How to show
$$\int_0^1 \frac{x^2}{\sqrt{x^4+1}} \, dx=\frac{\sqrt{2}}{2}-\frac{\pi ^{3/2}}{\Gamma \left(\frac{1}{4}\right)^2}$$
I tried hypergeometric expansion, yielding $\, _2F_1\left(\frac{1}{2},\frac{3}{4};\frac{7}{4};-1\right)$. Can this be evaluated analytically? Any help will be appreciated.
Prove that $\int_0^1 \frac{x^2}{\sqrt{x^4+1}} \, dx=\frac{\sqrt{2}}{2}-\frac{\pi ^{3/2}}{\Gamma \left(\frac{1}{4}\right)^2}$
closed-formdefinite integralsgamma functionintegration
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Best Answer
An alternative approach. By termwise integration of a Maclaurin series
$$ I=\int_{0}^{1}\frac{x^2}{\sqrt{1+x^4}}\,dx = \sum_{n\geq 0}\left[\frac{1}{4^n}\binom{2n}{n}\right]\frac{(-1)^n}{4n+3} $$ and due to the fact that $P_{2n}(0)=\left[\frac{1}{4^n}\binom{2n}{n}\right](-1)^n$, the RHS of the previous line is the value at $x=\frac{1}{2}$ of the following function: $$ f(x) = \sum_{n\geq 0}\frac{1}{2n+3} P_{n}(2x-1). $$ Now we may exploit three well known Fourier-Legendre expansions: $$ \sqrt{1-x} = \sum_{n\geq 0}\frac{2}{(1-2n)(2n+3)}P_n(2x-1), $$ $$ K(x) = \sum_{n\geq 0}\frac{2}{2n+1}P_n(2x-1), $$ $$ E(x) = \sum_{n\geq 0}\frac{4}{(1-2n)(2n+1)(2n+3)}P_n(2x-1) $$ (the argument $x$ stands for the elliptic modulus, like in Mathematica's notation) and deduce from partial fraction decomposition that
$$ I = \left.\frac{1}{2}K(x)-E(x)+\sqrt{1-x}\right|_{x=1/2}. $$ $K\left(\frac{1}{2}\right)$ and $E\left(\frac{1}{2}\right)$ are well known to be related to $\Gamma\left(\frac{1}{4}\right)^2$, and related to each other via Legendre's identity.