<s>[INST] Calculate $\int_{0}^{1}\frac{\arctan(x)}{x\sqrt{1-x^2}}dx$ I am preparing for a calculus exam and I was asked to calculate $$\int_{0}^{1}\frac{\arctan(x)}{x\sqrt{1-x^2}}dx$$
Using the hint that $$\frac{\arctan(x)}{x}=\int_{0}^{1}\frac{dy}{1+x^2y^2}$$
I ran into some trouble and would appreciate help.
What I did:
I used the hint, $$\int_{0}^{1}\frac{\arctan(x)}{x\sqrt{1-x^2}}dx=\int_{0}^{1}\int_{0}^{1}\frac{1}{(1+x^2y^2)\sqrt{1-x^2}}dydx$$
Since $x,y$ move between $0,1$ I thought maybe it is best to used the transform $x=r\cos\theta$ , $y=r\sin\theta$. $r \in [0,1]$, $\theta \in [0, \frac{\pi}{2}]$
But I didn't get anything meaningful, I didn't end up an something that is easy / possible to integrate. And I honestly can't think of a way to integrate this as it is.
 [/INST] The following is a different approach that doesn't use that hint.
$$ \begin{align} \int_{0}^{1} \frac{\arctan (x)}{x\sqrt{1-x^{2}}} \ dx &= \int_{0}^{1} \frac{1}{x \sqrt{1-x^{2}}} \sum_{n=0}^{\infty}  \frac{(-1)^{n}}{2n+1} x^{2n+1} \ dx \tag{1}\\ &= \sum_{n=0}^{\infty} \frac{(-1)^{n}}{2n+1} \int_{0}^{1} \frac{x^{2n}}{\sqrt{1-x^{2}}} \ dx  \\ &= \frac{1}{2} \sum_{n=0}^{\infty} \frac{(-1)^{n}}{2n+1} \int_{0}^{1} u^{(n+1/2)-1} (1-u)^{1/2-1} \ du \tag{2} \\ &= \frac{1}{2} \sum_{n=0}^{\infty} \frac{(-1)^{n}}{2n+1} B\left(n+\frac{1}{2}, \frac{1}{2} \right) \tag{3}\\ &=\frac{1}{2} \sum_{n=0}^{\infty} \frac{(-1)^{n}}{2n+1} \frac{\Gamma (n+\frac{1}{2}) \Gamma(\frac{1}{2})}{\Gamma (n+1)} \tag{4} \\ &= \frac{1}{2}\sum_{n=0}^{\infty} \frac{(-1)^{n}}{2n+1} \frac{\Gamma (\frac{1}{2})}{\Gamma (n+1)} \frac{\Gamma(2n) \Gamma (\frac{1}{2})}{\Gamma (n) 2^{2n-1}} \frac{2n}{2n}  \tag{5} \\ &= \frac{\pi}{2} \sum_{n=0}^{\infty} \frac{(-1)^{n} \Gamma(2n+1)}{2^{2n} \ \Gamma^{2}(n+1)} \frac{1}{2n+1} \tag{6} \\ &= \frac{\pi}{2} \text{arcsinh}(1) \tag{7}\end{align}$$
$ $
(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)
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