Evaluate the integral $\int_0^\infty x^{t-1}e^{-\beta x}dx$ I want to evaluate the following integral $$\int_0^\infty x^{t-1}e^{-\beta x}dx$$
where $\beta$ is a complex number.
Now, if $\beta$ was real, we could just set $y = \beta x$ and we will reduce to the Gamma function. Since $\beta$ is complex, though, when I set $y = \beta x$, I am integrating over the line with $\arg \beta$ on the complex plane, so I can't reduce directly to the Gamma function, can I?
I have found after some calculations that $$\int_0^\infty x^{t-1}e^{-\beta x}dx = \Gamma(t)\beta^{-t}$$ which is exactly what one would find if it didn't bother with the previous observation.
So my questions are:  Is my observation on the complex line correct? and 2) What is the best way to prove the result? 
My work
Write $\beta = a + ib$. Consider the integral as a function of $t,a,b$ to get $$I(t,a,b) = \int_0^\infty x^{t-1}e^{-a x}e^{-ibx}dx$$.
Notice that $$\frac{\partial I}{\partial a}(t,a,b) = -I(t+1,a,b)$$ and
$$\frac{\partial I}{\partial b}(t,a,b) = -iI(t+1,a,b)$$
Now since $\displaystyle I(t+1) = \frac t{a+ib}I(t)$, the previous two equations become $$\frac{\partial I}{\partial a} = -\frac t{a+ib}I$$ and
$$\frac{\partial I}{\partial b} = -\frac{it}{a+ib}I$$
which put together yield $I(t,a,b) = C(t) (a+ib)^{-t}$.
Also, since $\displaystyle I(t,1,0) = C(t) = \int_0^\infty x^{t-1}e^{-x}dx= \Gamma(t)$, we get 
$$I(t,a,b) = \Gamma(t) (a+ib)^{-t} = \Gamma(t)\beta^{-t}$$
which seems like too much work!
 A: You're correct in the result.  To verify this and to provide an instructive way forward, we present here a direct approach the relies on Cauchy's Integral Theorem.  
To that end, we let $I$ be the integral in the complex plane given by
$$I=\oint_C z^{t-1}e^{-\beta z}\,dz$$
where $\text{Re}(\beta)>0$, $t>0$, and $C=C_1+C_2+C_3$ is the contour comprised of the $4$ components
$(i)$ $C_1$ is the real-line segment from $(\epsilon,0)$ to $(R,0)$;
$(ii)$ $C_2$ is the circular arc centered at the origin with radius $R$ from $(R,0)$ to $(R\cos \theta_0,R\sin \theta_0)$ where $\beta e^{i\theta_0}=1 \implies e^{i\theta_0}=\frac{1}{\beta}$.
$(iii)$ $C_3$ is the line segment from $(R\cos \theta_0,R\sin \theta_0)$ to $(\epsilon \cos \theta,\epsilon \sin \theta)$;
$(iv)$ $C_4$ is the circular arc centered at the origin with radius $\epsilon$ from $(\epsilon \cos \theta,\epsilon \sin \theta)$ to $(\epsilon,0)$.
Since $z^{x-1}e^{-\beta z}$ is analytic in $C$, then from Cauchy's Integral Theorem, we have $I=0$.  Moreover, the contributions from $C_2$ and $C_4$ can easily be shown to vanish as $R\to \infty$ and $\epsilon \to 0$, respectively.  
Therefore, we have
$$\begin{align}
I&=\oint_C z^{t-1}e^{-\beta z}\,dz\\\\
&=\int_0^{\infty}(x)^{t-1}e^{-\beta x}\,dx+\int_\infty^0 (e^{i\theta_0}x)^{t-1}e^{-\beta e^{i\theta_0}x}e^{i\theta_0}dx=0\\\\
\implies \int_0^{\infty}(x)^{t-1}e^{-\beta x}\,dx&=\frac{1}{\beta^{t}}\int_0^\infty x^{t-1}e^{-x}\,dx\\\\
&=\frac{1}{\beta^{t}}\Gamma(t)
\end{align}$$
which was to be shown!
A: $$ f(\beta) = \int_0^\infty x^{t-1}e^{-\beta x}dx$$
is clearly an analytic function in $\beta$ if $\text{Re}(\beta) > 0$ (differentiate under the integral sign).
$$ g(\beta) = \Gamma(t) \beta^{-t} $$
is also analytic in $\beta$.  Since $f(\beta) = g(\beta)$ for all $\beta \in (0,\infty)$, it follows by the principle of isolated zeros that they must be equal on the intersections of their domains of definition.
