Limit with complex exponential: using the fact that it is bounded Consider the following integral
$$\int_{0}^{\infty} e^{(-a x)}e^{i(bx)}dx\,\,\,\,\,\,\,\,\,\,\,\,\,\, a,b \in \mathbb{R}\,\,\,\,\,\,\,\,\ a>0$$
When evaluated it becomes
$$\frac{1}{ib-a}\lim_{c\to \infty} \bigg[e^{(-a x)}e^{i(bx)}\bigg]\,\, \bigg|_{0}^{c}=\frac{1}{ib-a}\bigg\{-1+\lim_{c\to \infty}e^{(-a c)}e^{i(b c)}\bigg\}=\frac{1}{a-ib}$$ 
Now the following limit is zero
$$\lim_{c\to \infty}e^{(-a c)}e^{i(b c)}=0$$
But is it correct to say that it is zero because 
$$\lim_{c\to \infty}e^{(-a c)}=0 \,\,\,\,\,\mathrm{and} \,\,\,\,\, |e^{i(b c)}|<1 \,\,\,\,\, \forall c  \,\,\,\,?$$
i.e. because the limit of the real exponential is zero and the complex exponential is bounded?
 A: I apologize if I repeat things that you know, but for the sake of completeness of this answer, let's start from the beginning. If $z=x+iy$ is a complex number (where $x,y\in\mathbb{R}$), then
$$e^z=e^{x+iy}=e^xe^{iy}=e^x(\cos y+i\sin y),$$
and $\color{magenta}{|e^z|=e^x}$, because
$$|e^{iy}|=|\cos y+i\sin y|=\sqrt{\cos^2y+\sin^2y}=1.$$
So if $a,b,c$ are real numbers, then $|e^{i(bc)}|=1$ (not less than $1$) and $|e^{-ac}e^{i(bc)}|=|e^{-ac+ibc}|=e^{-ac}$. Assuming $a>0$, and knowing from the limit that $c>0$, we indeed can conclude that
$$\lim_{c\to+\infty}|e^{-ac+ibc}|=\lim_{c\to+\infty}e^{-ac}=0.$$
A: Let $a>0$
$e^{-ac}e^{ibc}=e^{-ac}\cos{bc}+ie^{-ac}\sin{bc}=P(c)+iQ(c)$
$$\limsup_{c \to +\infty}||P(c)+iQ(c)||=\limsup_{c \to +\infty}e^{-ac}=0$$
$$\liminf_{c \to +\infty}||P(c)+iQ(c)||=\liminf_{c \to +\infty}e^{-ac}=0$$
Thus $\lim_{c \to +\infty}|||P(c)+iQ(c)||=0\Rightarrow  \lim_{c \to +\infty} (P(c)+iQ(c))=0$
Note that $||.||$ denotes the complex absolute value.
A: I think a rather simple way to prove this which is related to your initial question is to use the squeeze theorem. We start from Euler's formula:
$$e^{ibc} = \cos (bc) +i\sin (bc) = x+iy $$ with $x$ and $y$ real.
We also have, by definition of the real trigonometric functions
$-1 \le \cos (bc) \le 1$ and similarly $-i \le i\sin (bc) \le i$
Thus for $a$ real and  $a\gt 0$, we find
$$ -e^{-ac}\le e^{-ac}\cos (bc) \le e^{-ac}$$
and 
$$ -ie^{-ac}\le ie^{-ac}\sin (bc) \le ie^{-ac}$$
As 
$$\lim_{c \to \infty}e^{-ac}=0$$ We can see that by product and the squeeze principle 
$$\lim_{c \to \infty}e^{-ac}x=\lim_{c \to infty}e^{-ac}iy=0$$
Thus by sum 
$$\lim_{c \to \infty}e^{-ac}e^{ibc}=0$$
