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It is known to me that Gamma-function $\Gamma(s)=\int_{0}^{\infty}x^{s-1}\exp(-x)$ defined on real numbers converges exactly when $s>0$. I am trying to generalize the convergence to complex numbers and show that $\Gamma(s)$ converges exactly when $\mathcal Re(s)> 0$.

When $\mathcal Re(s)> 0$, then $\Gamma(s)$ converges absolutely because $|x^{s-1}\exp(-x)|=x^{\mathcal Re(s)-1}\exp(-x)$ and $\int_{0}^{\infty}x^{\mathcal Re(s)-1}\exp(-x)$ converges.

I am not sure how to show that for $\mathcal Re(s) \le 0$ there is no convergence.

EDIT

With the help of one of the answers below (by Simply Beautiful Art) I have got that far: for $s \in \mathbb C \setminus \{0\}$ and $\mathcal Re (s) \in (-n-1,-n]$ with $n \in \mathbb N_0$ one can show that

$$\int^{a}_bx^{(s-1)}\exp(-x)=\bigg(\exp(-x)\sum^{n}_{i=0}x^{i+s}\prod^{i}_{j=0}\frac{1}{s+j}\bigg)\bigg|^a_b+\prod^{n}_{j=0}\frac{1}{s+j}\int^a_bx^{s+n}\exp(-x)$$

That can be shown using induction and integration by parts.

The integral on the right hand side converges (because $\mathcal Re(s+n+1)>0$). The limit of the sum as $a \to \infty$ is clearly $0$. I am not sure how to show that the limit of the sum as $b \to 0$ does not exist.

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  • $\begingroup$ Check the behaviour of the integrand at $0$. $\endgroup$ Aug 5, 2017 at 19:51
  • $\begingroup$ Sorry for not responding in forever. To show the left thing diverges, simply note that it equals$$x^s\underbrace{\exp(-x)\sum_{i=0}^nx^i\prod_{j=0}^i\frac1{s+j}}$$and the underlined part approaches $\frac1s$ while $|x^s|\to+\infty$. $\endgroup$ Aug 9, 2017 at 10:55
  • $\begingroup$ @SimplyBeautifulArt Looking at my question to you 3 days later I myself can not understand why I had asked it. Indeed, the underlined part goes to $\frac{1}{s}$ and as for $x^s$, its absolute value either goes to $\infty$ if $\mathcal Re (s) < 0$ or $x^s$ just spins on the unit circle if $\mathcal Re(s)=0$. So the whole thing can not converge. $\endgroup$
    – zesy
    Aug 9, 2017 at 12:14
  • $\begingroup$ Lol, no problem ;) $\endgroup$ Aug 9, 2017 at 12:19

1 Answer 1

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Note that it converges absolutely on $\Re(s)>0$. Let $s=a+bi$.

$$|\Gamma(s)|\le\int_0^\infty|x^{s-1}e^{-x}|~\mathrm dx=\int_0^\infty x^{a-1}e^{-x}~\mathrm dx=\Gamma(a)$$

For $-1<\Re(s)\le0$, we can see by integration by parts that

$$\int_0^\infty x^{s-1}e^{-x}~\mathrm dx=\lim_{(a,b)\to(0^+,\infty)}\frac1sx^se^{-x}\bigg]_a^b+\frac1s\int_a^bx^se^{-x}~\mathrm dx$$

The integral converges absolutely while the $u\cdot v$ part diverges. In general, for $-n<\Re(s)\le1-n$, integrate by parts $n$ times, observing the final remaining integral converges whilst the rest diverges (mainly as $x\to0^+$)

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  • $\begingroup$ Could you please take a look at the EDIT of my question below. I have summarized what I have understood from your answer? But I am struggling to complete the proof. $\endgroup$
    – zesy
    Aug 6, 2017 at 14:19

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