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

\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+\frac{\pi^2}{32} \end{align} where $\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 $t\geq 6.35$.

Answer 3

Using the below,

$$ \int_1^\infty \frac{x\ln x}{(x+1)(x^2+1)}dx=\int_0^\infty\frac{x\ln x}{(x+1)(x^2+1)}\,dx-\int_0^1\frac{x\ln x}{(x+1)(x^2+1)}\,dx$$

$$\implies I=I_1-I_2$$

Dealing with $I_1$,

$$I_1\underset{x\to\frac1{t}}=-\int_0^\infty \frac{\ln(x)}{(1+x)(1+x^2)}\,dx$$

Now consider,

$$I(a)=\int\limits_0^{\infty} \dfrac{ x^a }{(x+1)(x^2+1) } dx = \int\limits_0^{\infty} \dfrac{ x^a }{2(x+1) } dx + \int\limits_0^{\infty} \dfrac{ x^a }{2(x^2+1) } - \int\limits_0^{\infty} \dfrac{ x^{a+1} }{2(x^2+1) }dx$$

• Using $\color{red}{\int\limits_{0}^{\infty} \frac{x^{n-1}}{1+x^{m}} \ \text{dx} = \frac{\frac{\pi}{m}}{ \sin\left(\frac{n\pi}m\right)}}$

$$I(a)=\frac{\pi}{2}\frac{1}{1+\sqrt{2} \sin \left((1+2a)\frac{\pi}{4} \right)}$$

$$I'(a)=-\int_0^\infty \frac{x^a\ln(x)}{(1+x)(1+x^2)}\,dx=\frac{\sqrt{2} \, \pi^{2} \cos\left(\frac{2\pi a + \pi}{4}\right)}{\left(2\sqrt2 \sin\left(\frac{2\pi a + \pi}{4}\right) + 2\right)^{2}}$$

$$\color{blue}{\underset{\underset{a\to 0}{\lim}} I'(1)=I_1=-\int_0^\infty \frac{\ln(x)}{(1+x)(1+x^2)}\,dx=\frac{\sqrt{2} \, \pi^{2} \cos\left(\frac{ \pi}{4}\right)}{\left(2\sqrt2 \sin\left(\frac{ \pi}{4}\right) + 2\right)^{2}}=\frac{\pi^2}{16}\tag1}$$

Dealing with $I_2$,

$$I_2=\int_0^1\frac{x\ln x}{(x+1)(x^2+1)}\,dx$$

$$\int_0^1\frac{x\ln(x)}{\left(x + 1\right) \left(x^{2} + 1\right)}\,dx=- \frac{1}{2}\int_0^1\frac{\ln(x)}{x + 1}\,dx+\frac12\int_0^1\frac{(x+1)\ln(x)}{x^{2} + 1}\,dx$$

$$\int_0^1\frac{x\ln(x)}{\left(x + 1\right) \left(x^{2} + 1\right)}\,dx=- \frac{1}{2}\int_0^1\frac{\ln(x)}{x + 1}\,dx+\frac12\int_0^1\frac{x\ln(x)}{x^{2} + 1}\,dx+\frac12\int_0^1\frac{\ln(x)}{x^{2} + 1}\,dx$$

Two of the above integrals are already present in @Quanto's answer,

To evaluate the second integral,

$$\int_0^1\frac{x\ln(x)}{x^{2} + 1}\,dx=\sum_{n=0}^{\infty}(-1)^n \int_0^1 x^{2n+1}\ln(x)\,dx$$

$$=-\frac14\sum_{n=0}^{\infty}\frac{(-1)^n}{ \left(n + 1\right)^{2}}=-\frac{\pi^2}{48}$$

$$\color{blue}{I_2=\frac{\pi^2}{32}-\frac{G}{2}\tag2}$$

From $(1),(2)$,

$$\therefore I=\frac{\pi^2}{32}+\frac{G}{2}$$

Answer 4

$$\begin{align*} I &= \int_1^\infty \frac{x\log x}{(1+x)\left(1+x^2\right)} \, dx \\ &= \int_\tfrac12^1\frac{y\log\frac y{1-y}}{1-2y+2y^2} \, dy & x=\frac y{1-y} \\ &= \int_0^\tfrac12\frac{\left(\frac12+y\right) \log\frac{\frac12+y}{\frac12-y}}{\frac12+2y^2} \, dy & y\to y+\frac12 \\ &= -\int_0^1 \frac{\log z}{1+z+z^2+z^3} \, dz & z=\frac{1+2y}{1-2y} \\ &= \sum_{n\ge0} \int_0^1 \left(z^{4n+1}-z^{4n}\right) \log z \, dz \\ &= \sum_{n\ge0} \left(\frac1{(4n+1)^2} - \frac1{4(2n+1)^2}\right) & \text{by parts} \\ &= \frac1{16} \psi^{(1)} \left(\frac14\right) - \frac14 \cdot \frac14 \psi^{(1)}\left(\frac12\right) \\ &= \boxed{\frac G2 + \frac{\pi^2}{32}} \end{align*}$$

where $\psi^{(1)}$ is the first-order polygamma function. These particular values of $\psi^{(1)}$ are listed here.