Convergence of integral in complex analysis textbook

Question

In my complex analysis textbook it states that the following integral converges

$\int_{1}^{\infty}t^{\alpha - 1}e^{-t}dt$

where $\alpha$ is some real number such that $\alpha > 1$

Also, it seems that $t \rightarrow t^{\alpha - 1}e^{-t}$ is taken to be a real valued function of a real variable.

This integral appears in a demonstration of using a certain theorem. In the statement of the theorem, it states

"... and there exists a continuous non-negative $h: [0,\infty) \rightarrow [0,\infty)$ such that the integral $\int_{0}^{\infty}h(t)dt$ converges and ..." (Translated from Hebrew).

Is there more than one definition of convergence that could be applied to this? I do not think that my textbook previously defined such an (improper?) integral, so it may be a definition from "Calculus II" (Israeli Syllabus)

In the "most basic" or "most likely" sense of convergence, how can it be shown that this integral does in fact converge?

Thanks!

(Note: I tagged the question with "improper-integrals", but this should not be taken to imply that this it is assumed that this is the definition the author meant)

Answer 1

For each $A$, $\int_0^{A} t^{\alpha -1} e^{-t}\, dt$ exists by continuity of the integrand. If we show that $\int_A^{B} t^{\alpha -1} e^{-t}\, dt\to 0$ as $B>A \to \infty$ we can conclude that the improper integral exists. To show this, note that $e^{\frac t 2} \geq \frac {t^{k}} {k!}$ for any positive integer $k$ hence $ t^{\alpha -1} e^{-t} \leq Ce^{-t/2}$ for all $t>1$, for some finite constant $C$. Since $\int_A^{B} e^{-t/2} \, dt\to 0$ as $B>A \to \infty$ (as seen by direct computation of the integral) we are done.

Answer 2

When $\alpha>1$, for large $t>c$ the exponential $e^{-t}$ is greater than every power of $t$, say $$e^t>Kt^{2\alpha}$$ for a real constant $K$, then $$\int_c^\infty t^{\alpha-1}e^{-t}dt<\dfrac{1}{K}\int_c^\infty t^{-\alpha-1}dt=\dfrac{c^{-\alpha}}{K\alpha}<\infty$$

Answer 3

For any $p\in \mathbb R,$

$$\int_1^\infty t^p e^{-t}\,dt< \infty.$$

That's simply because $t^p$ is continuous on $[1,\infty),$ and exponential decay wipes out polynomial growth, bigtime, at $\infty.$ Perhaps you meant the integral from $0$ to $\infty?$