2
$\begingroup$

I have a function $f(|\boldsymbol{k}|,s,\theta)$ for which I am interested in its inverse Laplace transform. I am also interested in the function's mean value for constant $|\boldsymbol{k}|$, but technically I need to inverse Laplace transform first. I was wondering about the interchangeability of the inverse transform and the mean; that is, does:

\begin{equation} \frac{1}{2\pi}\int_0^{2\pi} \mathcal{L}^{-1}\left\{f(|\boldsymbol{k}|,s,\theta)\right\}\mathrm{d}\theta=\mathcal{L}^{-1}\left\{\frac{1}{2\pi}\int_0^{2\pi} f(|\boldsymbol{k}|,s,\theta)\mathrm{d}\theta\right\} \end{equation}
The inverse Laplace transform is w.r.t. $s$.

I could think of a few reasons why it may not hold; for example, would it hold if doing the mean integration first affects the poles of $s$ (if that could happen)?

Please forgive my lack of knowledge in math, I do not know much measure theory and am not sure of the formality of interchanging these two operators. Any help is greatly appreciated!

EDIT: The actual function looks like:

\begin{equation} f(|\boldsymbol{k}|,s,\theta)=\frac{\lambda_2 \left(D |\boldsymbol{k}|^2+2 \lambda_1+s\right)+\lambda_1 (i \boldsymbol{k}\cdot \boldsymbol{v}+\lambda_1+s)+\lambda_2^2}{(\lambda_1+\lambda_2) \left(\left(D |\boldsymbol{k}|^2+s\right) (i \boldsymbol{k}\cdot\boldsymbol{v}+\lambda_1+s)+\lambda_2 (s+i\boldsymbol{k}\cdot\boldsymbol{v} )\right)} \end{equation} All variables are real and positive except for $s$, of course. $\theta$ is the angle between $\boldsymbol{v}$ and $\boldsymbol{k}$. Carrying out the mean integration first gives: \begin{equation} f(|\boldsymbol{k}|,s)=\frac{\lambda_1}{(\lambda_1+\lambda_2) \left(D |\boldsymbol{k}|^2+\lambda_2+s\right)} \end{equation} which makes it easy to then find the inverse Laplace transform. Not sure if this is at all helpful.

$\endgroup$
  • 1
    $\begingroup$ Why can't you cancel the $|k|$ in the denominator and in the integral? $\endgroup$ – grdgfgr Jun 1 '15 at 8:35
  • $\begingroup$ Sorry that's right... I just wanted to show explicitly the expression for the mean. Let me change it.. $\endgroup$ – Simon Sehayek Jun 1 '15 at 8:36
  • 1
    $\begingroup$ I am not sure if this is something that would help you, but, F(s=0) is equal to the integral of all f(t). Insert s=0 into the laplace transform integral, it just becomes the area under f(t). $\endgroup$ – grdgfgr Jun 1 '15 at 8:44
  • $\begingroup$ Also, I see no reason why you should not be able to interchange the order. $\endgroup$ – grdgfgr Jun 1 '15 at 8:50
  • $\begingroup$ I just checked and the real part of my poles seem to be affected if I carry out the mean integration first. Wouldn't this change the limit in the definition of the inverse transform? $\endgroup$ – Simon Sehayek Jun 1 '15 at 8:52
0
$\begingroup$

I see no reason we should be able to change the order of integration (there might be convergence issues, but, in general, it should work). A simple example to demonstrate this:

$$\frac{s}{k^2\sin (\theta)^2+s^2}\rightarrow \cos(k\sin (\theta)t)$$

$$\int_0^{2 \pi } \frac{s}{s^2+\sin ^2(\theta)} \, d\theta=\frac{2\pi k}{\sqrt{k^2+s^2}}\rightarrow 2\pi k J_0(k t)$$

$$\int_0^{2 \pi } \cos (k t \sin (\theta)) \, d\theta=2 \pi k J_0(k t)$$

$\endgroup$

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.