Integration problem involving $f_{\theta}(x) = \frac{1}{\theta} e^{-\frac{x}{\theta}}$: $\int_0^{\infty} f_{\theta}^2(x) e^{\alpha f_{\theta}(x)}dx$

Question

Given the probability density function (p.d.f.) of an exponential distribution, i.e., $$f_{\theta}(x) = \frac{1}{\theta} e^{-\frac{x}{\theta}}, \; \; 0 < x < \infty; \;\; \theta > 0,$$ how do I solve the following integration ? $$\int_0^{\infty} f_{\theta}^2(x) e^{\alpha \,f_{\theta}(x)} \; \; dx, \; \; \alpha \in \mathbb{R}; \; \; \mathbb{R}\; \textrm{is the real number line.}$$

Answer 1

By change of variable $$ u=e^{\large -\frac{x}{\theta}},\quad du=-\frac1{\theta}e^{\large -\frac{x}{\theta}}dx, $$ one gets $$ \begin{align} \int_0^{\infty} f_{\theta}^2(x) e^{\alpha \,f_{\theta}(x)} \; dx&=\frac1{\theta^2}\int_0^{\infty} e^{\large -\frac{x}{\theta}}e^{\frac{\alpha}{\theta} \,e^{\large -\frac{x}{\theta}}} \; e^{\large -\frac{x}{\theta}}dx \\\\&=\frac1{\theta}\int_0^1 u\:e^{\frac{\alpha}{\theta}u} \; du \\\\&=\frac{e^{\alpha /\theta } (\alpha -\theta )+\theta }{\alpha ^2}, \end{align} $$ where we have made an integration by parts on the last step.