Let $f$ be the function defined by $$ \begin{array}{ccc} f : [-1,1] & \longrightarrow & {\mathbb R}\\ \quad x &\longmapsto & \sqrt{1-x^2} \end{array} $$ I would like to calculate the convolution product $f*f$ in terms of the elliptic integrals $$K(x)=\int_0^1 \frac{dt}{\sqrt{(1-t^2)(1 - xt^2)}}$$ and $$E(x)=\int_0^1 \frac{\sqrt{1 - xt^2}}{\sqrt{(1-t^2)}}dt$$ More precisely I would like to prove :$$f*f(x)=\left|\frac{2x}{3}\left((x^2+4)E\left(1-\frac{4}{x^2}\right)-8K\left(1-\frac{4}{x^2}\right)\right)\right|.$$
Can someone help me to prove the above formula?
Well, the Fourier transform of $\sqrt{1-x^2}$ (supported on $[-1,1]$) is given by $\sqrt{\frac{\pi}{2}}\,\frac{J_1(s)}{s}$ with $J_1(s)$ being the Bessel function $$ J_1(s) = \sum_{n\geq 0}\frac{(-1)^n}{4^{n+1}n!(n+1)!} s^{2n+1} = \frac{1}{\pi}\int_{0}^{\pi}\cos(s\sin\theta-\theta)\,d\theta $$ so the problem boils down to finding the inverse Fourier transform of $J_1(s)^2/s^2$. By Vandermonde's identity the square of a Bessel function of the first kind also have a nice power series (see page 26 of my notes). In our case $$ J_1(s)^2/s^2 = \sum_{n\geq 0}\frac{(-1)^n}{4^{n+1} n!(n+2)!}\binom{2n+2}{n+1}s^{2n}\tag{1}$$ and the Laplace transform of this function is clearly related to $K$ and $E$ (the complete elliptic integrals) since (according to Mathematica's notation) $$ K(x) = \frac{\pi}{2}\sum_{n\geq 0}\left(\frac{\binom{2n}{n}}{4^n}\right)^2 x^n,\qquad E(x) = \frac{\pi}{2}\sum_{n\geq 0}\left(\frac{\binom{2n}{n}}{4^n}\right)^2 \frac{x^n}{1-2n} $$ while, due to $(1)$,
$$\mathcal{L}\left(J_1(s)^2/s^2\right)(x) = \sum_{n\geq 0}\frac{(-1)^n}{4^{n+1}n!(n+2)!}\binom{2n+2}{n+1}\frac{(2n)!}{x^{2n+1}}=\frac{x}{2}\left(-1+\phantom{}_2 F_1\left(-\tfrac{1}{2},\tfrac{1}{2};2;-\tfrac{4}{x^2}\right)\right). $$
In order to finish the job it is enough to exploit the relation between the inverse Fourier transform and the Laplace transform, which in our case turns out to be just a matter of introducing/removing alternating signs.