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How do I prove that $$\int_{0}^{1}\frac{t^{a-1}}{e^t} dt$$ converges for $0<a,t<1$ I tried to bound it with couple of other integrals, for example: $$\int_{0}^{1}\frac{1}{te^t} dt$$ but it diverges.

My idea is that $t^{a-1}$ is bigger than 1 and that $$\int_{0}^{1}\frac{1}{e^t} dt$$ does converge..

Thanks for the help.

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    $\begingroup$ How about bounding with $t^{a-1}e^{-t} \le t^{a-1}$ ? Isn't that sufficient ? $\endgroup$ – Joel Cohen Dec 6 '11 at 12:52
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You basically answered the question already: $a > 0$ means that $a-1 > -1$. You should be able to convince yourself that $\int \limits_0^1 t^p\,dt$ converges when $p > -1$. As you already noted, the $e^{-t}$ term doesn't cause any trouble. (Here's a better hint: can you bound $e^{-t}$ by some constant $C$ on the interval, so that $\int \limits_ 0^1 t^{a-1}e^{-t}\,dt \leq C\int_0^1t^{a-1}\,dt$ ? )

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Later note: My first guess, below, as to what was desired, appears to be incorrect. See the further "later note" below.

If you mean "as $a\to\infty$" (the only guess I've got), then observe that $$ 0 < \int_0^1 \frac{t^{a-1}}{e^t}\;dt < \int_0^1 t^{a-1}\;dt $$ and let $a\to\infty$.

If you didn't mean as $a\to\infty$, then you'd better re-write your question to make it clear what you meant.

Later note: One of your other question suggests something about what you meant. Your phrase "for $0 < a,t<1$" is confusing. $a$ is a parameter that remains fixed as $t$ goes from $0$ to $1$. The variable $t$ is already "bound" by the expression $\int_0^1\cdots\cdots dt$, and anything that binds the variable $a$, such as "for $0

Let's try again: $$ 0<\int_0^1\frac{t^{a-1}}{e^t}\;dt < \int_0^1 t^{a-1}\;dt $$ and this is finite if $-1<a-1<0$.

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