Im reading a demonstration and there is a step Im not being able to work out.
So we have $\gamma:[-M,M]\to \Bbb{C}$ given by $\gamma(\varphi)=s+i\varphi$ where $s\in \Bbb{R}$ and the idea is to prove, using $\gamma$ that
$$\int_{-M}^MF(z)e^{zx}dz=\frac{1}{i}\int_{s-iM}^{s+iM}F(z)e^{zx}dz$$
Where $x \in \Bbb{R}$ and $F$ is an holomorphic function.
So the idea is:
$$\int_{-M}^MF(\gamma(\varphi))e^{\gamma(\varphi) x}\frac{\gamma'(\varphi)}{i}d\varphi=\frac{1}{i}\int_{\gamma}F(z)e^{zx}dz$$
Where we are using that $\gamma'(\varphi)=i$. So the next step says: Since the integrand $F(z)e^zx$ is holomorphic then
$$\frac{1}{i}\int_{\gamma}F(z)e^{zx}dz=\frac{1}{i}\int_{s-iM}^{s+iM}F(z)e^{zx}dz$$
And this last implication is the one I not see. Why does the integrand bein holomorphic implies this last equality?