<s>[INST] 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.
 [/INST] 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}
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