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

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}
