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?
Best Answer
You can use the great HolonomicFunctions package by Christoph Koutschan to prove this identity in Mathematica. It automatically proves for you that both sides of your identity satisfy the sixth order differential equation \begin{eqnarray} 0=&&x^2 \left(2 a^2-11 a+18 x^2+14\right) D_x^6 -2 x \left(2 a^3-19 a^2+18 a x^2+58 a-54 x^2-56\right) D_x^5 \\\\ &&+\left(2 a^4-25 a^3+28 a^2 x^2+115 a^2-133 a x^2-230 a+90 x^4+154 x^2+168\right) D_x^4 \\\\ &&-4 x \left(4 a^3-37 a^2+36 a x^2+115 a-99 x^2-114\right) D_x^3 \\\\ &&+4 \left(2 a^4-23 a^3+20 a^2 x^2+96 a^2-71 a x^2-172 a+18 x^4+71 x^2+112\right) D_x^2 \\\\ &&+8 x \left(4 a^2-34 a+36 x^2+43\right) D_x +8 \left(2 a^2-17 a+18 x^2+35\right). \end{eqnarray} Together with your check that the first six Taylor coefficients (with respect to x) agree, this proves your identity.