I would like to find the extrema of the following integral with respect to $u\left(s\right)$:
$\int_0^T\exp\left\{\int_0^s\left[\frac{1}{2\gamma^2}\left(u\left(v\right)+\lambda+\left(1-\gamma\right)\frac{1-e^{-\kappa\left(s-v\right)}}{\kappa}\sigma\right)^2+\frac{1-\gamma}{2\gamma\theta}u\left(v\right)^2-\frac{1}{2\gamma}\left(u\left(v\right)+\lambda\right)^2\right] dv\right\}ds$
where $\gamma$, $\lambda$, $\kappa$, $\sigma$ and $\theta$ are constants. The Euler-Lagrange equation is unfortunately not directly applicable here; I also tried with substituting $w\left(s\right)\triangleq u\left(s\right)^2$, but it does not simplify things either. Any hint is greatly appreciated!
Hints:
OP's functional is on the form $$ I[u;T]~:=~\int_0^T\! ds ~e[u;s], \tag{1} $$ $$ e[u;s]~:=~\exp(S[u;s]), \tag{2} $$ $$ S[u;s]~:=~\int_0^s \!dv~L(s,v,u(v)),\tag{3} $$ for some function $L: \mathbb{R}^3\to \mathbb{R}$.
Deduce that $$\forall v\in[0,T]:~~ \int_v^T\!ds ~e[u;s]\frac{\partial L(s,v,u(v))}{\partial u(v)}~=~0 \tag{4} $$ is a necessary condition for a stationary solution $u(v)$.
In OP's case, the $L$ function is quadratic in $u$. By scaling the $u$ variable, we may assume that $$L(s,v,u)~= \frac{1}{2}u^2+f(s,v)u+g(s,v),\tag{5}$$ for some functions $f,g:\mathbb{R}^2\to \mathbb{R}$.
The necessary condition (4) then becomes $$\forall v\in[0,T]:~~ u(v)~=~-\frac{\int_v^T\!ds ~e[u;s] f(s,v)}{\int_v^T\!ds ~e[u;s]} .\tag{6}$$