Special case of double integral where the upper bond of the inner integral is the integration variable of the out integral This is regarding a special type of double integral looking like the following:
$$\int_0^T  \exp\left(\int_0^\color{red}{t} f(s)\, ds\right) \,\color{red}{dt}. $$
Time being the integration variable here, one can think of $T$ as entire duration which sets the boundary of the outer integral, and the inner integral integrates from $0$ to each discrete time point $t$, and take the exponential of it. Here, the upper bond of the inner integral is the integration variable of the outer integral.
I suspect the expression can be simply to:
$$ \exp\left(\int_0^T f(t)\, dt\right). $$
but I'm not sure if this is correct. Can anyone enlighten me with the missing links in between and correct me if my suspicion is wrong.
You can look at it as
$$\int\limits_{0}^{T}e^{\int\limits_{0}^{t}f(s) \mathrm{d}s} \mathrm{d}t$$
for f$(s) = -\frac{s}{2}$, for example, you obtain well known
$$\int\limits_{0}^{T}e^{-\frac{t^2}{2}} \mathrm{d}t$$
You can look at it as
$$\int\limits_{0}^{T}e^{\int\limits_{0}^{t}f(s) \mathrm{d}s} \mathrm{d}t$$
for f$(s) = -\frac{s}{2}$, for example, you obtain well known
$$\int\limits_{0}^{T}e^{-\frac{t^2}{2}} \mathrm{d}t$$