# 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)

\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

• hopefully @Ritz will be nofitied of this solution to his original question of several years past Jun 8, 2017 at 5:09

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}

• This fact only gets us half way to the last line...it does not explain why the cos integral term equals the sin integral term in the next to last line. Perhaps the original (much respected) authors were simply wrong? This is a very important question, and still puzzling to me! Jun 7, 2017 at 18:43
• I've edited to show the coveted equality. Jun 8, 2017 at 1:34
• I understood that f(t) was real (and the implication)...I did not see the implication of it being CAUSAL! Bingo, that was the key,,,many thanks Jun 8, 2017 at 3:00
• You're welcome. My pleasure. Jun 8, 2017 at 3:03