When does the integral $$\int_0^1 x^q\left(\frac{1}{x} - \frac{1}{\sin(x)}\right)\,dx$$ converge?
So function $$x^q\left(\frac{1}{x} - \frac{1}{\sin(x)}\right)=x^{q-1}\left(1-\frac{x}{\sin(x)}\right)\le x^{q-1}\implies$$ integral converges when $q-1>-1\implies q>0$. But it doesn't right because Wolfram says it converges when $q=-1$.
2026-05-16 16:12:14.1778947934
When does the integral $\int_0^1x^q\left(\frac{1}{x} - \frac{1}{\sin(x)}\right)\,dx$ converge?
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Hint. There is only a potential issue as $x \to 0^+$, but in this case, by the Taylor series expansion of $\sin (\cdot)$, one gets $$ x^q\left(\frac{1}{\sin x}-\frac{1}{x}\right)=\frac{x^{q+1}}{6}+O\left(x^{q+3} \right) $$ which converges iff $-q-1<1$ that is iff $q>-2$.