I was asked to find the characteristic function of a pdf and differentiate it to get the expectation.
$p(x) = xe^{-x}$ for $x \ge 0$
I am doing this in the following way. Sorry that i don't know how to insert the correct symbol....
$$ϕw = \int_0^\infty e^{itx} x e^{-x} dx$$
$$ = \int_0^\infty e^{itx-x} x dx$$
Then exponent $ = -x(1-it)$
and let $ z = -x(1-it)$
with $dz/dx = -(1-it)$
Then we will have
$$ϕw = \int_0^\infty (\frac{z}{1-it})(e^z)(\frac{1}{1-it}) dz$$
$$ = \frac{1}{(1-it)^2}\int_0^\infty z e^zdz$$
$$ = \frac{1}{(1-it)^2} $$
Then i differentiate the above and i will get zero which make me not so sure about my answer. Can anyone have a look of my step above to see if i got enough mistake?
Thanks
$$\phi_X(t)=E(e^{itX})$$ $$\frac{d}{dt}(\phi_X(t))=E(iXe^{itX})$$ $$\frac{d}{dt}(\phi_X(0))=E(iX)=iE(X)$$ In other words, you found $\phi_X(t)$ correctly, but we must differentiate with respect to $t$ to find expectation. In your case we have:
$$\frac{d}{dt}(\phi_X(t))=\frac{d}{dt}\Bigg(\frac{1}{(1-it)^2}\Bigg)$$ $$=\frac{2i}{(1-it)^3}$$ $$ \frac{d}{dt}(\phi_X(0)) = 2 i=iE(X)$$ $$\therefore E(x)=2$$