Dividing the given integral into two improper integrals
$\int_{0}^{1} x^{n-1}e^{-x}dx$ and $\int_{1}^{\infty} x^{n-1}e^{-x}dx$
I) for $\int_{0}^{1} x^{n-1}e^{-x}dx$:
for $n-1 \ge 0$ integral (I) is proper.
for $n-1 \lt 0 $ converting the integral into $\int_{0}^{1} \frac{e^{-x}}{x^{1-n}}dx$ and taking $\phi(x) = \frac{1}{x^{m}}$ for the comparison test.
$\lim_{x \to 0}\frac{\frac{e^{-x}}{x^{1-n}}}{\frac{1}{x^{m}}}$ $\to 1$ when $m = 1-n$. Hence both f(x) and $\phi(x)$ converges or diverges together. When $m\lt1$, $\int_{0}^{1}\frac{1}{x^{m}}$ converges.
$\Rightarrow$ f(x) converges when $1-n \lt 1 \Rightarrow n \gt0$
(II) $\int_{1}^{\infty} x^{n-1}e^{-x}dx$
following the similar approach and using the fact that $\int_{a}^{\infty}\frac{dx}{x^{n}}$ converges when $n \gt 1$
$\lim_{x \to \infty} \frac{\frac{e^{-x}}{x^{1-n}}}{\frac{1}{x^{m}}} = 0 $ when $m = 1-n$
for convergence (as the limit is 0 we can judge only on convergence) $1-n \gt 1 \Rightarrow n \lt 0 $
What is wrong in this approach in proving the (II) improper integral is convergent for n >0.
Actual Proof:
$e^{x} \gt x^{n+1} \Rightarrow x^{n-1}e^{-x} \lt \frac{1}{x^{2}}$ as $\int_{1}^{\infty}\frac{1}{x^2}$ converges the improper integral (II) converges by comparison test.
Best Answer
If $n>0$, then the limit $$ \lim_{x\to \infty} \left(\frac{e^{-x}}{x^{1-n}}\right)\big/ \left(\frac{1}{x^{1-n}}\right) = 0, $$ so the limit comparison test for integrals says that if $\int_1^\infty \frac{1}{x^{1-n}} dx$ converges, then so too must the gamma integral. But $$\int_1^\infty \frac{1}{x^{1-n}} dx = \lim_{M\to\infty} \int_1^M \frac{1}{x^{1-n}} dx = \frac{1}{n}\left( \lim_{M\to\infty} M^{n} - 1\right)$$ does not converge for the $n$ under consideration, so the LCT does not apply.