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.

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    $\begingroup$ 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
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    $\begingroup$ 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

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:

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

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

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

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