Since $\exp(\cdot)$ is locally Lipschitz, the following SDE has a strong solution:
$$
\mathrm{d}X\_s=\exp(X\_s) \, \mathrm{d}B\_s,\quad X\_0=1,
$$
where $B$ is a standard Brownian motion. I wonder if the following expression holds:
$$\mathbb{E}\int\_0^T\exp(2X\_s) \, \mathrm d s<\infty.$$



Existence of solution for a non-linear SDE
Let us show that  

\begin{equation\*}
 E\int\_0^T e^{2X\_t}\,dt=\infty \quad\text{for real }T\ge T\_\*:=e^{-2}/2. \tag{1}\label{1}
\end{equation\*}


Indeed, letting $Y\_t:=e^{2X\_t}$, we have
$Y\_0=e^2$ and, by [Itô's lemma](https://en.wikipedia.org/wiki/It%C3%B4%27s_lemma#Mathematical_formulation_of_It%C3%B4%27s_lemma), for real $t\ge0$  

\begin{equation\*}
 dY\_t=2e^{4X\_t}dt+2e^{3X\_t}dB\_t= 2Y\_t^2dt+2e^{3X\_t}dB\_t. 
\end{equation\*}
So, for real $t$ and $u$ such that $0\le t\le u$ and $m(t):=EY\_t$ we have $m(0)=e^2$ and
\begin{equation\*}
 m(u)-m(t)=2\int\_t^u EY\_s^2ds\ge2\int\_t^u m(s)^2ds. 
\end{equation\*}
So, on any interval $[0,T)$ where the function $m$ is finite, $m\ge m(0)=e^2>0$ and $m$ is a continuous increasing function. Moreover, then for any $t\in[0,T)$
\begin{equation\*}
 m'\_+(t):=\liminf\_{u\downarrow t}\frac{m(u)-m(t)}{u-t}\ge 2m(t)^2. 
\end{equation\*}
So, for $h:=-1/m$ and any $t\in[0,T)$
\begin{equation\*}
 h'\_+(t)=\liminf\_{u\downarrow t}\frac{h(u)-h(t)}{u-t}
 =\liminf\_{u\downarrow t}\frac{m(u)-m(t)}{u-t}\frac1{m(u)m(t)}\ge2. 
\end{equation\*}
So,
\begin{equation\*}
 0>h(T)\ge h(0)+2T=-e^{-2}+2T,
\end{equation\*}
whence $T<T\_\*$. So, if $T\ge T\_\*$, then the increasing function $m$ is not finite on $[0,T)$. So, $m(t)=EY\_t=Ee^{2X\_t}=\infty$ for all $t$ in a left neighborhood of $T$ if $T\ge T\_\*$. So, we get \eqref{1}.
$\quad\Box$


