Calculate $\int_{1}^{\infty}\frac{x\ln x}{(1+x^2)^2}\text{dx}$

Question

$$\text{Calculate }\int_{1}^{\infty}\frac{x\ln x}{(1+x^2)^2}\text{dx}$$

This problem absolutely stumps me. What I tried (but didn't help me much) is to define an $F$, then differentiate, and then integrate $$F(a) = \int_{1}^{\infty}\frac{x\ln(ax)}{(1+x^2)^2}\text{dx}\Rightarrow F'(a) = \frac{1}{a}\int_{1}^{\infty}\frac{x}{(1+x^2)^2}\text{dx} = \frac{1}{2a}\int_2^\infty \frac{1}{t^2}\text{dt}=\frac{1}{4a} \Rightarrow\\ \Rightarrow F(a) = \frac{1}{4}\cdot \ln(a) + c$$ However, I do not know how to find $c$ (no value of $a$ seems to result a quite calculation of $F$). Other than this, my only speculation is that the beta or gamma functions may be involved down the line. Any help is much appreciated!

Answer 1

Set $u(x)=\frac{1}{2}\frac{1}{1+x^2}$ and $v(x)=\ln(x)$ for $x\geq 1$. Then

$\int_1^\infty \frac{x}{(1+x^2)^2}\ln(x) dx=\int_1^\infty -u'(x)v(x) dx=[-u(x)v(x)]_1^\infty+\int_1^\infty \frac{u(x)}{x} dx=\int_1^\infty \frac{1}{2}\frac{1}{x(1+x^2)}dx$.

Note that $\frac{1}{x(1+x^2)}=\frac{1}{x}-\frac{x}{1+x^2}$ for $x\geq 1$. Thus we have for any $a>1$

$\int_1^a \frac{1}{x(1+x^2)} dx=\int_{1}^a \frac{1}{x} dx-\int_1^\infty\frac{x}{(1+x^2)}dx=\ln(a)-\ln(\sqrt{1+x^2})|_1^a=\ln\left(\frac{a}{\sqrt{1+a^2}}\right)+\ln(\sqrt{2})$.

Thus we obtain

$\int_1^\infty \frac{x}{(1+x^2)^2}\ln(x) dx=\int_1^\infty \frac{1}{2}\frac{1}{x(1+x^2)}dx=\frac{1}{2}\lim_{a\to \infty} \left(\ln\left(\frac{a}{\sqrt{1+a^2}}\right)+\ln(\sqrt{2})\right)=\frac{1}{4}\ln(2)$.

Answer 2

In fact, $$\begin{eqnarray} &&\int_{1}^{\infty}\frac{x\ln x}{(1+x^2)^2}\text{dx}\\ &=&\int_{1}^{\infty}\frac x{(1+x^2)^2}\bigg(\int_1^x\frac1u\text{du}\bigg)\text{dx}\\ &=&\int_{1}^{\infty}\frac1u\bigg(\int_u^\infty\frac{x}{(1+x^2)^2}\text{dx}\bigg)\text{du}\\ &=&\int_{1}^{\infty}\frac1{2u(u^2+1)}\text{du}=\frac14\ln2. \end{eqnarray}$$

Answer 3

Letting $x\mapsto \frac{1}{x} $ transforms the integral into $$ \int_1^{\infty} \frac{x \ln x}{\left(1+x^2\right)^2} d x=-\int_0^1 \frac{x \ln x}{\left(1+x^2\right)^2} d x \\ $$ Differentiating the series for $|x|<1$, $$\frac{1}{1+x}=\sum_{n=0}^\infty(-1)^n x^n$$ and replacing $x$ by $x^2$ yields

$$\frac{1}{(1+x)^2}=\sum_{n=1}^\infty(-1)^n nx^{2n-2}$$ Plugging back gives us $$ \begin{aligned} I & =\sum_{n=1}^{\infty}(-1)^n n \int_0^1 x^{2 n-1} \ln x d x \\ & =\sum_{n=1}^{\infty}(-1)^n n\left(-\frac{1}{4 n^2}\right) \\ & =\frac{1}{4} \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} \\ & =\frac{1}{4} \ln 2 \end{aligned} $$