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StupidGenius
evaluate integral (sin t^2 )dt from 0 to infinite?
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Do you have to approximate the answer? Because its integrate indefinite is not an easy function...
First find the antiderivative of \(\sin^2t\):\[\int\sin^2t\,dt=\frac t2-\frac14\sin2t+c,\]then apply the limits:\[\left[\frac t2-\frac14\sin2t\right]_0^\infty.\]However, since\[\lim_{t\to\infty}\sin 2t\]doesn't converge, the integral doesn't either.
No, he wrote \[\mathrm{ \int_{0}^\infty \sin(x^2)dx}\]
http://www.wolframalpha.com/input/?i=+%5Cint_%7B0%7D%5E%5Cinfty+%5Csin%28x%5E2%29dx
\[\sum_{0}^{\infty} ((-1)^k/(2k+1))*(x^2)^{2k+1}\]
thanks for all but i want the steps of solution