While studying some dark matter related stuff, I came across to the following interesting identities: $$\int\limits_0^\infty\sqrt{\frac{y}{xp}}\,e^{-y}\left(K(p)-E(p)\right)dy= \frac{\pi x}{4} \left[I_0\left(\frac{x}{2}\right)K_0\left(\frac{x}{2}\right)-I_1\left(\frac{x}{2}\right)K_1\left(\frac{x}{2}\right)\right],$$ and $$\int\limits_0^\infty\sqrt{\frac{y}{xp}}\,\frac{K(p)-E(p)}{y^2}\,dy= \frac{\pi}{2x}.\tag{1}$$ Here $K$ and $E$ are complete elliptic integrals of the first and second kind, $I_0$,$I_1$ are modified Bessel functions of the first kind, $K_0$,$K_1$ are modified Bessel functions of the second kind, and $$p=\frac{x^2+y^2-\sqrt{(x^2-y^2)^2}}{2xy}=\left\{\begin{array}{c}\frac{y}{x},\,\,\mathrm{if}\,\,x\ge y, \\ \frac{x}{y},\,\,\mathrm{if}\,\,x\lt y. \end{array}\right . $$ are these identities known? I got them only indirectly by calculating a gravitational field strengths by two different methods. Is there any simple method to prove them?

Using $$\frac{y}{x}\left[\frac{K(m)}{x+y}+\frac{E(m)}{x-y}\right]=2\,\frac{\partial}{\partial y}\left [\sqrt{\frac{y}{xp}}\,\left(K(p)-E(p)\right)\right],$$ where $$m=\frac{2\sqrt{xy}}{x+y},$$ these integrals can be rewritten as follows $$\int\limits_0^\infty e^{-y}y\left[\frac{K(m)}{x+y}+\frac{E(m)}{x-y}\right]dy=\frac{\pi x^2}{2}\left[I_0\left(\frac{x}{2}\right)K_0\left(\frac{x}{2}\right)-I_1\left(\frac{x}{2}\right)K_1\left(\frac{x}{2}\right)\right],$$ and $$\int\limits_0^\infty \left[\frac{K(m)}{x+y}+\frac{E(m)}{x-y}\right]dy=\pi.$$ Eq. (1) can be rewritten in the (perhaps more interesting) form: $$\int\limits_0^1\frac{k+1}{k^2}\left[K(k)-E(k)\right ]dk=\frac{\pi}{2}.$$

  • $\begingroup$ Since the integrand and the right hand side are D-finite (holonomic) functions, I suspect these identities can be proved more or less algorithmically using creative telescoping. $\endgroup$ – Fredrik Johansson Nov 22 '16 at 7:54

I have not found the exact reference for your integral identity but you may like to check this out in the meantime.

Also, explore this table of integrals.

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