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Given $$ f(x)= \int \limits_0^x \sin(y^2) \cos(y^2) \mathrm{d}y $$

Anyone can help and guide me for this?I don't really have an idea of how to represent it as power series

Thank you!

My attempt: $$ 0.5\int_{0}^x \sum_{n=0}^\infty \frac{{(-1)}^n{({2{y}^{2})}}^{2n+1}}{(2n+1)!} \mathrm{d}y$$ after substituting Maclaurin series of sin x

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  • $\begingroup$ Why would you want to represent it (Who? Each of the functions or their integral?) as a power series? $\endgroup$
    – Timbuc
    Commented Apr 19, 2015 at 10:49
  • $\begingroup$ @Timbuc represent f(x) as power series $\endgroup$
    – UnusualSkill
    Commented Apr 19, 2015 at 10:53
  • $\begingroup$ Expand the integrand (after its simplification) as Taylor series and integrate. $\endgroup$
    – Claude Leibovici
    Commented Apr 19, 2015 at 11:39

2 Answers 2

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Hint:

$$\int_0^x\sin t^2\cos t^2\;dt=\frac12\int_0^x\sin 2t^2\;dt$$

and then you may want to check Fresnel Integrals

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  • $\begingroup$ Can I check answer with you? [(-1)^n (2x^2)^(2n+2)]/2(2n+2)! $\endgroup$
    – UnusualSkill
    Commented Apr 19, 2015 at 11:19
  • $\begingroup$ interval x in (-inf,inf) $\endgroup$
    – UnusualSkill
    Commented Apr 19, 2015 at 11:27
  • $\begingroup$ @UnusualSkill You've been a member of this site for looooong time: if you want to be properly understood please do begin learning the easy directions to properly write mathematics in this site. I can't understand completely what you wrote, though I'm afraid it is wrong. Check the following : meta.math.stackexchange.com/questions/5020/… $\endgroup$
    – Timbuc
    Commented Apr 19, 2015 at 11:31
  • $\begingroup$ $$\sum_{n=0}^\infty \frac{{(-1)}^{n}{(2x^2)}^{2n+2}}{2(2n+1)!} $$ $\endgroup$
    – UnusualSkill
    Commented Apr 19, 2015 at 11:43
  • $\begingroup$ It does not seem to be correct. Check your edit. $\endgroup$
    – Claude Leibovici
    Commented Apr 19, 2015 at 11:45
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Hint: use $\sin 2u = 2 \sin u \cos u$; the power series of $\sin$ and term by term integration.

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  • $\begingroup$ Can I check answer with you? [(-1)^n (2x^2)^(2n+2)]/2(2n+2)! $\endgroup$
    – UnusualSkill
    Commented Apr 19, 2015 at 11:18
  • $\begingroup$ interval x in (-inf,inf) $\endgroup$
    – UnusualSkill
    Commented Apr 19, 2015 at 11:27
  • $\begingroup$ it seems fine. edit your question with your attempt! $\endgroup$
    – mookid
    Commented Apr 19, 2015 at 12:12
  • $\begingroup$ really?becuz other said it is not right... $\endgroup$
    – UnusualSkill
    Commented Apr 19, 2015 at 12:39
  • $\begingroup$ my important attempt $\endgroup$
    – UnusualSkill
    Commented Apr 19, 2015 at 12:58

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