I am interested in knowing under which conditions the following integral has a closed-form solution:
$$ I_1 = \int_0^T{f(x)e^{-\left(F(x)+\alpha x\right)}dx} $$
where:
$$ \frac{dF}{dx}(x) = f(x) $$
Assume that both $F(x)$ and $f(x)$ are well defined over the interval $[0,\infty[$.
I have solved it when the function $f$ is constant $-$ $f(x) = \beta$ $-$ but for other functional forms $-$ e.g. $f(x) = x$, $f(x)=\sqrt{x}$, etc. $-$ I have been unable to find a solution $-$ I have tried integration by parts and change of variable.
Could anyone describe how this problem should be approached? Is there a general solution to this integral and, if so, under which conditions does that solution hold? I am looking for a closed-form solution.
FYI, the integral $I_1$ comes from the following integral:
$$ I_2 = \int_0^T{f(x)e^{-\int_0^x{(f(y)+\alpha) dy}}dx} $$
So I am really trying to solve $I_2$. Up to now I have only considered functions for which $F(0) = 0$, but if there is a general solution to $I_2$ I will be more than happy to hear it!
Edit: how about these cases?:
$$ I_1 = \int_0^T{xe^{-\left(\frac{1}{2}x^2+\alpha x\right)}dx} $$
$$ I_1' = \int_0^T{\sqrt{x}e^{-\left(\frac{2}{3}x^{3/2}+\alpha x\right)}dx} $$
$$ I_1'' = \int_1^T{\frac{1}{x}e^{-\left(\ln{x}+\alpha x\right)}dx} $$
Edit 2: as pointed out by @Nox, the integral $I_1$ has a solution involving the error function:
$$ e(T) = \frac{2}{\sqrt{\pi}}\int_0^Te^{-x^2}dx $$
To obtain it, it is necessary to add and subtract $\int_0^T\alpha e^{-(\frac{x^2}{2}+\alpha x)}dx$ to $I_1$ and make the following change of variable to this integral:
$$ u = \frac{\alpha+x}{\sqrt{2}}$$
Edit 3: for information, these integrals arise as the expectation of the following random variable:
$$ \mathbb{E} \left[e^{-\alpha \ \tau} \, \mathbb{I}_{\{\tau \ \leq \ T \, \}}\right] $$
The random variable $\tau$ is a stopping time defined as follows: let $N(t)$ be an inhomogeneous Poisson point process parameterised by the intensity function $f(x)$ such that for any $t>0$ and assuming $F(0)=0$:
$$ F(t) = \int_0^t f(x)dx < \infty $$
$$ \mathbb{P}(N(t) = n) = \frac{F(t)^n}{n!}e^{-F(t)}$$
It comes:
$$ \mathbb{P}(N(t) = 0) = e^{-F(t)} $$
The stopping time $\tau$ is then defined as:
$$ \tau = \min \{t>0 : N(t)>0\}$$
The distribution of $\tau$ is given by:
$$ \begin{align} & \mathbb{P}(\tau > t) = \mathbb{P}(N(t) = 0) = e^{-F(t)} \\[12pt] & \Rightarrow \mathbb{P}(\tau \leq t) = 1-e^{-F(t)} \\[12pt] & \Rightarrow \frac{d\mathbb{P}}{dt} = f(t)e^{-F(t)} \end{align} $$
Hence we have:
$$ \begin{align} \mathbb{E} \left[e^{-\alpha \ \tau} \, \mathbb{I}_{\{\tau \ \leq \ T \, \}}\right] & = \int_0^{\infty} e^{-\alpha \ t} \, \mathbb{I}_{\{t \ \leq \ T \, \}} f(t)e^{-F(t)} dt \\[12pt] & = \int_0^T f(t)e^{-(F(t)+\alpha t)} dt \end{align} $$
The question can't be answered for definite integrals because for some of them can have closed-form solutions for specific values of $T$ (such as $\infty$) and there is no general theory about this.
Now, by parts,
$$\int F'(x)e^{-F(x)}e^{-ax}dx=-e^{-F(x)}e^{-ax}+a\int e^{F(x)+ax}dx.$$
In general, the last integral has no closed-form, except for a constant, a multiple of $x$, or the logarithm of a function, which can make it tractable by pulling out of the exponential:
$$\int e^{\log g(x)+ax}dx=\int g(x)e^{ax}dx,$$ for example when $g$ is a polynomial or a trigonometric polynomial, or a combination of these.