Why does $\displaystyle\int_1^\infty\dfrac{\sin^2(x)}{x}$ diverge ?
Why does Dirichlet-Test not work ?
Define $f(x)=g(x)=\dfrac{\sin(x)}{x^{1/2}}$, Then $\forall b>1$ and $a>0;$ $$\Big\lvert\displaystyle\int_1^b\dfrac{\sin(x)}{x^{a}}\Big\rvert<+\infty$$
also for $a=1/2$
and since $\lim\limits_{x\to\infty}\dfrac{\sin(x)}{x^{1/2}}=0$
The integral should be convergent, did I overlook something ?
but with the same method of Dirichlet, one can show the divergence
$$\int\dfrac{\sin^2(x)}{x}=\int\dfrac{1-\cos(2x)}{x}=\underbrace{\displaystyle\int\dfrac{1}{x}}_{\text{divergent}}-\underbrace{\displaystyle\int\dfrac{\cos(2x)}{x}}_{\text{convergent}}$$
For example because, for every $n\geqslant1$, $$ \int_{n\pi+\pi/4}^{n\pi+3\pi/4}\frac{\sin^2x}x\mathrm dx\geqslant\int_{n\pi+\pi/4}^{n\pi+3\pi/4}\frac{\frac12}{(n+1)\pi}\mathrm dx=\frac1{4(n+1)}$$