Exact Answer
In indefinite integral, it is not integrable.
However in definite integral, by complex analysis,
\int_0^{\infty}\sin(x^2)dx=\sqrt{\pi}/(2\sqrt{2}).
BTW, the Fresnel integral is defined as
c(z)=\int_0^z\cos(\pi*t^2/2)dt.
So matlab and mathematica give solutions related with Fresnel integral.
However in definite integral, by complex analysis,
\int_0^{\infty}\sin(x^2)dx=\sqrt{\pi}/(2\sqrt{2}).
BTW, the Fresnel integral is defined as
c(z)=\int_0^z\cos(\pi*t^2/2)dt.
So matlab and mathematica give solutions related with Fresnel integral.
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