Is this proof of the following integrals fine $\int_{0}^{1} \frac{\ln(1+x)}{x} dx$?

$$\int_{0}^{1} \frac{\ln(1+x)}{x} dx = \int_{0}^{1} \frac1x\cdot (x-x^2/2 + x^3/3 -x^4/4 \ldots)dx = \int_{0}^{1}(1 - x/2 + x^2/3 - x^4/4 \dots)dx = 1-\frac1{2^2} + \frac1{3^3} - \frac{1}{4^4} \dots = \frac{\pi^2}{12}$$

Also,

$$\int_{0}^{1} \frac{\ln(1-x)}{x}dx = -\int_{0}^{1}\frac1x \cdot (x+x^2/2 +x^3/3 \ldots)dx =- \int_{0}^{1}(1+x/2+x^2/3 \dots )dx = -( 1 + \frac1{2^2} + \frac1{3^3} \dots) = -\frac{\pi^2}{6}$$

Using these integral, I evaluate $$\int_{0}^{1} \frac{\ln(x)}{1-x}$$

by letting $1-x=t$

$$\int_{0}^{1} \frac{\ln(x)}{1-x} =\int_{0}^{1} \frac{\ln(1-t)}{t} dx = -\frac{\pi^2}{6}$$

However, I'm facing problem while integrating

$$\int_{0}^{1} \frac{\ln(x)}{1+x}$$ using the same above way.

• The same trick of using the substitution $t=1-x$ won't work here, but one can expand in series though. Jun 10, 2018 at 8:53

Hint: integrate by parts.

Let $u = \ln(x)\implies u' = \dfrac1x$ and $v' = \dfrac1{1 + x}\implies v = \ln(1 + x)$. Therefore, $$\int\dfrac{\ln(x)}{1 + x}\,\mathrm dx = \ln(x)\ln(1 + x) - \int\dfrac{\ln(1 + x)}x\,\mathrm dx$$

• Thank you. Can you tell me if the method I used to integrate the first two integrals is fine or not Jun 10, 2018 at 8:51
• Yes, they seem fine. You can also verify using Wolfram Alpha. See this, this and this.
– an4s
Jun 10, 2018 at 9:00

$$\int_0^1 x^m\ln x\,dx=\int_0^1 ye^{-(m+1)y}\,dy=-\frac1{(m+1)^2}.$$ Then $$\int_0^1\frac{\ln x}{1+x}\,dx=\sum_{m=0}^\infty(-1)^m \int_0^1 x^m\log x\,dx=\sum_{m=1}^\infty\frac{(-1)^{m+1}}{(m+1)^2} =-\frac{\pi^2}{12}.$$

Let,

\begin{align}I&=\int_0^1 \frac{\ln x}{1-x}\,dx\\ J&=\int_0^1 \frac{\ln x}{1+x}\,dx\\ \end{align}

\begin{align}I-J&=\int_0^1 \frac{2x\ln x}{1-x^2}\,dx\end{align}

Perform the change of variable $y=x^2$,

\begin{align}I-J&=\frac{1}{2}\int_0^1 \frac{\ln x}{1-x}\,dx\\ &=\frac{1}{2}I\end{align}

Therefore,

$J=\dfrac{1}{2}I$

If $I=-\dfrac{\pi^2}{6}$ then $J=-\dfrac{\pi^2}{12}$