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Or, more generally, $$\Gamma (x+1)=\int_0^{\infty}t^{x}e^{-t}dt=p^x$$ with $p \in \mathbb{Z}^+$ and $x \in \mathbb{C}$.

Perhaps begin with $\large p^x=p^x \lim_{n \rightarrow \infty}[e^{-0}-e^{-n}]=\int_0^{\infty}p^xe^{-t}dt$.

Then

$$\int_0^{\infty}(t^{x}-p^x)e^{-t}dt=0$$

which I'm not sure how to solve. It smells of integration by parts, but I can't see anything feasible given that $x$ isn't a constant.

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Do you have any restrictions on $x$ or is $x=0$ ok? –  Fabian Jan 20 '13 at 14:55
    
@Alyosha Do you want a integral representation of solution or a numeric aproximation? –  Elias Jan 20 '13 at 14:56
    
@Fabian Yes, that's one of the solutions. –  Alyosha Jan 20 '13 at 14:57
    
@Alyosha: so you want all solutions? Is $x\in \mathbb{R}$ or $x\in\mathbb{C}$? –  Fabian Jan 20 '13 at 14:57
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You can use integration by parts ($x$ is constant w.r.t. $t$), but all it tells you is that $\Gamma(x+1)-p^x = x\Gamma(x)-p^x$, which is a tautology. –  John Moeller Jan 20 '13 at 21:25

1 Answer 1

Here's a plot of the complex roots of the equation $\Gamma(z+1) = 5^z$ near the origin.

enter image description here

Numerically it appears that the only other roots are those continuing along the three "rays" of roots pictured: the ray along the negative real axis (here the roots are near-integers) and the two symmetric arms which appear to tend to $\pi/3$ and $-\pi/3$ radians.

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