I have trouble following the derivation of equation $(2)$ in this paper.
The authors define the Laplace transform of a real-valued function $f(t)$ of a positive real variable $t$ as
\begin{equation} \hat{f}(s) = \int_{0}^\infty \mathrm{e}^{-st} f(t) \, \mathrm{d}t. \end{equation}
To invert the Laplace transform, the Bromwich integral is used, with $i^2 = -1$,
\begin{equation} f(t) = \frac{1}{2\pi i} \lim_{T \to \infty} \int_{a - iT}^{a + iT} \mathrm{e}^{st} \hat{f}(s) \, \mathrm{d}s. \tag{$*$} \end{equation}
In deriving a simplification of $(*)$, some steps are unclear to me. Is the integral in blue equal to zero because $f(t)$ is real-valued? How does one arrive at the result in red?
\begin{align} f(t) &= \frac{\mathrm{e}^{at}}{2\pi} \int_{-\infty}^{\infty} \bigl( \cos(ut) + i \sin(ut) \bigr) \hat{f}(a + iu) \, \mathrm{d}u \\ &= \frac{\mathrm{e}^{at}}{2\pi} \int_{-\infty}^{\infty} \bigl( \cos(ut) + i \sin(ut) \bigr) \bigl( \operatorname{Re}(\hat{f}(a + iu)) + i \operatorname{Im}(\hat{f}(a + iu)) \bigr) \, \mathrm{d}u \\ &= \frac{\mathrm{e}^{at}}{2\pi} \int_{-\infty}^{\infty} \bigl( \operatorname{Re}(\hat{f}(a + iu)) \cos(ut) - \operatorname{Im}(\hat{f}(a + iu)) \sin(ut) \bigr) \, \mathrm{d}u \\ &\quad + i\frac{\mathrm{e}^{at}}{2\pi} \color{blue}{\int_{-\infty}^{\infty} \bigl( \operatorname{Im}(\hat{f}(a + iu)) \cos(ut) + \operatorname{Re}(\hat{f}(a + iu)) \sin(ut) \bigr) \, \mathrm{d}u} \\ &= \frac{\mathrm{e}^{at}}{2\pi} \int_{-\infty}^{\infty} \bigl( \operatorname{Re}(\hat{f}(a + iu)) \cos(ut) - \operatorname{Im}(\hat{f}(a + iu)) \sin(ut) \bigr) \, \mathrm{d}u \\ &= \color{red}{\frac{2\mathrm{e}^{at}}{\pi} \int_{0}^{\infty} \operatorname{Re}(\hat{f}(a + iu)) \cos(ut) \, \mathrm{d}u} \end{align}
Edit Using Dr.MV's comments:
\begin{align} &\frac{\mathrm{e}^{at}}{2\pi} \int_{-\infty}^{\infty} \bigl( \operatorname{Re}(\hat{f}(a + iu)) \cos(ut) - \operatorname{Im}(\hat{f}(a + iu)) \sin(ut) \bigr) \, \mathrm{d}u \\ &= \frac{\mathrm{e}^{at}}{2\pi} \Bigl( \int_{0}^{\infty} \bigl( \operatorname{Re}(\hat{f}(a + iu)) \cos(ut) - \operatorname{Im}(\hat{f}(a + iu)) \sin(ut) \bigr) \, \mathrm{d}u \\ &\qquad\quad + \int_{-\infty}^{0} \bigl( \operatorname{Re}(\hat{f}(a + iu)) \cos(ut) - \operatorname{Im}(\hat{f}(a + iu)) \sin(ut) \bigr) \, \mathrm{d}u \Bigr) \\ &= \frac{\mathrm{e}^{at}}{2\pi} \Bigl( \int_{0}^{\infty} \bigl( \operatorname{Re}(\hat{f}(a + iu)) \cos(ut) - \operatorname{Im}(\hat{f}(a + iu)) \sin(ut) \bigr) \, \mathrm{d}u \\ &\qquad\quad - \int_{\infty}^{0} \bigl( \operatorname{Re}(\hat{f}(a - iv)) \cos(-vt) - \operatorname{Im}(\hat{f}(a - iv)) \sin(-vt) \bigr) \, \mathrm{d}v \Bigr) \\ &= \frac{\mathrm{e}^{at}}{2\pi} \int_{0}^{\infty} \Bigl( \cos(ut)\bigl( \operatorname{Re}(\hat{f}(a - iu)) + \operatorname{Re}(\hat{f}(a + iu)) \bigr) \\ &\hspace{4.5em}+ \sin(ut)\bigl( \operatorname{Im}(\hat{f}(a - iu)) - \operatorname{Im}(\hat{f}(a + iu)) \bigr) \Bigr) \, \mathrm{d}u. \end{align}
I do not see how this equals the part in red. Especially, the factor $\frac{2\mathrm{e}^{at}}{\pi}$ I cannot explain.