Given $f(x)=\lambda e^{-\lambda x}$, I want to find $\phi(t) = E(e^{itx})$ (characteristic function).
Classical way: \begin{align} \phi(t) &= \int_0^{\infty} e^{itx}\lambda e^{-\lambda x} dx \\ &= \lambda \int_0^\infty e^{-x(\lambda-it)} dx \\ &= -\frac{\lambda}{ \lambda-it} [e^{-x(\lambda-it)}]_0^\infty \end{align}
From here it seems trivial to find $\phi(t) = \frac{\lambda}{\lambda -it}$. The thing is that this holds only when $(\lambda -it)>0$. But how can I make sure this is true? Since we're talking about an exponential, $\lambda >0$. How can I prove that the entire complex number is actually positive?
Think about the modulus of a complex number. $\lambda - it$ is a constant complex number with real part $\lambda > 0$ and imaginary part $(-t)$. Then:$$e^{-x(\lambda -it)}=e^{-\lambda x}\cdot e^{i\cdot xt}$$
The second term in the product, $e^{i\cdot xt}$, is a complex number in the unit circle of the complex plane. While it does go around the circle as $x$ varies, its modulus is always $1$ throughout the whole process. So we're left with
$$\lVert e^{-x(\lambda -it)}\rVert = \lVert e^{-\lambda x} \rVert \xrightarrow{x \longrightarrow \infty}0$$
where the last limit follows from $\lambda > 0$. Now, of course, if the modulus goes to $0$, the whole number itself goes to $0$, so we're done.