Prove $\int_{0}^{1}{{2x+\ln{1-x\over 1+x}\over x^3(1-x^2)^{1/2}}}dx=-(\pi/2)^2$ and $\int_{0}^{1}{2x+\ln{1-x\over 1+x}\over x^2(1-x^2)^{1/2}}dx=-2$

Question

Motivated by this Question

Two similar integrals, but exhibit completely different closed forms

$$\int_{0}^{1}{{2x+\ln\left({1-x\over 1+x}\right)\over x^3\sqrt{1-x^2}}}\mathrm dx=-\color{blue}{\left({\pi\over 2}\right)^2}\tag1$$

and

$$\int_{0}^{1}{{2x+\ln\left({1-x\over 1+x}\right)\over x^2\sqrt{1-x^2}}}\mathrm dx=-\color{red}{2}\tag2$$

Here is my try:

I don't think substitution work here initially, but anyway I try

$u={1-x\over 1+x}\implies -{(1+x)^2\over 2}du=dx$, then $(1)$ becomes after a lengthy simplication to

$$\sqrt{2}\int_{0}^{1}{2\left({1-u\over 1+u}\right)+\ln u \over \sqrt{u}(1-u)^3}\cdot(1+u)^{11/2}\mathrm du\tag3$$

Or we split $(1)$ into

$$2\int_{0}^{1}{\mathrm dx\over x^2\sqrt{1-x^2}}+\int_{0}^{1}{\ln(1-x)\over x^3\sqrt{1-x^2}}\mathrm dx-\int_{0}^{1}{\ln(1+x)\over x^3\sqrt{1-x^2}}\mathrm dx=I_1+I_2-I_3\tag4$$

$I_1$ diverges.

How can we prove integrals $(1)$ and $(2)$?

Answer 1

HINT: prove that the antiderivative for your second indefinite integral is given by $$-\frac{\sqrt{1-x^2}\ln\left(\frac{1-x}{1+x}\right)}{x}+C$$

Answer 2

Note $$ \frac{1}{(1-x)^3}=\frac12\sum_{n=1}^\infty n(n+1)x^{n-1}, \int_0^1x^n\ln xdx=-\frac{1}{(n+1)^2}, \sum_{n=1}^\infty\frac1{(2n-1)^2}=\frac{\pi^2}{8}. $$ Letting $x=\frac{1-u}{1+u}$, one has \begin{eqnarray} I&=&\int_{0}^{1}{{2x+\ln{1-x\over 1+x}\over x^3(1-x^2)^{1/2}}}dx\\ &=&-\int_0^1\frac{(1+u)[2-u+(1+u)\ln u]}{\sqrt u(1-u)^3}du\\ &=&-\frac12\int_0^1\sum_{n=1}^\infty n(n+1)(1+u)[2-u+(1+u)\ln u]u^{n-\frac32}du\\ &=&-\frac12\sum_{n=1}^\infty n(n+1)\int_0^1(1+u)[2-u+(1+u)\ln u]u^{n-\frac32}du\\ &=&-\frac12\sum_{n=1}^\infty \bigg[\frac{3}{(2n-1)^2}+\frac{3}{(2n+3)^2}+\frac{1}{2n-1}-\frac{1}{2n+1}\bigg]\\ &=&-\frac{\pi^2}{4}. \end{eqnarray}

Answer 3

Hint: try the substitution with the hyperbolic tangent: $x=tanh(t)=\frac{e^{t}-e^{-t}}{e^{t}+e^{-t}}$
(the logarithmic part looks like the inverse of $tanh$ and the square root in the denominator looks like it might get simplified using some identity of hyperbolic trigonometric functions.)

Answer 4

Here's a possible path to #1 with some arguably important details left out. So I'm making this community wiki in case someone else wants to come along and fill in those details or just completely fix this.

Recall: $$ \ln(1+x) = \sum_{n=1}^{+\infty} \frac{(-1)^{n+1}}n x^n$$ So then: $$ \ln(1-x) = \sum_{n=1}^{+\infty} \frac{(-1)^{n+1}}n (-x)^n = \sum_{n=1}^{+\infty} \frac{-1}n x^n$$ And: \begin{align*} 2x + \ln\left(\frac{1-x}{1+x}\right) &= 2x + \ln(1-x) - \ln(1+x)\\[0.3cm] &= 2x + \sum_{n=1}^{+\infty} \frac{-1}n x^n - \sum_{n=1}^{+\infty} \frac{(-1)^{n+1}}n x^n\\[0.3cm] &= 2x + \left(-x -\frac12x^2 -\frac13x^3 -\cdots\right) - \left(x - \frac12 x^2 + \frac13 x^3 - \cdots\right)\\[0.3cm] &= 2x -2 \left(x + \frac13x^3 + \frac15x^5 + \cdots\right)\\[0.3cm] &= -2\sum_{n=1}^{+\infty} \frac{x^{2n+1}}{2n+1} \end{align*}

Then (sweeping some convergence details under the rug here) we have: \begin{align*} \int_0^1 \frac{2x + \ln\left(\frac{1-x}{1+x}\right)}{x^3\sqrt{1-x^2}} \, dx &= \int_0^1 \frac{\displaystyle -2\sum_{n=1}^{+\infty} \frac{x^{2n+1}}{2n+1}}{x^3\sqrt{1-x^2}} \, dx\\[0.3cm] &= -2\sum_{n=1}^{+\infty} \frac1{2n+1} \int_0^1 \frac{x^{2n-2}}{\sqrt{1-x^2}} \, dx \end{align*}

Let $x = \sin \theta$. Then $$ \int_0^1 \frac{x^{2n-2}}{\sqrt{1-x^2}} \, dx = \int_0^{\pi/2} \sin^{2n-2} \theta \, d\theta. $$

Wolfram Alpha says: $$\int_0^{\pi/2} \sin^{2n-2} \theta \, d\theta = \frac{\sqrt\pi \ \Gamma\left(n-\frac12\right)}{2\Gamma(n)}$$

I suspect this can be proved by induction but I didn't quite get there. Anyway, now we have:

$$ \int_0^1 \frac{2x + \ln\left(\frac{1-x}{1+x}\right)}{x^3\sqrt{1-x^2}} \, dx = -2\sum_{n=1}^{+\infty} \frac{\sqrt\pi \ \Gamma\left(n-\frac12\right)}{2(2n+1) \Gamma(n)}$$

Not quite sure how to evaluate that summation or if it's even possible. But it does give the correct value.