Using characteristic functions to find $E\cos(tX)$ and $E\sin(tX)$

Question

Let $X \sim \operatorname{Exp}(\lambda)$

How do we find $E\cos(tX)$ and $E\sin(tX)$?

I understand that a characteristic function can be derived as the following:

$\phi(t) = E(e^{itX}) = E\cos(tX) + iE(\sin(tX))$

Answer 1

I'm going to guess you mean the distribution is $e^{-\lambda x}(\lambda\,dx)$ for $x\ge0$ and not $e^{-x/\lambda}(dx/\lambda)$ for $x\ge0.$

\begin{align} & \operatorname{E}(\cos(tX) + i\sin(tX)) = \operatorname{E}(e^{itX}) \\[10pt] = {} & \int_0^\infty e^{itx} e^{-\lambda x}(\lambda\,dx) = \lambda \int_0^\infty e^{(it-\lambda)x} \,dx \\[10pt] = {} & \left.\lambda \frac{e^{(it-\lambda )x}}{it-\lambda} \right|_{x\,:=\,0}^{x\,:=\,\infty} = \frac \lambda {\lambda - it} \text{ since } \lambda >0\\[10pt] = {} &\frac{\lambda(it+\lambda)}{(\lambda-it)(\lambda+it)} = \frac{\lambda^2 + \lambda it}{\lambda^2+t^2} = \frac{\lambda^2}{t^2+\lambda^2} + i\frac{\lambda t}{t^2+\lambda^2}. \end{align} The real part of this last expression ie $\operatorname{E}(\cos(tX))$ and the imaginary part is $\operatorname{E}(\sin(tX)).$