A Sum that came up while solving a integral

Question

While evaluating $I$, I did the following-

$$\begin{align}I= \int_{0}^{1} \log \left(\dfrac{1+x}{1-x}\right) \dfrac{1}{x\sqrt{1-x^2}} \ \mathrm{d}x &= 2 \int_{0}^{1}\sum_{n=0}^{\infty} \dfrac{x^{2n+1}}{2n+1} \dfrac{1}{x\sqrt{1-x^2}} \ \mathrm{d}x\\ &=2\sum_{n=0}^{\infty} \int_{0}^{1} \dfrac{x^{2n}}{(2n+1)\sqrt{1-x^2}} \ \mathrm{d}x \end{align}$$ Then I used the substitution $x \mapsto \sin \theta $.

$$\begin{align} \therefore I &=2\sum_{n=0}^{\infty} \int_{0}^{\pi/2} \dfrac{\sin^{2n} {\theta}}{2n+1} \ \mathrm{d}\theta\\ &=\pi \sum_{n=0}^{\infty} \dfrac{(2n)!}{2^{2n}(n!)^2(2n+1)} \end{align}$$ The last step is due to Wallis' formula.

However, I couldn't solve the last series. My question is that how do we prove that $$\displaystyle\sum_{n=0}^{\infty} \dfrac{(2n)!}{2^{2n}(n!)^2(2n+1)}=\dfrac{\pi}{2} \ ?$$

Answer 1

Hint

For $|x| \leq 1$, $$\displaystyle\sum_{n=0}^{\infty} \dfrac{(2n)!}{2^{2n}(n!)^2(2n+1)}x^{2n}=\frac 1x\sum_{n=0}^{\infty} \dfrac{(2n)!}{2^{2n}(n!)^2(2n+1)}x^{2n+1}=\frac{\sin ^{-1}(x)}{x}$$

Answer 2

Instead of series expansion, you can do this:

$$\int_0^1 \ln\left(\frac{1+x}{1-x}\right)\frac{1}{x\sqrt{1-x^2}}\,dx=\frac{1}{2}\int_{-1}^1 \ln\left(\frac{1+x}{1-x}\right)\frac{1}{x\sqrt{1-x^2}}\,dx$$

$$=\int_{-1}^1 \frac{\ln(1+x)}{x\sqrt{1-x^2}}\,dx=\int_{-\pi/2}^{\pi/2} \frac{\ln(1+\sin\theta)}{\sin\theta}\,d\theta$$

Consider $$I(a)=\int_{-\pi/2}^{\pi/2} \frac{\ln(1+a\sin\theta)}{\sin\theta}\,d\theta \Rightarrow I'(a)=\int_{-\pi/2}^{\pi/2} \frac{1}{1+a\sin\theta}\,d\theta=\frac{\pi}{\sqrt{1-a^2}}$$

$$\Rightarrow I(a)=\sin^{-1}a+C$$

Since $I(0)=0$, we have $C=0$ and hence, $$\boxed{I(1)=\dfrac{\pi}{2}}$$

Answer 3

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$\sf\mbox{If you evaluates the sum you'll return to the original integral.}$

$\sf\mbox{In any case the evaluation requires to consider the original integral.}$

So, \begin{align}&\color{#66f}{\large% {1 \over \pi}\int_{0}^{1}\log\pars{1 + x \over 1 - x}\,{1 \over x\root{1 - x^{2}}}\,\dd x} ={2 \over \pi}\ \overbrace{\int_{0}^{1}\,{\rm arctanh}\pars{x}\,{1 \over x\root{1 - x^{2}}}\,\dd x} ^{\dsc{x}\ \equiv\ \dsc{\tanh\pars{t}}} \\[5mm]&={2 \over \pi}\int_{0}^{\infty}t\,{1 \over \tanh\pars{t}\root{1 - \tanh^{2}\pars{t}}}\,\sech^{2}\pars{t}\,\dd t ={2 \over \pi}\int_{0}^{\infty}{t \over \sinh\pars{t}}\,\dd t \\[5mm]&={4 \over \pi}\int_{0}^{\infty}{t\expo{-t} \over 1 - \expo{-2t}}\,\dd t ={4 \over \pi}\sum_{n\ =\ 0}^{\infty}\int_{0}^{\infty}t\expo{-\pars{2n + 1}t}\,\dd t ={4 \over \pi}\sum_{n\ =\ 0}^{\infty}{1 \over \pars{2n + 1}^{2}}\ \overbrace{\int_{0}^{\infty}t\expo{-t}\,\dd t}^{\ds{=}\ \dsc{1}} \\[5mm]&={4 \over \pi}\bracks{\sum_{n\ =\ 1}^{\infty}{1 \over n^{2}} -\sum_{n\ =\ 1}^{\infty}{1 \over \pars{2n}^{2}}} ={4 \over \pi}\pars{{3 \over 4}\ \overbrace{\sum_{n\ =\ 1}^{\infty}{1 \over n^{2}}}^{\ds{=}\ \dsc{\pi^{2} \over 6}}} =\color{#66f}{\Large{\pi \over 2}} \end{align}