Characteristic function of a exponential random variable, problems with complex integral.

Question

I tried to compute the characteristic function of a random variable, which is exponential distributed with parameter $\lambda$: \begin{align*} \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. \end{align*} This is an integral of a function $f: \mathbb R \to \mathbb C$, $f(x) := e^{x(it-\lambda)}$ over a real segment $[0,\infty)$. In complex analysis we had that in general the integral of such a function is defined as (with $f(x) = u(x)+iv(x)$): \begin{align*} \int_0^\infty f(x) \, dx = \int_0^\infty u(x) \, dx + i \int_0^\infty v(x) \, dx. \end{align*} In this case we have that \begin{align*} f(x) = e^{x(it-\lambda)} = \frac{\cos(tx)}{e^{\lambda x}}+i\frac{\sin(tx)}{e^{\lambda x}}, \end{align*} but \begin{align*} \int_0^\infty\frac{\cos(tx)}{e^{\lambda x}} \, dx +i \int_0^\infty \frac{\sin(tx)}{e^{\lambda x}} \, dx \end{align*} doesn't converge. Thus I don't know how to go on.