Compute expectation of integral with respect to Brownian motion $\mathbb E(|\int_0^te^sdB_s|)$

Question

Let $B_s$ the standard Brownian motion. Is it possible to compute $$\mathbb E(|\int_0^te^sdB_s|)$$

or at least is it possible to find un upper bound of this quantity?

Answer 1

By Jensen inequality and using that $\langle B \rangle_s = s$ $$\Bbb E [ | \int_0^t e^s \text d B_s| ] = \sqrt{(\Bbb E [ | \int_0^t e^s \text d B_s| ])^2} \leq \sqrt{\Bbb E [ ( \int_0^t e^s \text d B_s)^2 ]} = \sqrt{\Bbb E [\int_0^te^{2s}\text d s]} = \sqrt{\frac{e^{2t} - 1}{2}}$$

Answer 2

Note that $\int_0^te^sdB_s$ is of a normal distribution $N(0,\sigma)$, with $\sigma = \sqrt{\frac{e^{2t} - 1}{2}}$. Thus,

$$\mathbb E\left[\left|\int_0^te^sdB_s \right|\right]=\frac{2}{\sqrt{2\pi\sigma^2}}\int_0^\infty x e^{-\frac{x^2}{2\sigma^2}}dx = \sqrt{ \frac{e^{2t} - 1}{\pi} } $$