I read a solution for some exercise in Hilbert spaces theory, which included some improper integrals convergence which I did not compeletly understand:
If it matters, the original question was -
And what I don't understand is, why is the second integral, i.e. $\int_1^\infty \frac{dt}{t^{kp}}=\infty$?
Your question is
As the exponent is constant, you can just note the anti-derivative of $\cfrac{1}{x^{kp}}$ is $\cfrac{x^{1-kp}}{1-kp}$. Earlier, it states that $pk \lt 1$, so $1 - pk \gt 0$ and, thus, the anti-derivative is unbounded above.
Alternatively, for $t \gt 1$, $t^{pk} \lt t$ so $\cfrac{1}{t^{pk}} \gt \cfrac{1}{t}$. This gives that
$$\int_1^\infty \frac{dt}{t^{kp}} \gt \int_1^\infty \frac{dt}{t} = \infty \tag{1}\label{eq1}$$
because the anti-derivative of $\frac{1}{x}$ is $\ln\left(x\right)$ which, of course, is unbounded above.