What happened to interchange Laplace transform and one definite integral (Fubini's theorem is not satisfied)?

Question

I have been long time puzzled by the correct application of Fubinis theorem. Is it true that the bounds need to be constant if the integrals can be interchanged? Please correct if not.

Let's consider the following Laplace transform pair.

\begin{align} \mathcal{L}^{-1}\Bigl[\frac{2(\cos(at)-\cos(bt))}{t}\Bigr] = \color{red}{-}\ln\Bigl[\frac{s^2+a^2}{s^2+b^2}\Bigr]\tag{1}\label{e:1} \end{align}

Since I observe that

\begin{align*} \int_a^b\sin(xt) \, \mathrm{d}t = \frac{\cos(at)-\cos(bt)}{t} \end{align*}

I tried to first do Laplace tranform and then integral as follows

\begin{align*} 2\int_a^b \int_0^\infty \sin(xt)e^{-st} \,\mathrm{d}t\mathrm{d}x = 2\int_a^b \frac{x}{x^2+s^2} \, \mathrm{d}x =\ln\Bigl[\frac{s^2+b^2}{s^2+a^2}\Bigr] \end{align*}

which is negative to $\eqref{e:1}$. I might have made silly mistakes I was not aware of.

EDIT: It turned out that is another error in Schaum's book p169 (1968, Spiegel).

Answer 1

Fubini–Tonelli Theorem tells that, if $f$ is a function on the product set $X \times Y$¹⁾, then the followings hold:

1. (Tonelli's Theorem) We always have $$ \int_{X}\int_{Y} |f(x, y)| \, \mathrm{d}y\mathrm{d}x = \int_{Y} \int_{X} |f(x, y)| \, \mathrm{d}x\mathrm{d}y = \iint_{X\times Y} |f(\mathbf{z})| \, \mathrm{d}\mathbf{z} $$ regardless of whether they are finite or not.

2. (Fubini's Theorem) Moreover, if the above integrals are finite, then we also have $$ \int_{X}\int_{Y} f(x, y) \, \mathrm{d}y\mathrm{d}x = \int_{Y} \int_{X} f(x, y) \, \mathrm{d}x\mathrm{d}y = \iint_{X\times Y} f(\mathbf{z}) \, \mathrm{d}\mathbf{z}. $$

So, that $f$ is bounded by a constant is neither a sufficient nor necessary condition for Fubini's Theorem to be applicable. (There are counterexamples in both directions.)

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In OP's case,

\begin{align*} \int_{0}^{\infty} \int_{a}^{b} \left| \sin(xt)e^{-st} \right| \, \mathrm{d}x\mathrm{d}t &\leq \int_{0}^{\infty} \int_{a}^{b} e^{-st} \, \mathrm{d}x\mathrm{d}t = \frac{b-a}{s} < \infty, \end{align*}

hence Fubini's Theorem is applicable. Also, OP's computation turns out to be correct, yielding

$$ \mathcal{L}\left[ \frac{2(\cos(at) - \cos(bt))}{t} \right](s) = \log\left[ \frac{s^2 + b^2}{s^2 + a^2} \right] $$

Below is a verification of this identity using Mathematica:

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^{1) The most well-known version requires that $(X, \mu_X)$ and $(Y, \mu_Y)$ are $\sigma$-finite measure spaces and $f : X \times Y \to \mathbb{R}$ is measurable. Then by replacing $\mathrm{d}x$, $\mathrm{d}y$, and $\mathrm{d}\mathbf{z}$ by $\mu_X(\mathrm{d}x)$, $\mu_Y(\mathrm{d}y)$, and $(\mu_X\otimes\mu_Y)(\mathrm{d}\mathbf{z})$, respectively, the statement in the above answer holds. Sine measurability is rarely an issue in preliminary analysis, we can sweep this under the rug for beginners.}