How is the simplified version (below) of the Bromwich inverse Laplace transform integral derived?

Question

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

Answer 1

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 $u$.

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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

$$\begin{align} f(t)&=2f_o(t)\\\\ &=f(t)-f(-t)\\\\ &=-\frac{e^{at}}{\pi}\int_{-\infty}^\infty \text{Im}(\hat f(a+iu))\sin(ut)\,du\\\\ &=-\frac{2e^{at}}{\pi}\int_{0}^\infty \text{Im}(\hat f(a+iu))\sin(ut)\,du \end{align}$$