# Closed form of Integral of ellipticK and log using Mellin transform? $\int_{0}^4 K(1-u^2) \log[1+u z] \frac{du}{u}$

I am trying to evaluate the integral: $\mathcal{I}(z)=\int_{0}^a K(1-u^2) \log[1+u z] \frac{du}{u}$, $\qquad$ ($a$ fixed, $a>0$ and $K$ is the complete elliptic integral of the first kind)

in closed form, and I believe that it can be done using Mellin Transform, but I have been unsuccessful so far... any help would be appreciated!

First, I rewrite my integral as:

$\mathcal{I}(z)=\int_{0}^{\infty} \Theta(a-u) K(1-u^2) \log[1+u z] \frac{du}{u}$

to make the integration regime agree with the Mellin transform. I have then tried several lines of attack (see below). Ideas/help on how I might be able to proceed are very welcome :)

1) View the integral as a Mellin Convolution, and try to evaluate it directly for example with Mathematica's MellinConvolve[] (the precise code would be, where y=1/z):

MellinConvolve[HeavisideTheta[4 - u] EllipticK[1 - u^2], Log[1 + 1/u], u, y]

2) Use the convolution property of the Mellin transform to find the Mellin transform of the integral, and the try to transform back:

$\mathcal{M}\left(\mathcal{I}(z)\right)[s]= \mathcal{M}\left(\frac{1}{u}\Theta(a-u) K(1-u^2) \right)[1-s] \mathcal{M}\left(\log(1+u)\right)[s]$

This I can find in a straight-forward way using mathematica to get: $\mathcal{M}\left(\frac{1}{u}\Theta(a-u) K(1-u^2) \right)[s] =\frac{\left(\frac{1}{a}\right)^{1-s}}{4 \pi } G_{3,3}^{2,3}\left(a^2| \begin{array}{c} \frac{1}{2},\frac{1}{2},\frac{3-s}{2} \\ 0,0,\frac{1-s}{2} \\ \end{array} \right)$

and

$\mathcal{M}(\log(1+u))[s]= \frac{\pi \csc (\pi s)}{s}$

but I can't get the inverse transform...

3) View the integral of a special case (s=0) of:

$\int_{0}^{\infty} \Theta(a-u) K(1-u^2) \log[1+u z] u^{s-1} du$

so it would be given by the Mellin Transform of $\Theta(a-u) K(1-u^2) Log[1+u z]$, but I cannot seem to find this (I don't get it to work if I rewrite them in terms of MeijersG-functions either, but maybe this can help).