frac{t}{a}\right)-\text{Li}_2\left(-\frac{1}{a}\right)$$

Recombine all pieces together and use the limit when $t \to \infty$. If you look for an asymptotics, $$\int_1^t \frac{x\log(x)}{(x+1)(x^2+1)}dx=\frac{16 C+\pi ^2}{32} -\frac{1+\log (t)}{t}+\frac{2 \log (t)+1}{4 t^2}+O\left(\frac{\log(t)}{t^5}\right)$$

which is in a relative error smaller than $0.1$% as soon as $t\geq 4.06$ and smaller than $0.01$% as soon as $t\geq 6.35$.