# Proving a hypergeometric function identity

While playing around with the fractional calculus, I got stuck trying to show that two different ways of differintegrating the cosine give the same result. DLMF and the Wolfram Functions site don't seem to have this "identity" or something that can obviously be transformed into what I have, so I'm asking here.

The "identity" in question is

$(\alpha-1)\left({}_1 F_2 \left(1;\frac{1-\alpha}{2},\frac{2-\alpha}{2};-\frac{x^2}{4}\right)-{}_1 F_2 \left(-\frac{\alpha}{2};\frac12,\frac{2-\alpha}{2};-\frac{x^2}{4}\right)\cos(x)\right)\stackrel{?}{=}\alpha x \sin(x)\,{{}_1 F_2 \left(\frac{1-\alpha}{2};\frac32,\frac{3-\alpha}{2};-\frac{x^2}{4}\right)}$

Expanding the LHS minus the RHS in a Taylor series shows that the coefficients up to the 50th power are 0; trying out random complex values of $\alpha$ and $x$ seems to verify the identity. I would however like to see a way to confirm the identity analytically. How do I go about it?

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The usual approach to such identities is to find a linear differential equation satisfied by both sides. Here you might get away without finding one explicitly; if you know one of a certain form exists and both sides are the same modulo a high power of $x$, possibly that would be enough to ensure that both sides are the same. – Robin Chapman Sep 5 '10 at 12:55

Awesome stuff, thanks a lot! This takes care of my particular problem, though I still can't help but feel there's a more general addition theorem for ${}_1 F_2$ lurking behind the scenes, of which what I have is a mere special case. I wonder how one might find this? – J. M. Sep 6 '10 at 1:26