Does the improper integral $\int_0^1 \frac{1}{x}\sin\frac{1}{x^2}$ exist?

Question

Does the improper integral $\displaystyle\int_0^1 \frac{1}{x}\sin\frac{1}{x^2}$ exist?

I don't know how to use the comparison test, and I cannot find a proper comparison function.

Answer 1

By a change of variables $u=\frac1{x^2},\frac{du}{dx}=-\frac2{x^3}$: $$\int_0^1\frac{\sin(1/x^2)}x=\frac12\int_1^\infty\frac{\sin u}u\,du$$ Thus the original integral converges, since $\int_0^\infty\frac{\sin u}u\,du=\frac\pi2$ (see here for example). Indeed, its value is $$\frac12\left(\frac\pi2-\operatorname{Si}(1)\right)=0.31235\dots$$ where $\rm{Si}$ is the sine integral.

Answer 2

If you set $y = 1/x^2$ i.e. $x = y^{-1/2}$ then $dx = -\frac12y^{-3/2}dy$, so $dx/x = -\frac12 dy/y$, and the integral becomes $$ -\frac12\int_1^\infty \frac{\sin (y)}ydy$$ the convergence of the original integral is the same question as the convergence of $$ -\frac12\lim_{a\to\infty} \int_1^a \frac{\sin (y)}ydy$$ The answer is yes, the improper integral exists, one can follow the proofs in Evaluating the integral $\int_0^\infty \frac{\sin x} x \,\mathrm dx = \frac \pi 2$?, but you won't get as clean a result. This integral evaluates to $-\frac12(\operatorname{Si}(1)-\frac{\pi}{2})$.