Why $x$ should be **real**

Question

Show that if $x$ is real and positive ($x \in \mathbb{R}^{> 0}$) then: $$Γ(x)= ∫_0^1(- \ln ⁡t )^{x-1} dt $$

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Solution :

First recall the gamma function:

$$Γ(x)= ∫_0^∞ e^{-t} t^{x-1} dt \; \; \;\;\;\;\;\;\;\; \;\;\;\;\;\;\;\; (1) $$

Let $u = e^{-t}$

$$\ln u = -t \rightarrow t = - \ln u \;\;\; $$and$$ \;\;\; dt = - \frac{1}{u} du$$

then we can rewrite equation $(1)$ in the new form of below:

$$Γ(x)= ∫_0^1(-\ln⁡ u )^{x-1} du $$

and the solution is completed.

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I can justify why $x$ must be positive but I don't know where realness of $x$ is needed?

Any Ideas?

Answer 1

As far as I can remember (but I shall not be able to prove it),$x$ could be any real or complex number provided that $\Re(x)>0$.

Using a CAS, I computed the integral using $x=a+i b$. The result was $\Gamma (a+i b+1)$ under the conditions $\Re(a)+1>\Im(b)$.