Suppose $X$ has a density $f_X$ and let $$Y := c e^{-\alpha X}\chi_{\{X > 0\}}$$ for some $\alpha, c > 0$. Find $f_Y$ in terms of $f_X$.
So let $f : \mathbb{R} \to \mathbb{R}$ continuous and bounded. We have $$\begin{align*} E(f(Y)) =& \int_{-\infty}^\infty f(c e^{-\alpha x}\chi_{\{x > 0\}}(x))f_X(x)dx \\ =& \int_{0}^\infty f(c e^{-\alpha x}\chi_{\{x > 0\}}(x))f_X(x)dx\\ &+\int_{-\infty}^0 f(c e^{-\alpha x}\chi_{\{x > 0\}}(x))f_X(x)dx\\ =& \int_{0}^\infty f(c e^{-\alpha x})f_X(x)dx\\ &+f(0)\int_{-\infty}^0 f_X(x)dx\\ =& \frac{1}{\alpha}\int_0^c \frac{f(s)}{s}f_X\left(-\frac{1}{\alpha}\log\frac{s}{c}\right)ds + f(0)\int_{-\infty}^0 f_X(x)dx \end{align*}$$ As comparing with my book we should get $$f_Y(y) = \frac{1}{\alpha}\frac{1}{y}f_X\left(-\frac{1}{\alpha}\log\frac{y}{c}\right) \chi_{(0,c)}(y)$$ which can be read off from the first term. However, what does happen with the second term? Does it just not appear in the density of $Y$ because $f(0)$ is constant?