I have a problem solving differential equation, where I think there should be a Cauchy Principal value involved, but I do not see how it it should emerge. Let's say we have a differential equation:
$$ \mathrm{i} \dot{C}_1(t)=\int \limits_0^{\infty} d \omega \, \nu(\omega) e^{\mathrm{i}(\omega_{10}-\omega)t}, $$ where all the variables are real and positive. If we now solve the differential equation, it needs only to carry out the time integral
$$ \mathrm{i} C_1(t)=\int dt\int \limits_0^{\infty} d \omega \, \nu(\omega) e^{\mathrm{i}(\omega_{10}-\omega)t}=\int \limits_0^{\infty} d \omega \, \frac{\nu(\omega)}{\mathrm{i}(\omega_{10}-\omega)} e^{\mathrm{i}(\omega_{10}-\omega)t}. $$ So the problem is that if the last integral is not taken as principal value, it will give $\infty$, so I would rather need it to be $$ \mathcal{P}\int \limits_0^{\infty} d \omega \, \frac{\nu(\omega)}{\mathrm{i}(\omega_{10}-\omega)} e^{\mathrm{i}(\omega_{10}-\omega)t}, $$ where $\mathcal{P}$ is the principal value. Would anyone have any idea where the principal value would emerge from, if at all? I was thinking maybe it has something to do when swapping the order of integrals in order to carry out the time integral first, but I am not sure.
Yes, your observation is precisely that Fubini does not apply, because one of the iterated integrals is (sometimes) infinity. But you can cut off a small ball around $\newcommand{\o}{\omega_{10}}\o>0$ before interchanging the integrals: \begin{align} \int_0^t \int_0^\infty v(\omega)e^{i(\o-\omega)t}d\omega dt &= \lim_{\epsilon\to 0}\int_0^t\int_{\omega\in[0,\infty),|\omega-\o|>\epsilon} v(\omega)e^{i(\o-\omega)t}d\omega dt \\ &= \lim_{\epsilon\to 0}\int_{\omega\in[0,\infty),|\omega-\o|>\epsilon} \int_0^tv(\omega)e^{i(\o-\omega)t}d\omega dt \\ &= \lim_{\epsilon\to 0}\int_{\omega\in[0,\infty),|\omega-\o|>\epsilon} \frac{v(\omega)e^{i(\o-\omega)t}}{i(\o-\omega)}d\omega \\&=:\mathcal P\!\!\!\int_0^\infty \frac{v(\omega)e^{i(\o-\omega)t}}{i(\o-\omega)}d\omega \end{align} The above works only when $\o\neq 0$. Also, if $v(\o)=0$ and is $C^1$ then the above still works but then the integral does exist, so $\mathcal P\!\!\int_0^\infty=\int_0^\infty$.