# Convergence of Integral of gamma function

The reason for finiteness of the integral definition of the $$\Gamma$$-function in $$\operatorname{Re}z>1$$ is,

$$\left|\int\limits_0^{\infty}t^{z-1}e^{-t}\mathrm{d}t\right|\leq \int_0^{\infty}\left|t^{z-1}e^{-t}\right|\mathrm{d}t=\int_0^{\infty}t^{\alpha}e^{-t}dt,$$ where $$\alpha>0$$. Now, for lower value of $$t$$ close to zero, $$t^{\alpha}$$ is small, and as $$t$$ increases, we can see by L' Hospital rule that $$\lim_{t\rightarrow \infty} t^{\alpha}/e^t=0$$. Hence, intuitively the integral converges. But can anyone give a more formal explanation of this?

• What is your question? Are you trying to understand why the integral converges for $\operatorname{Re}z>1$? – vitamin d Feb 20 at 18:50
• Yes, I gave a intuitive explanation for this, but I formal explanation would be helpful – roydiptajit Feb 20 at 19:40
• I would appreciate an upvote and a tick :) – vitamin d Feb 20 at 20:22

Splitting the integrals up $$\int\limits_0^\infty t^{\alpha-1}e^{-t}dt=\int\limits_0^1 t^{\alpha-1}e^{-t}dt+\int\limits_1^\infty t^{\alpha-1}e^{-t}dt$$

Now, we have that

$$0\le t\le 1\;\implies\;\; t^{\alpha-1}e^{-t}\le t^{\alpha-1}\;,\;\;\text{and}\;\;\int\limits_0^1t^{\alpha-1}dt=\left.\frac{t^\alpha}\alpha\right|_0^1=\frac1\alpha$$

and we also have that for sufficiently big $$N$$, with $$N\leq t$$

$$t^{\alpha-1}e^{-t}\stackrel{\text{}}\le e^{-t/2}\;,\;\;\text{and}\;\;\int\limits_1^\infty e^{-t/2}dt=\left.-2e^{-t/2}\right|_1^\infty=2 e^{-1/2}.$$

• For your second inequality, there has to be some condition on $\alpha$. It is not true for all $\alpha>0$ (otherwise $\Gamma(\alpha)$ would be bounded when $\alpha$ is bounded away from zero). – Gary Feb 20 at 20:31
• It is $\alpha>1$ in the second inequality – vitamin d Feb 20 at 20:33
• So $\Gamma (\alpha ) \le \frac{1}{\alpha } + 2e^{ - 1/2}$? In particular $\Gamma(\alpha)<3$ for $\alpha\geq 1$? – Gary Feb 20 at 20:33
• You can use for example $$e^{t/2} = 1 + t + \cdots + \frac{{t^{\left\lceil {\alpha - 1} \right\rceil } }}{{2^{\left\lceil {\alpha - 1} \right\rceil } \left\lceil {(\alpha - 1)!} \right\rceil }} + \cdots > \frac{{t^{\alpha - 1} }}{{2^{\left\lceil {\alpha - 1} \right\rceil } \left\lceil {(\alpha - 1)!} \right\rceil }},$$ i.e., $$2^{\left\lceil {\alpha - 1} \right\rceil } \left\lceil {(\alpha - 1)!} \right\rceil e^{ - t/2} > t^{\alpha - 1} e^{ - t}$$ for all $t\geq 1$. – Gary Feb 20 at 20:40
• You may remove $1\leq t$ and integrate form $N$. – Gary Feb 20 at 20:45