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**Mathstud28** Assuming I understand what you are saying correctly, since this integral is uniformly convergent (I leave that to you to prove) then we have that

$\displaystyle \begin{aligned}F\left(\infty\right)&=\lim_{x\to\in fty}\int_0^{\infty}x^{a-1}e^{-x}dx\\

&=\int_0^{\infty}\lim_{x\to\infty}x^{a-1}e^{-x}dx\\

&=\int_0^{\infty}0dx\\

&=0\quad\blacksquare

\end{aligned}$

For proof of the limit we see that our limit is infinity over infinity, an undeterminate form for which L'hopital's may be applied. But we see that this yields another indeterminate form. So we may keep taking limits until

$\displaystyle \lim_{x\to\infty}\frac{x^{a-1}}{e^x}=\lim_{x\to\infty}\frac{(a-1)x^{a-2}}{e^x}=\underbrace{\cdots}_{a-3\text{ number of times}}=\lim_{x\to\infty}\frac{(a-1)!}{e^x}=0$