Does this equality hold? I think it is connected to Hankel’s contour integral in which the direction of the contour is slightly deformed.
The exp(ib) means infinity in the direction of exp(ib) on the complex plane.
Asked
Active
Viewed 74 times
0
-
Sorry, it is not readable! – Z Ahmed Aug 15 '19 at 14:14
-
I proved it with Laplace transform. Thank you – Mike Park Aug 15 '19 at 16:25
-
$\Gamma(z)=\int_0^{\infty}e^{-u}u^{z-1}du$ provided that $Re(z) > 0$. I have no idea what you mean by introducing exp(ib). – Axion004 Aug 15 '19 at 20:20
-
The usual integration path is from 0 to infinity on the positive real axis. What I wanted to know was will the result not change if I take the integration path from 0 to any other complex infinity other than real positive infinity (e.g. imaginary infinity). And the answer was yes if b is an acute angle. – Mike Park Aug 16 '19 at 01:41
-
It is perhaps connected to the deformation of path in complex integral but I was not sure if it is valid to deform the contour when the integrand contains a branch cut (branch point). This is why I wanted to check this by direct computation. – Mike Park Aug 16 '19 at 01:44