# Reference for summaton formula

1. May 27, 2010

### Pere Callahan

Hi all, I am desperately looking for a reference for a summation formula, which I have obtained with Mathematica.
$$\sum_{k=1}^\infty{\frac{(-z)^k}{[(k+1)!]^2}H_{k+2}}=\frac{1}{2z^{3/2}}\left[\sqrt{z}\left[2-3z+\pi\operatorname{Y}_0\left(2\sqrt{z}\right)\right]-2\operatorname{J}_1\left(2\sqrt{z}\right)-\sqrt{z}\operatorname{J}_0\left(2\sqrt{z}\right)\left[2\gamma+\log z\right]\right]$$

where $H_k=\sum_{n=1}^k{1/n}$ is the k-th harmonic number, J and Y are Bessel functions and $\gamma$ is the Euler-Mascheroni constant. I couldn't find anything resembling the formula in any of the standard books. Of course, any hints as to how to prove the relation from scratch are also highly appreciated.

Thank you,
Pere

Last edited: May 27, 2010
2. May 27, 2010

### Pere Callahan

I now managed to show that the formula is equivalent to showing that
$$\int_0^1 \frac{J_0\left(2 \sqrt{z}\right)-x J_0\left(2 \sqrt{x} \sqrt{z}\right)}{(x-1) z} \, dx$$
$$=\frac{\sqrt{z} \left(\pi Y_0\left(2 \sqrt{z}\right)+(\log (z)+2 \gamma ) J_2\left(2 \sqrt{z}\right)\right)-(\log (z)+2 \gamma +2) J_1\left(2 \sqrt{z}\right)}{2 z^{3/2}}.$$

Maybe any hints on how to do this integral?

Thanks,
Pere