Laplace Transform of Power I've come across the following derivation, for example here, for the following Laplace transform of a power:
$$\mathcal L \{ t^a \} = \frac {\Gamma(a+1)}{s^{a+1}}$$
For $\operatorname{Re}(a) > -1, \operatorname{Re}(s)>0$. The proof I've seen does the following:
$$\mathcal L\{t^a\} := \int_0^{\to +\infty}t^a e^{-st} \, \mathrm dt$$
$$ = s^{-1} \int_0^? \left({\frac u s}\right)^a e^{-u}\, \mathrm du$$
and so on. 
The upperbound of the integral is what I'm questioning. The proofs I've seen have $\infty$ in the upper bound, but I don't see how you can replace the complex number $st$ with a real number $u$ without making the bounds of the integral complex. But that would turn the Riemann integral into a contour integral! How is that allowed? Is the proof wrong?
 A: Let $s=\sigma+i\omega \in \mathbb{C}$, where $\sigma>0$ and $\omega \in \mathbb{R}$.  
Let $I(a,s)$ be given by the integral

$$\bbox[5px,border:2px solid #C0A000]{\int_0^\infty t^ae^{-st}\,dt=\lim_{L\to \infty}\int_0^L t^ae^{-st}\,dt}\tag 1$$

Now, enforcing the substitution $z=st$ in the integral on the right-hand side of $(1)$ reveals
$$\int_0^L t^ae^{-st}\,dt=\frac{1}{s^{a+1}}\int_0^{L\sigma+iL\omega}z^ae^{-z}\,dz \tag2$$

Next, we move to the complex plane.  Let $f(z)=z^ae^{-z}$.  If $a$ is not an integer, then $z^a$ is multivalued and has a branch point at $z=0$.  If we cut the plane along the negative real axis, then $z^a$ is holomorphic on $\mathbb{C}\setminus \{z|\text{Re}(z)\le 0, \text{Im}(z)=0\}$.

Using Cauchy's Integral Theorem, we have
$$\oint_C z^ae^{-z}\,dz =0 \tag 3$$
where $C$ is a closed contour, comprised of $(i)$ the line segment from $0$ to $L\sigma$, $(ii)$  the line segment from $L\sigma$ to $L\sigma+iL\omega$, and $(iii)$ the line segment from $L\sigma+iL\omega$ to $0$.  (We also need to deform the contour around the branch point, but the contribution to the integral can be made arbitrarily small).  
Therefore, we can write $(3)$ as 
$$\int_0^{L\sigma}z^ae^{-z}\,dz+\int_{L\sigma}^{L\sigma+iL\omega}z^ae^{-z}\,dz+\int_{L\sigma+iL\omega}^0 z^ae^{-z}\,dz=0 \tag 4$$
As $L\to \infty$, the second integral on the left-hand side of $(4)$ approaches zero.  Therefore, letting $L\to \infty$ in $(4)$ yields
$$\int_0^{\infty}t^ae^{-t}\,dt=\lim_{L\to \infty}\int_0^{L(\sigma+i\omega)}z^ae^{-z}\,dz \tag 5$$
Putting together $(1)$, $(2)$, and $(5)$ yields the coveted equality

$$\bbox[5px,border:2px solid #C0A000]{\int_0^\infty t^ae^{-st}\,dt=\frac{1}{s^{a+1}}\int_0^\infty t^ae^{-t}\,dt}$$

And we are done!
