Consider the following integral : $$\int_{-\infty}^{+\infty}\Big | \frac{e^{i(\omega+\omega_0)\frac{N\pi}{\omega_0}}e^{i\omega t}}{\omega+\omega_0}\Big |d\omega$$
I have to show that this integral does not exists. My attempt : I have $$\int_{-\infty}^{+\infty}\Big | \frac{e^{i(\omega+\omega_0)\frac{N\pi}{\omega_0}}e^{i\omega t}}{\omega+\omega_0}\Big |d\omega=\int_{-\infty}^{+\infty}\frac{1}{\mid \omega+\omega_0\mid}d\omega\geq \int_{-\infty}^{\infty}\frac{1}{\mid \omega\mid}d\omega$$ But then, I don't really know what to do with the last integral. Can I split it ? Like this : $$\int_{-\infty}^{\infty}\frac{1}{\mid \omega\mid}d\omega=\int_{0}^{\infty}\frac{1}{ \omega}d\omega-\int_{-\infty}^{0}\frac{1}{ \omega}d\omega$$ Before splitting an integral, don't we have to show that it exists ? Here I am not sure if $\int_{-\infty}^{+\infty}1/(\mid \omega \mid)d\omega$ exists... Any help would be appreciated,
For complex $\omega_0$, it can be not trivial to get $\int_{-\infty}^{+\infty}\frac{1}{\mid \omega+\omega_0\mid}d\omega\geq \int_{-\infty}^{\infty}\frac{1}{\mid \omega\mid}d\omega$ directly.
However, you can definitely say $\int_{-\infty}^{+\infty}\frac{1}{\mid \omega+\omega_0\mid}d\omega\geq \int_{2|\omega_0|}^{\infty}\frac{1}{\mid \omega\mid}d\omega$ (we reduce integration area, and function is non-negative), then say $\int_{2|\omega_0|}^{\infty}\frac{1}{\mid \omega\mid}d\omega \geqslant \int_{2|\omega_0|}^{+\infty} \frac{1}{2\omega}d\omega$ (replace function with strictly lower), and the last integral obviously diverges (we can explicitly find it).
In general, if your function is non-negative (or non-positive), you can split the integral - the original integral converges iff all parts converges. Equally - integral converges absolutely iff all parts converges absolutely
If function changes sign, then it's possible for integral to converge but parts to diverge - for example, $\int_0^\infty\frac{\sin x}{x}\, dx$ converges, but $\int_{\{x: \sin(x) > 0\}} \frac{\sin x}{x}\, dx$ and $\int_{\{x: \sin(x) \leq 0\}} \frac{\sin x}{x}\, dx$ both diverge.