How is the simplified version (below) of the Bromwich inverse Laplace transform integral derived? I do not understand how the last equality is derived from the previous.
Apparently the first term in the integral (involving cos) is equivalent to the second (involving sin)!! How so??  I DO understand how the integral range is halved (since F(s)* =F(s*);where F(s) is the Laplace transform of f(t), and *=complex conjugate.  And the imaginary part must equal zero, and is dropped in the last 2 lines...OK.
Any help would be appreciated since this form is used often in numerical inverse Laplace transform algorithms. Thanks... M D Mill
[Note:f"hat"(s)= the Laplace transform of f(t), s=(a+iu)
[Note: This question has been asked before but not answered adequately, and I     have added some new specificity.]
\begin{align}
f(t) &= \frac{\mathrm{e}^{at}}{2\pi j} \int_{-\infty}^{\infty} \bigl( \cos(ut) + i \sin(ut) \bigr) \hat{f}(a + iu) \, j\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}
These equations are from the web source:Abate and Whitt, 1995
http://www.columbia.edu/~ww2040/LaplaceInversionJoC95.pdf
 A: The inverse Laplace Transform $\hat f(a+iu)$ is given by
$$\hat f(a+iu)=\int_{-\infty}^\infty f(t)e^{-at}\cos(ut)\,dt-i \int_{-\infty}^\infty f(t)e^{-at}\sin(ut)\,dt$$
Hence, the real part of $\hat f(a+iu)$ is an even function of $u$ while the imaginary part of $\hat f(a+iu)$ is an odd function of $u$.

Moreover, since $f(t)=0$ for $t<0$, then $f(t)=2f_e(t)=2f_0(t)$ where $f_e(t)=\frac{f(t)+f(-t)}{2}$ is the even part of $f(t)$ and $f_o(t)=\frac{f(t)-f(-t)}{2}$ is the odd part of $f(t)$.
Therefore, since $f(t)$ is given by
$$f(t)=\frac{e^{at}}{2\pi}\int_{-\infty}^\infty \left(\text{Re}(\hat f(a+iu))\cos(ut)-\text{Im}(\hat f(a+iu))\sin(ut) \right)\,du$$
then we see that 
$$\begin{align}
f(t)&=2f_e(t)\\\\
&=f(t)+f(-t)\\\\
&=\frac{e^{at}}{\pi}\int_{-\infty}^\infty \text{Re}(\hat f(a+iu))\cos(ut)\,du\\\\
&=\frac{2e^{at}}{\pi}\int_{0}^\infty \text{Re}(\hat f(a+iu))\cos(ut)\,du
\end{align}$$
and 
$$\begin{align}
f(t)&=2f_o(t)\\\\
&=f(t)-f(-t)\\\\
&=-\frac{e^{at}}{\pi}\int_{-\infty}^\infty \text{Im}(\hat f(a+iu))\sin(ut)\,du\\\\
&=-\frac{2e^{at}}{\pi}\int_{0}^\infty \text{Im}(\hat f(a+iu))\sin(ut)\,du
\end{align}$$
