I want to calculate the following integral:
$\int e^{a+bx} \ f(x)\ dx$
where $f(x)$ is the probability distribution of $X$. Is there a way to calculate this integral?
P.S. I don't want to use any specific density function, but rather I prefer to have an answer with $F(x)$ in it.
Thank you.
By parts,
$$\int e^{a+bx} f(x)\,dx=\frac1be^{a+bx} f(x)-\int \frac1be^{a+bx} f'(x)\,dx.$$
Then by induction
$$\int e^{a+bx} f(x)\,dx=\sum_{k=0}^\infty\frac{(-1)^{k}}{b^{k+1}}e^{a+bx} f^{(k)}(x).$$
Or
$$\int e^{a+bx} f(x)\,dx=e^{a+bx}f^{(-1)}(x)-b\int e^{a+bx} f^{(-1)}(x)\,dx=\cdots\\ =e^{a+bx}\sum_{k=0}^\infty (-b)^kf^{(-k-1)}(x)$$
provided the integral terms vanish in the limit. ($f^{(-k)}$ denotes the $k^{th}$ antiderivative.)
There is not much better that you can do.
We don't have a general closed form for all density functions.
One may observe that considering any integrable function $g$ one may write $$ \begin{align} \int g(x)\:dx =\int e^{a+bx} \cdot e^{-a-bx} g(x)\:dx =\int e^{a+bx} \cdot f(x)\:dx \end{align} $$ with $$ f(x):=e^{-a-bx} g(x). $$ Thus having a general answer would mean that we have a general closed form of any antiderivative of $g$.
We do have a general closed form only for some families: $f(x):=cx^\alpha, f(x):=e^{\alpha x+\beta},\cdots$.