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StupidGenius

  • 2 years ago

evaluate integral (sin t^2 )dt from 0 to infinite?

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  1. StupidGenius
    • 2 years ago
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    |dw:1355422740273:dw|

  2. darthjavier
    • 2 years ago
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    Do you have to approximate the answer? Because its integrate indefinite is not an easy function...

  3. across
    • 2 years ago
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    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. darthjavier
    • 2 years ago
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    No, he wrote \[\mathrm{ \int_{0}^\infty \sin(x^2)dx}\]

  5. darthjavier
    • 2 years ago
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    http://www.wolframalpha.com/input/?i=+%5Cint_%7B0%7D%5E%5Cinfty+%5Csin%28x%5E2%29dx

  6. dorito
    • 2 years ago
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    \[\sum_{0}^{\infty} ((-1)^k/(2k+1))*(x^2)^{2k+1}\]

  7. StupidGenius
    • one year ago
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    thanks for all but i want the steps of solution

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