# How is the simplified version 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 $\mathrm{cos}$) is equivalent to the second (involving $\mathrm{sin}$)!! How so??

I DO understand how the integral range is halved (since $F(s)^*=F(s^*)$; where $F(s)$ is the Laplace transform of $f(t)$. Any help would be appreciated since this form is used often in numerical inverse Laplace transform algorithms.

[Note: $\hat{f}(s)$ below represents the Laplace transform of $f(t)$]

This quote is from the web source Abate and Whitt, 1995.

I agree that the derivation is unclear, yet the final result is correct (for $t>0$, see below). There are two conditions that are necessary for the final result to be true:

1. $f(t)$ is real-valued
2. $f(t)$ is causal

The step from line 2 to line 3 in the derivation assumes that $f(t)$ is real-valued, i.e., only the real part of the integrand is considered. The last step leading to the final result assumes causality of $f(t)$, i.e., $f(t)=0$ for $t<0$.

From the third line in the derivation we have for real-valued $f(t)$

$$f(t)=\frac{e^{at}}{2\pi}\int_{-\infty}^{\infty}\left[\text{Re}\left(\hat{f}(a+iu)\right)\cos(ut)-\text{Im}\left(\hat{f}(a+iu)\right)\sin(ut)\right]du\tag{1}$$

and, consequently,

$$f(-t)e^{2at}=\frac{e^{at}}{2\pi}\int_{-\infty}^{\infty}\left[\text{Re}\left(\hat{f}(a+iu)\right)\cos(ut)+\text{Im}\left(\hat{f}(a+iu)\right)\sin(ut)\right]du\tag{2}$$

Since for causal $f(t)$ we have $f(-t)=0$ for $t>0$, we can write

$$f(t)=f(t)+f(-t)e^{2at},\qquad t>0\tag{3}$$

Consequently, adding $(1)$ and $(2)$ gives

$$f(t)=\frac{e^{at}}{\pi}\int_{-\infty}^{\infty}\text{Re}\left(\hat{f}(a+iu)\right)\cos(ut)du,\qquad t>0\tag{4}$$

And since for real-valued $f(t)$ the integrand in $(4)$ is even, we finally obtain

$$f(t)=\frac{2e^{at}}{\pi}\int_{0}^{\infty}\text{Re}\left(\hat{f}(a+iu)\right)\cos(ut)du,\qquad t>0\tag{5}$$

q.e.d.

Note that this is result is only valid for $t>0$, which is not stated in the paper you quoted. Of course, for $t<0$ we have $f(t)=0$.

Also note that instead of $(3)$ we could have written

$$f(t)=f(t)-f(-t)e^{2at},\qquad t>0\tag{6}$$

from which we can conclude that

$$f(t)=-\frac{e^{at}}{\pi}\int_{-\infty}^{\infty}\text{Im}\left(\hat{f}(a+iu)\right)\sin(ut)du,\qquad t>0\tag{7}$$

and, taking into account that for real-valued $f(t)$ the integrand is even, we get

$$f(t)=-\frac{2e^{at}}{\pi}\int_{0}^{\infty}\text{Im}\left(\hat{f}(a+iu)\right)\sin(ut)du,\qquad t>0\tag{8}$$

Comparing $(5)$ with $(8)$ we see the equivalence

$$\int_{0}^{\infty}\text{Re}\left(\hat{f}(a+iu)\right)\cos(ut)du=-\int_{0}^{\infty}\text{Im}\left(\hat{f}(a+iu)\right)\sin(ut)du\qquad t>0\tag{9}$$