Question about double integrale and Fubini's theorem

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Question, Am I right about what I state down here:


I want to find the Laplace transform, of an integral:

$$\int_0^\infty e^{-\text{s}t}\left\{\int_0^\infty\text{y}\left(t,\text{x}\right)\space\text{d}\text{x}\right\}\space\text{d}t$$

So, in general I am studying the double integral:

$$\mathcal{I}\left(\text{s}\right)=\int_0^\infty\text{f}\left(\text{s},t\right)\left\{\int_0^\infty\text{y}\left(t,\text{x}\right)\space\text{d}\text{x}\right\}\space\text{d}t=\int_0^\infty\int_0^\infty\text{f}\left(\text{s},t\right)\text{y}\left(t,\text{x}\right)\space\text{d}\text{x}\space\text{d}t$$

In order to switch the order of integration, we can apply Fubini's theorem:

$$\color{red}{\mathcal{I}\left(\text{s}\right)=\int_0^\infty\int_0^\infty\text{f}\left(\text{s},t\right)\text{y}\left(t,\text{x}\right)\space\text{d}t\space\text{d}\text{x}}$$

The red part is equal when:

$$\int_0^\infty\int_0^\infty\left|\text{f}\left(\text{s},t\right)\text{y}\left(t,\text{x}\right)\right|\space\text{d}\text{x}\space\text{d}t\space<\text{n}\to\infty$$

But when I look at [this website][1], they only talk about when the integral equals a definite value, and my integral contains the variable $\text{s}$, how do I deal with that?

  [1]: https://en.wikipedia.org/wiki/Fubini%27s_theorem