My professor went over an example in class and I am kind of confused. We are giving a proof of the theorem that $\forall a \gt 0, \int_{0}^{\infty} e^{-(a+ib)t} \, dt \, \colon= \lim_{R \to \infty} \int_{0}^{R} e^{-(a+ib)t} \, dt = \frac{1}{a+ib}$.
So, to prove, we integrated over the triangle with sides $\gamma_1 = Ra$, $\gamma_2 = R(a+ib)$, and hypotenuse $\gamma_3$. Then, we parameterized them: $$\gamma_1 = t, t \in [0, Ra]$$ $$\gamma_2 = aR + ibt, t \in [0, 1]$$ $$\gamma_3 = t(a+ib), t \in [0, \mathbb{R}]$$
Note: hypotenuse travels outwards from origin is intended.
Then, by Cauchy's Theorem, $$\int_{\gamma_1}e^{-z}\,dz + \int_{\gamma_2}e^{-z}\,dz - \int_{\gamma_3}e^{-z}\,dz = 0$$
So, $$\int_{\gamma_3}e^{-z} \, dz = (a+ib) \int_{0}^{R}e^{-t(a+ib)} \, dt$$
This follows from $\int f(z) \, dz = \int \gamma (t) \cdot \gamma ' (t) \, dt$.
$$\Rightarrow \int_{\gamma_1}e^{-z} \, dz = \int_{0}^{Ra} e^{-t} \, dt \rightarrow \int_{0}^{\infty}e^{-t} \, dt = 1$$
So, $$\text{if} \space z = Ra + ibt, t \in [0, R] \Rightarrow \vert e^{-z} \vert = e^{-Ra}$$
This is the step I'm confused about. What is the motivation behind this substitution? Where does it come from? How does that imply RHS? I think I am missing something relatively simple, but I just cannot see it.
I will now finish the proof, just for sake of completeness:
$$\Bigg| \int_{\gamma_2} e^{-z} \, dz \Bigg| \le e^{-Ra} L(\gamma_2) = e^{-Ra} \vert b \vert R = 0$$
$$\Rightarrow (a+bi) \int_{0}^{\infty} e^{-(a+ib)t} \, dt = 1 + \lim_{R \rightarrow \infty} \Bigg| \int_{\gamma_2} e^{-z} \, dz \Bigg| = 1$$
$$\Rightarrow (a+bi) \int_{0}^{\infty} e^{-(a+ib)t} = \frac{1}{a+ib}$$
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