Prove the improper integral of the Gamma function $\Gamma(t)$ converges for $z \in \mathbb C$ with $Re(z) > 0$: The gamma function $\Gamma(t)$ is defined by the following improper integral $$\Gamma(t) = \int^{\infty}_0 x^{t-1}e^{-x}dx$$

It is said that the integral converges for $z \in \mathbb C$ with $Re(z) >0$, and I've been trying to find a way of proving this.

Using integration by parts we know $\Gamma(t+1) = t\Gamma(t)$. However this doesn't really prove anything does it ? Because here we assume the improper definite integral converges (correct me if I'm wrong). Also if $t$ is not a positive integer, but some positive real number, we must evaluate $\Gamma(t)$ for some $t \in (0,1)$.

Could someone tell me how to determine that this integral actually converges ?

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    $\begingroup$ First, for complex $z$, take the modulus of the integrand. Since $\lvert x^{z-1}\rvert = x^{\operatorname{Re} z - 1}$ for $x > 0$, that takes you to the real case, and all you need to show is $$\int_0^\infty x^{t-1}e^{-x}\,dx < \infty$$ for $t > 0$. $\endgroup$ – Daniel Fischer Jan 5 '14 at 16:24

Hints using the comment of Daniel:

$$\int\limits_0^\infty x^{t-1}e^{-x}dx=\int\limits_0^1 x^{t-1}e^{-x}dx+\int\limits_1^\infty x^{t-1}e^{-x}dx$$

Now, we have that

$$0\le x\le 1\;\implies\;\; x^{t-1}e^{-x}\le x^{t-1}\;,\;\;\text{and}\;\;\int\limits_0^1x^{t-1}dx=\left.\frac{x^t}t\right|_0^1=\frac1t$$

and we also have that

$$1\le x\;\implies\;\;x^{t-1}e^{-x}\stackrel{\text{Prove this!}}\le e^{-x/2}\;,\;\;\text{and}\;\;\int\limits_1^\infty e^{-x/2}dx=\left.-2e^{-x/2}\right|_1^\infty=2 e^{-1/2}$$

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    $\begingroup$ @NicolasLykkeIversen, review the tests for convergence of improper integrals of non-negative functions. $\endgroup$ – DonAntonio Jan 5 '14 at 19:16
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    $\begingroup$ You are missing a constant (depending on $t$) in the estimate for $x \geqslant 1$. At the moment, you say that $\Gamma(t)$ remains bounded for $t\to \infty$. (Nevertheless, +1, I trust you will fix the omission.) $\endgroup$ – Daniel Fischer Jan 5 '14 at 21:02
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    $\begingroup$ @NicolasLykkeIversen, try the following argument (note the double implications!):$$x^{t-1}e^{-x}\le x^{-x/2}\iff x^{t-1}\le e^{x/2}\iff \frac{x^{t-1}}{e^{x/2}}\le 1$$ and the last inequality is true for $\;x\;$ big enough since $\;\frac{x^{t-1}}{e^{x/2}}\xrightarrow[x\to\infty]{}0\;$ (use l'Hospital, for example) $\endgroup$ – DonAntonio Jan 7 '14 at 18:23
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    $\begingroup$ No, of course not @NicolasLykkeIversen: it is enough that it is true for all $\;x\ge N\;$ ,since then you can split the integral $$\int\limits_1^\infty=\int\limits_1^N+\int\limits_N^\infty$$ $\endgroup$ – DonAntonio Jan 7 '14 at 18:33
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    $\begingroup$ So when you say $1\le x\;\implies\;\;x^{t-1}e^{-x}\stackrel{\text{Prove this!}}\le e^{-x/2}$, you really mean for some big enough $N$, $$N\le x\;\implies\;\;x^{t-1}e^{-x}\stackrel{\text{Prove this!}}\le e^{-x/2}$$ don't you? $\endgroup$ – Jack Nov 22 '17 at 1:01

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