How do I integrate $\int_0^1\arctan(x)\log(\frac{1-x}{1+x})\mathrm{d}x$?

Question

I've recently come across an interesting integral, which is of the form:

$$\int_0^1\arctan(x)\log\left(\frac{1-x}{1+x}\right)\mathrm{d}x$$

To start, I expanded the arctangent into its series expansion, then utilized the Weierstraß substitution in order to remove the fractional term from the logarithm:

$$t = \frac{1-x}{1+x}$$

Finally, I'm left with this integral:

$$2 \sum_{k \geq 0} \frac{(-1)^k}{2k+1} \int_0^1 \frac{(1-t)^{2k+1}}{(1+t)^{2k+3}} \log(t)\mathrm{d}t$$

Which looks an awful lot like the beta function, namely:

$$B(x, y) = (1-a)^y \int_0^1 \frac{(1-t)^{x-1} t^{y-1}}{(1-at)^{x+y}} \mathrm{d}t, \quad a \leq 1$$

For the following values, the integrals are nearly identical:

$$a=-1,$$ $$x=2k+2,$$ $$y=1$$

However, this is the bit where I fail to make progress.

I see that the integrals are clearly just off by that logarithm, but I cannot find a relation between them in order to progress with this integral.

I've tried differentiating with respect to the parameter $y$ in order to bring in that logarithm, but that obviously doesn't do much - as the parameter also lies in the denominator and causes unwanted trouble.

I've also tried constructing integrals which are similar to this one, but only have the parameter $y$ in the numerator; however, I haven't been able to make much progress doing that either. These integrals end up looking nothing like the beta function.

Answer 1

Integrate by parts with $dx=d(x-1)$

\begin{align} &\int_0^1\tan^{-1}x\ln\frac{1-x}{1+x}\ {dx}\\ \overset{ibp}= & \int_0^1 \frac{(1-x)\ln \frac{1-x}{1+x}}{1+x^2}\ \overset{\frac{1-x}{1+x}\to x}{dx} -2\int_0^1 \frac{\tan^{-1}x}{1+x}dx\\ =& \ \int_0^1 \frac{\ln x}{1+x^2}dx -\frac34 \int_0^1\frac{\ln x}{1+x}dx -2\int_0^1 \frac{\tan^{-1}x}{1+x}dx\\ =& -G-\frac34 \left(-\frac{\pi^2}{12}\right)-2\cdot \frac\pi8\ln2 =-G +\frac{\pi^2}{16} -\frac\pi4\ln2 \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

Noting $$ d\bigg[(x-1)\ln\bigg(\frac{1-x}{1+x}\bigg)-2\ln(1+x)\bigg]=\ln\bigg(\frac{1-x}{1+x}\bigg) $$ one has, by IBP, \begin{eqnarray} &&\int_0^1\tan^{-1}x\ln\bigg(\frac{1-x}{1+x}\bigg)\ {dx}\\ &=&-\int_0^1\bigg[(x-1)\ln\bigg(\frac{1-x}{1+x}\bigg)-2\ln(1+x)\bigg]\frac{1}{1+x^2}dx\\ &\overset{x\to\frac{1-x}{1+x}}{=}&\int_0^1\bigg(\frac{\ln x}{1+x^2}+\frac{x\ln x}{1+x^2}-\frac{\ln x}{1+x}+\frac{2\ln(\frac2{1+x})}{1+x^2}\bigg)dx\\ &=&-G-\frac{\pi^2}{48}+\frac{\pi^2}{12}+\frac14\pi\ln2\\ &=&-G+\frac{\pi^2}{16}+\frac14\pi\ln2. \end{eqnarray} Update: Let \begin{eqnarray} I&=&\int_0^1\frac{\ln(1+x)}{1+x^2}\\ &=&\int_0^1\int_0^1\frac{x}{(1+tx)(1+x^2)}dtdx\\ &=&\int_0^1\int_0^1\frac{x}{(1+tx)(1+x^2)}dxdt\\ &=&\int_0^1\bigg(\frac\pi4\frac{t}{1+t^2}+\frac{2\ln2}{1+x^2}-\frac{1+t}{1+t^2}\bigg)dt\\ &=&\frac14\pi\ln2-I \end{eqnarray} and hence $$ I=\frac18\pi\ln2. $$ \begin{eqnarray} &&\int_0^1\frac{2\ln(\frac2{1+x})}{1+x^2}\\ &=&\frac{\pi}2\ln2-2I=\frac{\pi}4\ln2. \end{eqnarray}

