I need to understand how to establish two identities. The first is
$$ \int_{C} z^{-1-q}(1-z)^{-1-\lambda } dz=\frac{2 \pi \Gamma (q+\lambda +1)}{\Gamma (\lambda +1) \Gamma (q+1)}, q\geq 0, \lambda >0$$ where C is a contour $z=1/2+ i t, t \in (-\infty,\infty) $
Mathematica believes this
With[{z = l + I t},
Integrate [
z^(-q - 1) (1 - z)^(-λ - 1), {t, -Infinity, Infinity},
Assumptions -> {q > 0, λ > 0, 0 < l < 1}]]
When q=0 (RHS is just 2 \pi), this comes from the pole at 0, by using a half circle contour which surrounds it, and blowing it to $\infty$. With q>0, my first idea was to use a rectangle excluding the branch point at 0, but this cannot work since the integrand explodes there
The second identity is a contour integral representation of the HypergeometricU function with powers integrand (unlike the ones given at http://functions.wolfram.com/HypergeometricFunctions/HypergeometricU/07/02/ )
$$ \int_{C} e^{-x z} z^{-1-q}(1-z)^{-\lambda } dz= \frac{2 \pi\ \ e^{-x} x^{q+\lambda} \ U(q+1 , q+\lambda+1,x)}{\Gamma(\lambda ) }, q\geq 0, \lambda >0, x>0$$
Mathematica's NIntegrate
confirms this
cnS = {q -> 5/Pi, \[Lambda] -> 4 Pi, l -> 1/3};
int[x_] := Exp[ -x z] z^(-q - 1) (1 - z)^(-\[Lambda]) /. cnS;
RBr[x_] :=
Chop[NIntegrate [
int[x] /. z -> I t + l /. cnS, {t, -Infinity, Infinity}], 10^(-9)];
R[z_] := 2 Pi Exp[- z] z^
n HypergeometricU[q + 1, q + \[Lambda] + 1, z]/Gamma[\[Lambda]];
Print["R Bromwich=", RBr[1], " R exact= ", Chop[R[1] //. cnS // N]]
but the hard question for me is how to prove this by providing the right integration contour.
I also had some intreaguing problems when $\lambda$ is not an integer, but they disappeared on my minimal example above:) I am still wondering though whether NIntegrate
could give a wrong answer in cases when a parameter switches from integer to noninteger.