Let us define an auxiliary functions: $$f_i(x)=x^{c_i-1}\Phi^{(2)}_2(b_i,b'_i;c_i;v_i x,w_i x)$$ with $\Phi^{(2)}_2(\cdot)$ - standard Humbert function of two variables and $i=1,2$.
The question is: Is it possible to compute the following integral $$\int_{\alpha}^{\infty}f_1(x)f_2(x-\alpha){\rm d}x.$$ At a first glance it looks like a convolution integral. I can find the Laplace transform of the functions $f_i(x)$ but will this approach be correct since it only "looks like" but not a convolution. The series representation of Humbert functions does not give any solution since the integral blows up.
I know that the integral is finite, since it represents some combination of probability distributions and the numeric solution can be obtained without any problem. What I am seeking is the closed-form solution.
Addendum: the parameter $x\geq 0$, but $v_i, w_i$ can be negative.
Not an answer, but if you want your cross-correlation integral to look like a convolution, define:
$$g_2(x) = f_2(-x)H(-x) $$
where $H(x)$ is the Heaviside unit step function.
Then
$$\begin{align*} I(\alpha) &= \int_\alpha^\infty f_1(x)f_2(x-\alpha) dx\\ \\ &= \int_{-\infty}^\infty f_1(x)f_2(x-\alpha)H(x-\alpha) dx\\ \\ &= \int_{-\infty}^\infty f_1(x)g_2(\alpha-x) dx\\ \\ &= (f_1*g_2)(\alpha) \\ \end{align*}$$
is your integral expressed as a convolution.