Computing: $\int_0^1\frac{x^2\sin(x^2)+\sin(\frac{1}{x^2})}{x^2}dx$

Question

I would like to compute the exact value of the integral below. $$\int_0^1\frac{x^2\sin(x^2)+\sin(\frac{1}{x^2})}{x^2}dx$$

I have proved the convergence already. but failed to the residues theorem in other to get the exact value. It will be great if somebody could provide with some hint.

Answer 1

In fact \begin{eqnarray} &&\int_0^1\frac{x^2\sin(x^2)+\sin(\frac{1}{x^2})}{x^2}dx\\ &=&\int_0^1\sin(x^2)dx+\int_0^1\frac{\sin(\frac{1}{x^2})}{x^2}dx. \end{eqnarray} For the second integral, under $x\to\frac1x$, one has \begin{eqnarray} &&\int_0^1\frac{x^2\sin(x^2)+\sin(\frac{1}{x^2})}{x^2}dx\\ &=&\int_0^1\sin(x^2)dx+\int_0^1\frac{\sin(\frac{1}{x^2})}{x^2}dx\\ &=&\int_0^1\sin(x^2)dx+\int_1^\infty\sin(x^2)dx\\ &=&\int_0^\infty\sin(x^2)dx\\ &=&\frac{\sqrt{2\pi}}{4}. \end{eqnarray}