Prove $\int_{0}^{\infty}{\ln x\ln\left(x\over 1+x\right)\over (1+x)^2}dx$ and $\int_{0}^{\infty}{\ln x\ln\left(x\over 1+x\right)\over 1+x}dx$

Question

Two well known constants

$$\int_{0}^{\infty}{\ln x\ln\left(x\over 1+x\right)\over (1+x)^2}=\zeta(2)\tag1$$

$$\int_{0}^{\infty}{\ln x\ln\left(x\over 1+x\right)\over 1+x}=\zeta(3)\tag2$$

An attempt:

Applying IBP: $$\int{\ln x\over x}\mathrm dx={1\over 2}\ln^2 x+C\tag3$$

Rewrite $(2)$ as

$$\int{\ln^2 x\over 1+x}\mathrm dx-\int{\ln(x)\ln(1+x)\over 1+x}\mathrm dx=I_1-I_2\tag4$$

Let integrate $I_1$, applying IBP

$$I_1=\ln^2(x)\ln(1+x)-2\int{\ln(x)\ln(1+x)\over x}\mathrm dx\tag5$$

$$I_1=-\ln^2(x)\ln(1+x)+2\ln(x)\ln(1+x)-2\int{\ln(x)\ln(1+x)\over x}\mathrm dx\tag6$$

$I_1$ is not going down any further.

Let try $I_2$ applying IBP

$$I_2={1\over 2}\ln(x)\ln(1+x)+{1\over 2}\int{\ln^2(x)\over x}\mathrm dx\tag7$$

$$\int{\ln^2(x)\over x}\mathrm dx={1\over 3}\ln^3(x)+C\tag8$$

$$I_2={1\over 2}\ln(x)\ln(1+x)+{1\over 6}\ln^3(x)+C\tag9$$

This seem too complicate, what I am doing here

How can we tackle $(1)$ and $(2)$ in a less cumbersome way?

Answer 1

Hint. By the change of variable $$ u=\frac{x}{x+1},\quad du=\frac{dx}{(1+x)^2}, $$ one gets $$ \begin{align} \int_{0}^{\infty}{\ln x\ln\left(x\over 1+x\right)\over (1+x)^2}&=\int_{0}^{1}\left(\ln u - \ln(1-u)\right)\ln u \:du \\\\&=\int_{0}^{1}\ln^2 u\:du - \int_{0}^{1}\ln(1-u)\ln u \:du \\\\&=2 +\sum_{n=1}^\infty \frac1n\int_{0}^{1}u^{n}\ln u \:du \\\\&=2 -\sum_{n=1}^\infty \frac1{n(n+1)^2} \\\\&=2 +\frac{\pi^2}6-2 \\\\&=\zeta(2). \end{align} $$Similarly, one has $$ \begin{align} \int_{0}^{\infty}{\ln x\ln\left(x\over 1+x\right)\over 1+x}&=\int_{0}^{1}\left(\ln u - \ln(1-u)\right)\frac{\ln u}{1-u} \:du \\\\&=\int_{0}^{1}\frac{\ln^2 u}{1-u} \:du - \int_{0}^{1}\frac{\ln(1-u)\ln u }{1-u} \:du \\\\&=2\zeta(3)- \int_{0}^{1}\frac{\ln(1-v)\ln v }{v} \:dv \\\\&=2\zeta(3)-\zeta(3) \\\\&=\zeta(3), \end{align} $$ where we have used some standard results.

Answer 2

Splitting the first integral into two and for the second one using $x\to\frac1x$, one has \begin{eqnarray} \int_{0}^{\infty}{\ln x\ln\left(x\over 1+x\right)\over (1+x)^2}dx&=&\int_{0}^{1}{\ln x\ln\left(x\over 1+x\right)\over (1+x)^2}dx+\int_{1}^{\infty}{\ln x\ln\left(x\over 1+x\right)\over (1+x)^2}dx\\ &=&\int_{0}^{1}{\ln x\ln\left(x\over 1+x\right)\over (1+x)^2}dx+\int_{0}^{1}{\ln \frac1x\ln\left(\frac1x\over 1+\frac 1x\right)\over (1+\frac1x)^2}\frac1{x^2}dx\\ &=&\int_{0}^{1}{\ln x\ln\left(x\over 1+x\right)\over (1+x)^2}dx+\int_{0}^{1}{\ln x\ln\left(1+x\right)\over (1+x)^2}dx\\ &=&\int_{0}^{1}{\ln^2 x\over (1+x)^2}dx\\ &=&\zeta(2). \end{eqnarray} Splitting the second integral into two and for the second one using $x\to\frac1x$, one has \begin{eqnarray} \int_{0}^{\infty}{\ln x\ln\left(x\over 1+x\right)\over 1+x}dx&=&\int_{0}^{1}{\ln x\ln\left(x\over 1+x\right)\over 1+x}dx+\int_{1}^{\infty}{\ln x\ln\left(x\over 1+x\right)\over 1+x}dx\\ &=&\int_{0}^{1}{\ln x\ln\left(x\over 1+x\right)\over 1+x}dx+\int_{0}^{1}{\ln \frac1x\ln\left(\frac1x\over 1+\frac 1x\right)\over 1+\frac1x}\frac1{x^2}dx\\ &=&\int_{0}^{1}{\ln x\ln\left(x\over 1+x\right)\over 1+x}dx+\int_{0}^{1}{\ln x\ln\left(1+x\right)\over x(1+x)}dx\\ &=&\int_{0}^{1}{\ln^2 x\over 1+x}dx+\int_{0}^{1}{\frac1x\ln x\ln(1+x)}dx-2\int_{0}^{1}{\ln x\ln\left(1+x\right)\over1+x}dx. \end{eqnarray} It is easy to see that $$ \int_{0}^{1}{\ln^2 x\over 1+x}dx+\int_{0}^{1}{\frac1x\ln x\ln(1+x)}dx=\frac34\zeta(3) $$ by using series. Note \begin{eqnarray} 2\int_{0}^{1}{\ln x\ln\left(1+x\right)\over1+x}dx&=&\ln x\ln^2(1+x)\bigg|_0^1-\int_0^1\frac1x\ln^2(1+x)dx\\ &=&-\frac{1}{4}\zeta(3). \end{eqnarray} So $$ \int_{0}^{\infty}{\ln x\ln\left(x\over 1+x\right)\over 1+x}dx=\zeta(3). $$