Showing that $\overline{\Gamma (z)}=\Gamma (\overline{z})$

Question

I am tring to show that:

$\overline{\Gamma (z)}=\Gamma (\overline{z})$

for z $\in$ $\mathbb C$.

The only thing I found is: "This is immediate from, say, the Euler form of $\Gamma$ and the fact that complex conjugation preserves products and sums." (source: https://proofwiki.org/wiki/Complex_Conjugate_of_Gamma_Function).

I see what is meant by this, but I want to show the equation without using alternative forms of the gamma function like the Euler form or the Weierstrass form. I want to use the usual form:

$\Gamma(z)=\int _0^{\infty } \text{dt} \text{ } t^{z-1} \text{ } e^{-t}, \text{ }\text{Re} (z)>0$

Any ideas?

Answer 1

Ok lets just write out the definition of $\Gamma(z)$ and $\Gamma(\overline z)$ and compare the results (we set $z=a+ib$ and $\Re(z)>0$):

Lets start with $\Gamma(z)$:

We have that: \begin{align*} \Gamma(z) &= \int_0^\infty t^{z-1} e^{-t} dt = \int_0^\infty e^{(a-1+ib)\ln(t) }e^{-t}dt \\ &=\int_0^\infty t^{a-1}e^{-t} \left(\cos(b\ln(t)) +i \sin(b \ln(t))\right) dt \end{align*}

Now for $\Gamma(\overline z)$: \begin{align*} \Gamma(\overline{z}) &= \int_0^\infty t^{\overline{z}-1} e^{-t} dt = \int_0^\infty e^{(a-1-ib)\ln(t) }e^{-t}dt \\ &=\int_0^\infty t^{a-1}e^{-t} \left(\cos(b\ln(t)) -i \sin(b \ln(t))\right) dt\\ &= \overline{\Gamma(z)}. \end{align*}

As we wanted.