Proof of $\Gamma(1-z) \Gamma (z) = \frac{\pi}{\sin \pi z}$

Question

In the proof of the above result from the book by Stein and Shakarchi, they have used the substitution $vt = u$ where $t>0$ to get $$\Gamma (1-z) = \int_{0}^{\infty} e^{-u} u^{-z} du = t\int_{0}^{\infty} e^{-vt} (vt)^{-z} dv.$$ Well, up to this part it was okay. But I am not understanding the following mixing of integrals: $$ \begin{align} \Gamma (1-z) \Gamma (z) & = \int_{0}^{\infty} e^{-t} t^{z-1} \Gamma (1-z) dt \\ &= \int_{0}^{\infty} e^{-t} t^{z-1} \left(t \int_{0}^{\infty} e^{-vt}(vt)^{-z}dv \right) dt \\ &=\int_{0}^{\infty} \int_{0}^{\infty} e^{-t(1+v)}v^{-z} dvdt. \end{align}$$
I want to understand how the introduced variable $t$ is taken to be the same with the dummy variable $t$ while mixing the integrals.

Answer 1

The value of the second integral (including the $t$ prefactor) in your first line is independent of $t$, as its equation to the expression $\Gamma(1 - z)$, which contains no $t$s, shows.

Thus you can plop it in under

$$\int_{0}^{\infty} e^{-t} t^{z-1} \Gamma(1 - z)\ dt$$, and "hook" its $t$ to the $t$ therein without changing the value of this integral because while $t$ is changing "underneath" the integral, the whole thing you've substituted does not change with it.

Effectively, it's just a more complex manipulation in the same vein as the more elementary technique of "multiplying and dividing/adding and subtracting the same thing" under the integral sign. You are adding a term which doesn't change the value of the integral and which doesn't change with the change in the variable of integration, yet which allows you to modify the expression inside algebraically in a useful manner.

To see that it is independent of $t$, note that it can be obtained from

$$\Gamma(1 - z) = \int_{0}^{\infty} e^{-u} u^{-z}\ du$$

by substitution of $u = vt$, $du = t\ dv$, treating for the purposes of integration $t$ as a constant, and the whole point of a change of variables is that it does not change the value of the integral.