# Solving a general integral (expectation of some variant of exponential distribution)

Suppose $X$ is distributed exponentially with parameter $\lambda$. Its pdf is $\lambda e^{-\lambda x}$, and the calculation of its expectation is straight forward: $\mathbb{E}(X) = \int_0^\infty x\lambda e^{-\lambda x} dx=\lambda^{-1}$ (using integration by parts).

I now introduce this variant of the exponential distribution: let $H:[0,\infty)\to[0,\infty)$ be an increasing continuous bijection, and define $Y=H^{-1}(X)$. If I am not mistaken, the pdf of $Y$ is $\lambda e^{-\lambda H(x)}$. The idea behind the distribution of $Y$ is that the time, a term which is natural to the exponential distribution, moves not linearly.

Calculating the expectation of $Y$, in general, looks impossible. It is the result of $\int_{0}^\infty x\lambda e^{-\lambda H(x)}dx$. However, with some assumptions on $H$ I guess this can be solved.

The difficult question is this: can this integral be solved for $H(x)=\int_0^x h(t)dt$ such that $h$ is a rational function with $h(0)=0$?

The simpler question is this: can this integral be solved for $H(x)$ which is a polynomial?

Actually the PDF of $Y$ is the function $g$ defined by $$g(y)=\lambda h(y)\mathrm e^{-\lambda H(y)},$$ where $h$ is the derivative of $H$, if $H$ is differentiable. In the general case, the CDF of $Y$ is the function $G$ defined by $$G(y)=P(Y\leqslant y)=P(X\leqslant H(y))=1-\mathrm e^{-\lambda H(y)}.$$ In particular, $$E(Y)=\int_0^\infty P(Y\gt y)\mathrm dy=\int_0^\infty\mathrm e^{-\lambda H(y)}\mathrm dy.$$