Transforming a Weibull Distribution X follows a Weibull distribution (k, $\lambda$), but I want to find the expected value of Y where
$$y = A e^{(-Bx)}$$
So I'm trying to solve:
$$E[Y] = {\int_0^\infty}A e^{(-Bx)}{f(x)}{dx}$$
where $f(x)$ is the pdf of the Weibull distribution.
$$f(x)=\frac {k}{\lambda} (\frac {x}{\lambda})^{k-1}e^{-(x/\lambda)}$$
This is as far as I've gotten:
$$E[Y] = A \frac {k}{\lambda^k} {\int_0^\infty} x^{k-1}(e^{-x})^{(\frac {1}{\lambda}+B)}{dx}$$
I can see the form of the Gamma function in there, but I'm not sure how to isolate it.
Note: I actually know the expected value of Y that I want (as well as the values of A, k, and $\lambda$). I'm hoping the solution here will be easy enough to invert and then solve for B.

For the time being, let $a=B+\frac 1 \lambda$
$$\int_0^\infty x^{k-1}\,e^{-a x}\,dx=a^{-k} \Gamma (k)\quad \text{if} \quad \Re(k)>0\land \Re(a)>0$$ So, you want to solve for $a$ the equation
$$a^{-k} \Gamma (k)=\frac{\lambda ^k}{A k}\mathbb E[Y]\implies a=\frac 1\lambda\left(\frac{A k \Gamma (k)}{\mathbb E[Y]}\right)^{\frac{1}{k}}$$ then $B=a-\frac 1 \lambda$
