Proof

Let $x > 0$ be a real number and the improper integral $\int_0^\infty \exp(-t)t^{x-1}dt$ be given.
By definition of improper integral, $$\int_0^\infty \exp(-t)t^{x-1}dt=\underbrace{\int_{0}^1 \exp(-t)t^{x-1}dt}_{=:J_1}+\underbrace{\int_{1}^\infty \exp(-t)t^{x-1}dt}_{=:J_2}.$$
Because $\exp(0)=1$, we get $$\lim_{t\searrow 0}\frac{\exp(-t)t^{x-1}}{\frac{1}{t^{1-x}}}=\lim_{t\searrow 0}\frac{\exp(-t)t^0}{1}=1.$$
Using the limit comparizon test, the Riemann integral $J_1$ is convergent if and only if this one does: $$\int_{0}^1 \frac{1}{t^{1-x}}dt,$$ which is the case for any fixed $x > 0.$
On the other hand, for all $x$ we have $$\lim_{t\to\infty }\frac{\exp(-t)t^{x-1}}{\frac{1}{t^2}}=\lim_{t\to\infty}\frac{\exp(-t)t^{x+1}}{1}=0.$$
Using the limit comparizon test once again, the improper integral $J_2$ is convergent since this one does: $$\int_{1}^\infty \frac{1}{t^2}dt,$$ which is convergent for all $x$ (in fact, this integral does not depend on $x$ at all).
Altogether, the improper integral of the sum $J_1+J_2$ is convergent if and only if $x > 0.$
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Heuser Harro: "Lehrbuch der Analysis, Teil 1", B.G. Teubner Stuttgart, 1994, 11th Edition