I'm trying to prove that the following integral diverges: $$ \int_0^1\frac{\sin(\frac{1}{x})}{x^2}dx $$ I tried to use Dirichlet theorm but without any success. How can I show it without using taylor?

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    $\begingroup$ Have you tried substituting $y=1/x$? $\endgroup$ – h3fr43nd Jul 22 at 15:09


Substitute $\frac1x\rightarrow t$ to get $\int_1^{\infty}\sin(t)dt$

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