Is $\mathbb{E}\exp \left( k \int_0^T B_t^2 \, dt \right)<\infty$ for small $k>0$? Suppose that $B$ is a Brownian motion. Does it hold that
\begin{equation}
\mathbb{E}\left[\exp\left(k\int_0^T[B(t)]^{2}\,dt\right)\right] <\infty\text{ ?}
\end{equation}
for some positive constant $k$?
my idea: I think that $\int_{0}^{T}[B(t)]^{2}dt$ is actually a Normal random variable $X\sim N(\frac{T^{2}}{2},\sigma(T))$ where $\sigma(T)$ is some function of $T$. Then, we know that $\mathbb{E}[\exp(\theta X)]$ is finite for such $X$. Am I correct?

Obviously,
$$\int_0^T B_t^2 \, dt \leq T \cdot \sup_{t \leq T} B_t^2$$
Since
$$\sup_{t \leq T} B_t^2 \leq \left(\sup_{t \leq T} B_t \right)^2 + \left( \inf_{t \leq T} B_t \right)^2$$
and
$$\sup_{t \leq T} B_t \sim - \inf_{t \leq T} B_t \sim |B_T|$$
(this is a direct consequence of the reflection principle), we get by the Cauchy-Schwarz inequality
$$\begin{align*} \mathbb{E}\exp \left(k \int_0^T B_t^2 \, dt \right) &\leq \sqrt{\mathbb{E}\exp \left(2kT \left[ \sup_{t \leq T} B_t \right]^2 \right)} \sqrt{\mathbb{E}\exp \left(2kT \left[ \inf_{t \leq T} B_t \right]^2 \right)}\\ &= \mathbb{E}\exp \left(2kT B_T^2 \right). \end{align*}$$
Finally, as $B_T \sim N(0,T)$, we observe that
$$\mathbb{E}\exp (2kT B_T^2) = \int_{\mathbb{R}} e^{2kT x^2} \frac{1}{\sqrt{2\pi T}} \exp \left(- \frac{x^2}{2T} \right) \, dx$$
is finite if $2k T< \frac{1}{2T}$, i.e. if $k < \frac{1}{4T^2}$. 
