Let $I_n(z) = \int_n^{n+1} e^{-t}t^{z-1} dt$. For $0 < \epsilon \leq Re(z) \leq K < \infty$, show that $|I_0(z)| \leq \epsilon^{-1}$

Question

In a previous part to this problem, I showed that $|e^{-t}t^{z-1}| \leq e^{-t}t^{Re(z)-1}$. So I got:

$|I_0(z)| = |\int_0^1 e^{-t}t^{z-1} dt| \leq \int_0^1 |e^{-t}t^{z-1}| dt \leq \int_0^1 e^{-t}t^{Re(z)-1} dt$.

I assume that since $t^x$ is decreasing for $t \in (0, 1)$, $\epsilon \leq Re(z)$ implies $t^{Re(z)} \leq t^{\epsilon}$. But how does that help me evaluate the integral? I must be missing something pretty basic. Integration by parts shouldn't do anything, since we can't guarantee that $\epsilon-1$ is an integer.

I will also need to show that $|I_n(z)| \leq (n+1)^Ke^{-n}$, but I might be able to figure it out from help with the other question.

Answer 1

$t^{Re(z)} \leq t^{\epsilon}$ is correct for $0 \le t \le 1$, and integration by parts does help, together with the estimate $t^\epsilon \le 1$: $$ |I_0(z)| \le \int_0^1 e^{-1}t^{\epsilon-1} \, dt = \left. \frac 1 \epsilon e^{-t}t^\epsilon\right|_{t=0}^{t=1} + \frac 1 \epsilon \int_0^1 e^{-t} t^\epsilon \, dt \\ = \frac 1 \epsilon e^{-1} + \frac 1 \epsilon \int_0^1 e^{-t} t^\epsilon \, dt \\ \le \frac 1 \epsilon e^{-1} + \frac 1 \epsilon \int_0^1 e^{-t} \, dt = \frac 1 \epsilon \, . $$