Calculate $\int_{0}^{1}\frac{\arctan(x)}{x\sqrt{1-x^2}}dx$

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\int_{0}^{1}{\arctan\pars{x} \over x\root{1 - x^{2}}}\,\dd x:\ {\large ?}}$

Let's consider $\ds{\fermi\pars{\mu} \equiv \int_{0}^{1}{\arctan\pars{\mu x} \over x\root{1 - x^{2}}}\,\dd x\,,\quad\fermi\pars{0} = 0\,,\qquad\fermi\pars{1}={\large ?}}$:

\begin{align} \fermi'\pars{\mu}&= \int_{0}^{1}{\dd x \over \pars{\mu^{2}x^{2} + 1}\root{1 - x^{2}}} =\int_{0}^{\pi/2}{\dd\theta \over \mu^{2}\cos^{2}\pars{\theta} + 1} =\int_{0}^{\pi/2}{\sec^{2}\pars{\theta}\,\dd\theta\over \mu^{2} + 1 + \tan^{2}\pars{\theta}} \\[3mm]&=\int_{0}^{\infty}{\dd t \over t^{2} + 1 + \mu^{2}} ={1 \over \root{1 + \mu^{2}}}\int_{0}^{\infty}{\dd t \over t^{2} + 1} ={\pi \over 2}\,{1 \over \root{1 + \mu^{2}}} \end{align}

\begin{align} \fermi\pars{1}&={\pi \over 2}\ \overbrace{\int_{0}^{1}{\dd\mu \over \root{1 + \mu^{2}}}} ^{\ds{\mbox{Set}\ \mu \equiv \sinh\pars{\theta}}} ={\pi \over 2}\,{\rm arcsinh}\pars{1} \end{align}

$$\color{#66f}{\large% \int_{0}^{1}{\arctan\pars{x} \over x\root{1 - x^{2}}}\,\dd x ={\pi \over 2}\,{\rm arcsinh}\pars{1}} \approx 1.3845 $$

$$ \begin{align} \int_{0}^{1}\int_{0}^{1}\frac{1}{(1+x^2y^2)\sqrt{1-x^2}}dydx &=\int_{0}^{1}\int_{0}^{1}\frac{1}{(1+x^2y^2)\sqrt{1-x^2}}dxdy\\ &=\int_{0}^{1}\int_{0}^{\pi/2}\frac{1}{1+y^2 \sin^2 t}dtdy\\ &=\int_{0}^{1}\int_{0}^{\pi/2}\frac{1}{\cos^2 t+ (1+y^2)\sin^2 t}dtdy\\ &=\int_{0}^{1}\int_{0}^{\pi/2}\frac{1}{1 + (1+y^2)\tan^2 t}\frac{1}{\cos^2 t}dtdy\\ &=\int_{0}^{1}\int_{0}^{\infty}\frac{1}{1 + (1+y^2)s^2}dsdy\\ &=\int_{0}^{1}\frac{1}{1+y^2}\int_{0}^{\infty}\frac{1}{\left((1+y^2)^{-1/2}\right)^2+s^2}dsdy\\ &=\int_{0}^{1}\frac{1}{1+y^2}\sqrt{1+y^2}\arctan \left(s\sqrt{1+y^2}\right)\Biggl|_0^\infty dy\\ &=\int_{0}^{1}\frac{\pi}{2\sqrt{1+y^2}}dy\\ &=\frac{\pi}{2}\operatorname{arcsinh} y\Biggl|_0^1\\ &=\frac{\pi}{2}\operatorname{arcsinh} 1 \end{align} $$

The following is a different approach that doesn't use that hint.

$$ \begin{align} \int_{0}^{1} \frac{\arctan (x)}{x\sqrt{1-x^{2}}} \ dx &= \int_{0}^{1} \frac{1}{x \sqrt{1-x^{2}}} \sum_{n=0}^{\infty} \frac{(-1)^{n}}{2n+1} x^{2n+1} \ dx \tag{1}\\ &= \sum_{n=0}^{\infty} \frac{(-1)^{n}}{2n+1} \int_{0}^{1} \frac{x^{2n}}{\sqrt{1-x^{2}}} \ dx \\ &= \frac{1}{2} \sum_{n=0}^{\infty} \frac{(-1)^{n}}{2n+1} \int_{0}^{1} u^{(n+1/2)-1} (1-u)^{1/2-1} \ du \tag{2} \\ &= \frac{1}{2} \sum_{n=0}^{\infty} \frac{(-1)^{n}}{2n+1} B\left(n+\frac{1}{2}, \frac{1}{2} \right) \tag{3}\\ &=\frac{1}{2} \sum_{n=0}^{\infty} \frac{(-1)^{n}}{2n+1} \frac{\Gamma (n+\frac{1}{2}) \Gamma(\frac{1}{2})}{\Gamma (n+1)} \tag{4} \\ &= \frac{1}{2}\sum_{n=0}^{\infty} \frac{(-1)^{n}}{2n+1} \frac{\Gamma (\frac{1}{2})}{\Gamma (n+1)} \frac{\Gamma(2n) \Gamma (\frac{1}{2})}{\Gamma (n) 2^{2n-1}} \frac{2n}{2n} \tag{5} \\ &= \frac{\pi}{2} \sum_{n=0}^{\infty} \frac{(-1)^{n} \Gamma(2n+1)}{2^{2n} \ \Gamma^{2}(n+1)} \frac{1}{2n+1} \tag{6} \\ &= \frac{\pi}{2} \text{arcsinh}(1) \tag{7}\end{align}$$

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(1) Maclaurin expansion of arctan(x)

(2) let $u = x^{2}$

(3) integral representation of the beta function

(4) defintion of the beta function in terms of the gamma function

(5) gamma function duplication formula

(6) $\Gamma(z+1) = z \Gamma(z)$ and $\Gamma(\frac{1}{2}) = \sqrt{\pi}$

(7) Maclaurin expansion of arcsinh(x)

May I show another way?. It does not use the given hint though.

As Norbert done, make the sub $\displaystyle x=\sin(t), \;\ dx=\cos(t)dt$

$$\int_{0}^{\frac{\pi}{2}}\frac{\tan^{-1}(\sin(t))}{\sin(t)}dt$$

Now, use the series for arctan: $\displaystyle \tan^{-1}(x)=\sum_{n=0}^{\infty}\frac{(-1)^{n}x^{2n+1}}{2n+1}$

only, $x=\sin(t)$ and we get:

$$\int_{0}^{\frac{\pi}{2}}\sum_{n=0}^{\infty}\frac{\sin^{2n}(t)(-1)^{n}}{2n+1}dt$$

But, using the famous and known result: $\displaystyle \int_{0}^{\frac{\pi}{2}}\sin^{2n}(t)dt=\frac{\pi (2n)!}{2^{2n+1}(n!)^{2}}$, one obtains the series:

$$\frac{\pi}{2}\sum_{n=0}^{\infty}\frac{(-1)^{n} (2n)!}{(2n+1)2^{2n}(n!)^{2}}$$

Now, notice what that series is?. It is the Taylor series for $\displaystyle \sinh^{-1}(1)$

which is $$\sinh^{-1}(x)=\sum_{n=0}^{\infty}\frac{(-1)^{n}(2n)!x^{2n+1}}{2^{2n}(n!)^{2}(2n+1)}$$

so, we have $$\frac{\pi}{2}\sinh^{-1}(1)$$