Calculating the integral expression $\int_0^{\infty}\lambda^{t-1}e^{-\lambda z}d\lambda$ for complex-valued z

Question

The problem is computing the integral expression $$f(t)=\int_0^{\infty}\lambda^{t-1}e^{-\lambda z}d\lambda$$ where z is a complex variable with $Re(z)> 0$ and t is a real variable. Is it correct to substitute $w=\lambda z$ in the integral and write down $f(t)$ as $$f(t)=z^{-t}\int_0^{\infty}w^{t-1}e^{-w}dw=z^{-t}\Gamma(t)$$ where w is a complex variable?

Note that while in the first integral $\lambda$ is a real variable which approaches positive infinity of real numbers, in the second integral $w$ is a complex variable and the integral is only meaningful when its value is independent from how $w$ approaches the infinity point of the extended complex plane.

Answer 1

Consider the finite integral from $0$ to $\Lambda$ instead. When you do the substitution $w = \lambda z$ with $dw = z d\lambda$ the limits of integration become $0$ and $\Lambda/z$, so in fact your new integral actually has a very specific contour - the half-line from the origin to infinity along the complex number $1/z$.

Answer 2

If you can persuade yourself that the integral exists, and that it has to be analytic in $z$ for $\Re(z) > 0$, then you can argue that you have two analytic functions that agree on the positive real axis. Hence they must agree on $\Re(z) > 0$.