I have the following integral to evaluate:
$\int_{A_0,0}^{y(t),t} e^te^{y}\dot{y}dt$
where $y$ is time dependent or $y=y(t)$.
Here are my two different attempts and neither gave me a correct answer,
$\int_{A_0,0}^{y(t),t} e^te^{y}\dot{y}dt=e^te^y|^{y(t),t}_{A_0,0}-\int_{A_0,0}^{y(t),t} e^te^{y}dt=e^te^y|^{y(t),t}_{A_0,0}-e^te^y|^{y(t),t}_{A_0,0}+\int_{A_0,0}^{y(t),t} e^te^{y}\dot{y}dt$
then,
$0=0$
Another attempt is by using Gamma function,
$\int_{A_0,0}^{y(t),t} e^te^{y}\dot{y}dt=\int_{A_0,0}^{y(t),t}e^t\Gamma(1,-y)\dot y dt=\int_{A_0,0}^{y(t),t}e^t\int_{-y}^{\infty}e^{-t}\dot y dt dt=\int_{A_0,0}^{y(t),t}\int_{-y}^{\infty}e^te^{-t}\dot y dt dt=\int_{A_0,0}^{y(t),t}\int_{-y}^{\infty}\dot y dt dt=\infty$
I have a third idea of treating the integral as a transform integral but how to do the transform its not clear to me. Is it possible to evaluate this integral analytically and/or numerically? Any ideas or suggestions to try?
I don't understand your gamma attempt. Anyway, here goes mine. $\int_{A_0,0}^{y(T),T} e^t e^y \dot y dt = e^{T+y(T)}-e^{A_0} - \int_{A_0,0}^{y(T),T} e^t e^y dt$. Now, for that integral, try the new variable $x=e^t$, so $dx=e^tdt$ and let $\bar y(x) = y(t(x))=y(log(x))$, and substituting, we get $\int_{A_0,0}^{y(T),T} e^t e^y dt = \int_{A_0,1}^{\bar y(e^T),e^T} e^{\bar y(x)} dx$.
So, being able to (analitically) solve your integral seems to be equivalent to being able to solve this last one which, generically, is impossible (take $\bar y(x) = x^2$, for example).
For a numerical approach, many softwares could help you if you know the function y(x), but if you don't, that integral seems to strongly depend on the function $y(t)$.
Also, if you call $z(T)$ your integral, you get $\dot z = e^t e^y \dot y$, with the obvious initial conditions (becomes linear with $m=e^{y(t)}$. So if you happen to have an ODE for $y$, you could solve both simultaneously, or even more easily numerically.
I hope that helps. Where did you come across that integral?