Problem calculating the characteristic function of the exponential distribution

Question

I was trying to calculate the characteristic function of the exponential distribution $$\varphi(t) = \mathbb E[e^{itX}] = \int_{-\infty}^\infty e^{itx} \lambda e^{-\lambda x} \cdot 1_{[0,\infty)}(x) \, dx = \lambda \int_0^\infty e^{itx-\lambda x} \: dx = \lambda \int_0^\infty e^{x(it-\lambda)} \: dx$$ Now here I substituted $u = x(it-\lambda)$and got $$\frac{\lambda}{it-\lambda}[e^{x(it-\lambda)}]_0^{\infty}$$ and I know the solution should be $$\frac{\lambda}{it-\lambda}[0-1] = \frac{\lambda}{\lambda - it}$$ But I don't understand why $$\lim_{x \rightarrow \infty}(e^{x(it-\lambda)}) = \lim_{x \rightarrow \infty} (\frac{e^{xit}}{e^{x \lambda}})= 0$$

Answer 1

Note that we have $\lambda>0$. We obtain \begin{align*} \color{blue}{\lim_{x\to\infty}}\color{blue}{e^{x(it-\lambda)}} &=\lim_{x\to\infty}\left(e^{-\lambda x}e^{itx}\right)\\ &=\left(\lim_{x\to \infty}e^{-\lambda x}\right)\left(\lim_{x\to \infty} e^{itx}\right)\\ &=0\cdot\left(\lim_{x\to \infty} e^{itx}\right)\\ &\,\,\color{blue}{=0} \end{align*} since $e^{itx}$ has modulus $1$.