In Abramowitz/Stegun you read that
$$Q_s(x) := \int_{0}^\infty ( x + \sqrt{x^2-1} \cosh t )^{-s} dt, \quad x > 1, \Re(s)>0$$
is a solution to the differential equation $$(1-x^2)y''-2xy'+s(s-1)y=0.$$
How do you prove that? I differentiated under the integral sign and wrote down $(1-x^2)y''-2xy'+s(s-1)y$ with $Q_s$ for $y$ but couldn't see how it vanishs. Do you need to use some fancy functional equations of $\cosh$ to obtain the result?
Incomplete answer:
Note that your Dif-EQ $$(1-z^2)\,y'' -2zy' + s(s+1)y = 0 \tag{1}$$ Is a special case of Legendre Equation, which we can solve by transforming it into a sub-case of the Hypergometric Dif-Eq, $$z(1-z)y'' + \left[c-(a+b+1)z \right] y' - ab\,y = 0$$ which can be solve completely via the Hypergeometric function $_2F_1$, yielding the following solution to $(1)$: $$y(z)=c_1P_s(z)+c_2Q_s(z)$$ Concerning ourselves with the function $P_s(z)$, (which is simpler to express but very similar) we get the definition $$P_s(z) = {_2}F_1\left(-s,s+1;1;\frac{1-z}{z}\right)$$ Now you will need to apply the right integral identity. As I mentioned in a deleted comment, one can apply the Barnes Integral formula here with a lot of brute force, but I think there should be a cleaner method.