Answer 3

$$\begin{align*} I &= \int_0^1 \arctan(x)\log\left(\frac{1-x}{1+x}\right) \, dx \\ &= 2 \int_0^1 \arctan\left(\frac{1-y}{1+y}\right) \log(y) \, \frac{dy}{(1+y)^2} \tag1 \\ &= \frac\pi2 \int_0^1 \frac{\log(y)}{(1+y)^2} \, dy - 2 \underbrace{\int_0^1 \frac{\arctan(y)\log(y)}{(1+y)^2} \, dy}_J \tag2 \\[2ex] J &= \int_0^1 \left(\frac{\log(y)}{(1+y^2)(1+y)} + \frac{\arctan(y)}{y(1+y)}\right) \, dy \tag3 \\ &= \frac12 \int_0^1 \frac{\log(y)}{1+y} \, dy + \frac12 \int_0^1 \frac{1-y}{1+y^2} \log(y) \, dy \tag4 \\ &\quad\quad + \int_0^1 \frac{\arctan(y)}y \, dy - \int_0^1 \frac{\arctan(y)}{1+y} \, dy \\ &= - \frac\pi4\log(2) + \frac12 \int_0^1 \frac{\log(y)}{1+y} \, dy \\ &\quad\quad - \frac12 \int_0^1 \frac{1+y}{1+y^2} \log(y) \, dy + \int_0^1 \frac{\log(1+y)}{1+y^2} \, dy \tag3 \end{align*}$$

---

• $(1)$ : substitute $x=\dfrac{1-y}{1+y}$
• $(2)$ : $\arctan\left(\dfrac{1-y}{1+y}\right) = \dfrac\pi4-\arctan(y)$ for $y>-1$
• $(3)$ : integrate by parts
• $(4)$ : partial fractions

---

The remaining integrals can all be computed by invoking the series expansion of $\dfrac1{1-y}$ and integrating by parts:

$$\begin{align*} \int_0^1 \frac{\log(y)}{1+y} \, dy &= \sum_{n=0}^\infty (-1)^n \int_0^1 y^n \log(y) \, dy \\ &= \sum_{n=0}^\infty \frac{(-1)^{n+1}}{(n+1)^2} \\ &= \sum_{n=1}^\infty \frac{(-1)^n}{n^2} \\ &= -\frac{\pi^2}{12} \\[2ex] \int_0^1 \frac{\log(y)}{(1+y)^2} \, dy &= \sum_{n=1}^\infty (-1)^{n+1} n \int_0^1 y^n \log(y) \, dy \\ &= \sum_{n=1}^\infty \frac{(-1)^{n+1}}n \\ &= \log(2) \\[2ex] \int_0^1 \frac{1+y}{1+y^2} \log(y) \, dy &= \sum_{n=0}^\infty (-1)^n \int_0^1 \left(y^{2n} + y^{2n+1}\right) \log(y) \, dy \\ &= \sum_{n=0}^\infty (-1)^{n+1} \left(\frac1{(2n+1)^2} + \frac1{(2n+2)^2}\right) \\ &= -G - \frac{\pi^2}{48} \end{align*}$$

The last integral requires a bit of massaging first; to that end, reuse/undo the substitution from step $(1)$.

$$\begin{align*} \int_0^1 \frac{\log(1+y)}{1+y^2} \, dy &= \int_0^1 \frac{\log(2)-\log(1+y)}{1+y^2} \, dy \tag{$-1$} \\ \implies 2 \int_0^1 \frac{\log(1+y)}{1+y^2} \, dy &= \log(2) \int_0^1 \frac{dy}{1+y^2} \\ \implies \int_0^1 \frac{\log(1+y)}{1+y^2} \, dy &= \frac\pi8\log(2) \end{align*}$$

---

Putting everything together, we ultimately find

$$I = \boxed{\frac{\pi^2}{16} - G - \frac\pi4 \log(2)}$$