$$| \int_x^\infty \frac{e^{-t}}{t^c} dt| \lt \frac{e^{-x}}{x^c} $$
Why is this true (for $c \ge 2 \in \mathbf N$ and $x\gt 0$)? I get that the graph is decreasing above $x$ and moreover that $| \int_x^\infty \frac{1}{t^c} dt|$ is convergent by p-test and hence so is $| \int_x^\infty \frac{e^{-t}}{t^c} dt|$ because multiplying by $e^{-t}$ which is less than 1 won't hurt that, but how do I find the specific bound?
On
By integration by parts $$ \int_{x}^{+\infty}\frac{e^{-t}}{t^c}\,dt = -\left.e^{-t}t^{-c}\right|_{x}^{+\infty}-\int_{x}^{+\infty}\frac{c e^{-t}}{t^{c+1}}\,dt=\frac{e^{-x}}{x^c}-c\int_{x}^{+\infty}\frac{e^{-t}}{t^{c+1}}\,dt$$ where the last integral is positive and bounded by $\frac{c}{x^{c+1}}\int_{x}^{+\infty}e^{-t}\,dt = \frac{ce^{-x}}{x^{c+1}}.$
Multiply the integrand by $(t/x)^c \ge 1$. \begin{align} \left|\int_x^\infty \frac{e^{-t}}{t^c} \mathop{dt}\right| \le \frac{1}{x^c} \left|\int_x^\infty e^{-t} \mathop{dt}\right| = \frac{e^{-x}}{x^c}. \end{align}