Requesting Clarification on how Expectation for fbm with H = 1 is being computed in a paper

I am having some trouble following how the expectation is being computed in the proof of Lemma 2.3 in this paper. For your convenience I am reproducing the relevant part here. Let $B(t)$ denote an fBm process with $H=1$, then since $B(t) = t \mathcal{N}$, where $\mathcal{N}$ is a standard Normal random variable, then \begin{eqnarray} \mathrm{E}\left[\exp\left(\sup_{t\in[0,\sqrt{2}aT]}\left(\mathcal{N}t-\frac{1}{2}t^2\right)\right)\right] &=& \int_{-\infty}^{0}\exp(0)\frac{1}{\sqrt{2\pi}}\mathrm{e}^{-\frac{x^2}{2}}dx\\ &+& \int_{0}^{\sqrt{2}aT}\exp(\frac{x^2}{2})\frac{1}{\sqrt{2\pi}}\mathrm{e}^{-\frac{x^2}{2}}dx\\ &+& \int_{\sqrt{2}aT}^{\infty}\exp(T\sqrt{2}ax - a^2T^2)\frac{1}{\sqrt{2\pi}}\mathrm{e}^{-\frac{x^2}{2}}dx. \end{eqnarray} I can follow that $B(t) = t \mathcal{N}$, but after this, I am not able to follow, the following 2 things about how the expectation is being computed.

1) How the integral has been broken into three regions with different $\exp\left(*\right)$ parts multiplying the standard normal density. 2) How can the integration in this case be done with just the standard normal density over the entire real line.

It would be of great help if anyone can please shed some light why this is true. Thank you :-)

The function $xt-\frac{1}{2}t^2$ looks like a parabola opening down, with vertex at $x$. If $x \le 0$ then the function is decreasing on $[0, \sqrt{2}aT]$ so its maximum occurs at $t=0$. If $x \ge \sqrt{2}aT$ then its maximum occurs at $t=\sqrt{2}aT$. Otherwise it occurs at $x$.