Given:
$$f_X(x) = \lambda e^{-\lambda x},\; x\in X$$
Wanted: The corresponding characteristic function $\phi(ju)$.
\begin{align} \phi(ju)&=\mathbb{E}(e^{j^2ux})\\ &= \lambda \int^{\infty}_0 e^{-ux}e^{-\lambda x}dx\\ &= \lambda \int^{\infty}_0 e^{(-u-\lambda) x}dx \\ &= \lambda \left [ \frac{1}{-u-\lambda} e^{-u-\lambda)x} \right]^{\infty}_0\\ &= \lambda \left [ 0 - \frac{1}{-u-\lambda} \right] \\ &= \frac{1}{\lambda - u } \end{align}
The definition of the characteristic function is
$$ \phi(t) = \mathbb{E}(e^{jtx})~. $$
Can I set $t$ equal to $ju$? So so that the definition is
$$ \phi(t=ju) = \mathbb{E}(e^{jjux})~?$$