Proving that the Gamma function $\Gamma(y)$ converges for $y>0$.

Question

How can I justify that $$\Gamma(y)=\int_0^\infty t^{y-1}e^{-t} \, \mathrm{d}t$$ exists for all $y>0$? I'm struggling to compare it to a known convergent integral.

Answer 1

$$\Gamma(y) = \int\limits_0^\infty y^{t-1}e^{-y} \, \mathrm{d}y=\int\limits_0^1 y^{t-1}e^{-y} \, \mathrm{d}y+\int\limits_1^\infty y^{t-1}e^{-y} \, \mathrm{d}y$$

We have that for $0 < y \leq 1 \implies y^{t-1}e^{-y}\le y^{t-1}$ and that $$\int\limits_0^1y^{t-1} \, \mathrm{d}y=\left.\frac{y^t}t\right|_0^1=\frac1t$$

Hence the integral converges for the interval $0 < y \leq 1$. We now need to show convergence for the other interval, which we do so below.

We also have for $y \geq 1$ that the since the exponential grows faster than any polynomial, for every $y$ we can take $N \in \mathbb{N}$ so big that for $t\geq N \implies e^{t/2} > t^{y-1}$ and hence \begin{align}\int_1^{\infty} e^{-t}t^{y-1} \, \mathrm{d}t &= \int_1^{N} e^{-t}t^{y-1} \, \mathrm{d}t + \int_N^{\infty} e^{-t}t^{y-1} \, \mathrm{d}t \\ &< \int_1^{N} e^{-t}t^{y-1} \, \mathrm{d}t + \int_N^{\infty} e^{-t}e^{t/2} \, \mathrm{d}t \\ &= \int_1^{N} e^{-t}t^{y-1} \, \mathrm{d}t + \int_N^{\infty} e^{-t/2} \, \mathrm{d}t < \infty \end{align}