Showing convergence of integral I need to proof that the following integral exists:
$$\int_1^\infty t^{a+\sigma-1} e^{-t} \, dt \text{ for every }0<\sigma<a$$
However, I can´t find a proper upper bound for that integral. Intutively, it makes sense since $e^t$ converges faster to $\infty$ than $t^{a+\sigma-1}$
Any tips ?
 A: The integral $\int_1^\infty t^pe^{-t}\,dt$ converges for every $p\in \mathbb R.$ Proof: Fix $p$ and show $t^pe^{-t} \le e^{-t/2}$ for large $t.$ (Not sure where your $0<\sigma < a$ is coming from, but we don't need it.) 
A: $$
\int_1^\infty t^{a+\sigma-1} e^{-t} \, dt < \int_0^\infty t^n e^{-t} \, dt.
$$
where $n = \lceil a + \sigma-1\rceil = \big(\text{the smallest integer}\ge a+\sigma-1\big)$.
Integrating by parts yields
$$
\int_0^\infty t^n \Big( e^{-t} \, dt\Big) = \left. -t^n e^{-t} \vphantom{\frac 1 1} \, \right|_0^\infty + n \int_0^\infty t^{n-1} e^{-t} \, dt. \tag 1
$$
L'Hopital's rule shows the first term on the right is $0$ (and there is also an intelligent way to see that).
Then iterating the integration by parts, we get
\begin{align}
\int_0^\infty t^n e^{-t} \, dt & = n \int_0^\infty t^{n-1} e^{-t} \, dt \\[10pt]
& = n(n-1) \int_0^\infty t^{n-2} e^{-t} \, dt \\[10pt]
& = n(n-1)(n-2) \int_0^\infty t^{n-3} e^{-t} \, dt \\[10pt]
& = n(n-1)(n-2)(n-3) \int_0^\infty t^{n-4} e^{-t} \, dt \\[10pt]
& = \cdots \cdots \\[10pt]
& = n(n-1)(n-2)(n-3) \cdots 1 \int_0^\infty e^{-t} \, dt.
\end{align}
