## anonymous 3 years ago evaluate integral (sin t^2 )dt from 0 to infinite?

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1. anonymous

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2. anonymous

Do you have to approximate the answer? Because its integrate indefinite is not an easy function...

3. across

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.

4. anonymous

No, he wrote $\mathrm{ \int_{0}^\infty \sin(x^2)dx}$

5. anonymous
6. anonymous

$\sum_{0}^{\infty} ((-1)^k/(2k+1))*(x^2)^{2k+1}$

7. anonymous

thanks for all but i want the steps of solution