# Evaluate: $\int_{-1}^{1}\ln\left(\frac{1+t}{1-t}\right)\frac{1}{1-at}dt$

The value of $$\int_{-1}^{1}\ln\left(\frac{1+t}{1-t}\right)\frac{1}{1-at}dt$$ (where $$0)

is equal to

(A) $$\frac{1}{2a}\left(\ln(\frac{1-a}{1+a})\right)^2$$

(B) $$\frac{1}{2a}\ln(\frac{1+a}{1-a})$$

(C) $$\frac{1}{a}\ln(\frac{1+a}{1-a})$$

(D) $$\frac{1}{2a}\left(\ln(\frac{1+a}{1-a})\right)^2$$

(E) $$\frac{1}{a}\left(\ln(\frac{1-a}{1+a})\right)^2$$

My Attempt

$$I=\int_{-1}^{1}\frac{\ln(1+t)}{1-at}dt-\int_{-1}^{1}\frac{\ln(1-t)}{1-at}dt=\int_{-1}^{1}\frac{\ln(1+t)}{1-at}dt-\int_{-1}^{1}\frac{\ln(1+t)}{1+at}dt$$

$$I=\int_{-1}^{1}\ln(1+t)\left(\frac{1}{1-at}-\frac{1}{1+at}\right)dt=\int_{-1}^{1}\ln(1+t)\left(\frac{2at}{1-a^2t^2}\right)dt$$

After this I am not able to do. Is this approach correct I wonder

• consider two substitutions, one for $u=1+t$, and one for $u = 1-t$. Solve the integrals separately $ln(1+t) - ln(1-t)$ and then cancel terms Commented Mar 13 at 0:57
• (A) and (D) are the same.
– Gary
Commented Mar 13 at 3:58

Letting $$x=\frac{1-t}{1+t}$$ transforms the integral into \begin{aligned} I & =-2 \int_0^{\infty} \frac{\ln x}{[(1-a)+(1+a) x](1+x)} d x \\ & =-\frac{2}{1+a} \underbrace{ \int_0^{\infty} \frac{\ln x}{\left(x+k\right)(1+x)} d x}_{J} \end{aligned} where $$k= \frac{1-a}{1+a}$$.
Putting $$x\mapsto\frac{k}{x}$$ changes \begin{aligned} J & =\int_0^{\infty} \frac{\ln k-\ln x}{\left(1+x\right)(x+k)} d x \\ & =\ln k \int_0^{\infty} \frac{d x}{\left(1+x\right)(x+k)}-J\\&= \frac{ \ln k}{2} \int_0^{\infty} \frac{d x}{\left(1+x\right)(x+k)}\\&=\frac{\ln k}{2} \cdot \frac{1}{k-1} \int_0^{\infty} \left( \frac{1}{1+x}-\frac{1}{x+k}\right) d x\\&= \frac{\ln ^2 k}{2(k-1)}\\&= \frac{\ln ^2\left(\frac{1-a}{1+a}\right)}{\frac{-4a}{1+a}} \end{aligned} Hence $$\boxed{I=\frac{1}{2 a} \ln ^2\left(\frac{1-a}{1+a}\right)}$$
\begin{align} &\int_{-1}^1\ln\left(\frac{1+t}{1-t}\right)\frac{1}{1-at}\ \overset{t\to \frac{1-t}{1+t}}{dt}\\ =& -\frac2{1+a}\int_0^\infty \frac{\ln t}{(t+1)(t+ \frac{1-a}{1+a})} \overset{t\to \frac{1-a}{1+a}\frac1t}{dt}\\ =& -\frac2{1+a}\int_0^\infty \frac{\ln \frac{1-a}{1+a}-\ln t}{(t+1)(t+ \frac{1-a}{1+a})} {dt}\\ =& -\frac{\ln \frac{1-a}{1+a}}{1+a}\int_0^\infty \frac{1}{(t+1)(t+ \frac{1-a}{1+a})} {dt}\\ =& -\frac{\ln \frac{1-a}{1+a}}{1+a} \bigg( -\frac{1+a}{2a}{\ln\frac{1-a}{1+a}}\bigg)= \frac{1}{{2a}} \ln^2\frac{1-a}{1+a} \end{align}
\begin{aligned} \int_{-1}^1\ln\left(\frac{1+t}{1-t}\right)\frac{1}{1-at}\mathrm dt&=\int_{-1}^1\ln(1+t)\frac{2at}{1-a^2t^2}\mathrm dt\\ &=-2at\int_{-1}^1\sum_{k\geq 1}\frac{(-1)^kt^{k}}{k}\sum_{n\geq 0}a^{2n}t^{2n}\mathrm dt\\ &=-2\sum_{k\geq 1}\sum_{n\geq 0}\frac{(-1)^ka^{2n+1}}{k}\int_{-1}^1t^{2n+k+1}\mathrm dt\\ &=2\sum_{k\geq 0}\sum_{n\geq 0}\frac{(-1)^ka^{2n+1}}{k+1}\int_{-1}^1t^{2n+k+2}\mathrm dt \end{aligned} In view that \int_{-1}^1t^{2n+k+2}=\left\{ \begin{aligned} &0,\ 2m+k+2\ \text{odd}\\ &\frac{2}{2m+2k+3},\ 2m+k+2\ \text{even} \end{aligned} \right. , this means that only the terms with even $$k$$ will "survive" because $$2n+2+k=$$even$$+k=$$even. So we'll transform $$k\to 2k$$: \begin{aligned} 2\sum_{k\geq 0}\sum_{n\geq 0}\frac{a^{2n+1}}{2k+1}\int_{-1}^1t^{2(n+k+1)}\mathrm dt&=4\sum_{k\geq 0}\sum_{n\geq 0}\frac{a^{2n+1}}{(2k+1)(2k+2n+3)}\\ &= (...)\\ &=2\sum_{i\geq 0}\sum_{j\geq 0}\frac{a^{2i+2j+1}}{(2i+1)(2j+1)}\\ &=\frac{1}{2a}\left(\ln\left(\frac{1+a}{1-a}\right)\right)^2\implies \mathbf{(A)=(D)}\ \text{is the solution} \end{aligned}
Though I kinda cheated in this last part, I'd need help on finding the steps in $$(...)$$ to convert the double summation I got into the other one since I don't really know how to do that...