0
$\begingroup$

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 $f(t)$]

enter image description here

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

$\endgroup$
2
$\begingroup$

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

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.