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Math Help - integrallll

  1. #1
    Member vernal's Avatar
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    integrallll



    please help me. thanks
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  2. #2
    MHF Contributor chisigma's Avatar
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    Re: integrallll

    Quote Originally Posted by vernal View Post


    please help me. thanks
    A primitive of \cos x^{2} cannot be written in term of elementary functions... what You can try is to expand the \cos x^{2} in McLaurin series and then integrate 'term by term'...

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  3. #3
    Member vernal's Avatar
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    Re: integrallll

    I do not understand.

    Please tell some of it
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  5. #5
    MHF Contributor chisigma's Avatar
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    Re: integrallll

    An integral of the type...

    C(x)=\int_{0}^{x} \cos u^{2}\ du (1)

    ... is known as 'Fresnel Integral' and some informations about them You can find in...

    Fresnel Integrals -- from Wolfram MathWorld

    The function C(*) as defined in (1) cannot be expressed in term of elementary functions but a way to pratically compute the (1) consists in using the McLaurin expansion...

    \cos u^{2}= \sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2 n)!}\ u^{4n} (2)

    ... and integrating 'term by term' obtaining...

    C(x)= \sum_{n=0}^{\infty} \frac{(-1)^{n}}{(4 n+1)\ (2 n)!}\ x^{4n+1} (3)

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    \chi \sigma
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  6. #6
    Member vernal's Avatar
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    Re: integrallll

    chisigma

    thanks for your help
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  7. #7
    Member vernal's Avatar
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    Re: integrallll

    one of may friend said:

    we have:


    and
    .

    then we know:






    then






    Now my question is: Which is correct?
    Both are probably true
    What is the relationship between them?
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  8. #8
    Super Member Random Variable's Avatar
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    Re: integrallll

    Your friend didn't justify that  \int_{0}^{\infty} e^{-\alpha x^{2}} \ dx = \sqrt{\frac{\pi}{\alpha}}} is valid for imaginary \alpha.



     \int_{-\infty}^{\infty} e^{-ix^{2}} \ dx = \frac{1}{\sqrt{i}} \int^{\sqrt{i} \infty}_{-\sqrt{i} \infty} e^{-u^{2}} \ du $


    or more approriately  \frac{1}{\sqrt{i}} \lim_{R \to \infty} \int^{\sqrt{i} R}_{-\sqrt{i} R} e^{-u^{2}} \ du $


     = \sqrt{\frac{\pi}{i}} \lim_{R \to \infty} \text{erf}( \sqrt{i} R) where erf(x) is the error function


    We need to show that the limit evaluates to 1. The value of the error function at infinity is 1. But we have the error function at complex infinity. The limit can be evaluated by expanding the error function in a series near real infinity, substituting  \sqrt{i} R for x, and then taking the limit.
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  9. #9
    MHF Contributor chisigma's Avatar
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    Re: integrallll

    What RV says is true: the result obtained by vernal's friend is correct but he didn't previously demonstrate that both the integrals...

    \int_{0}^{\infty} \cos x^{2}\ dx\ ,\ \int_{0}^{\infty} \sin x^{2}\ dx (1)

    ... do exist. An example will clarify the concept: starting from the well known formula...

    \int_{0}^{\infty} e^{- \alpha x}\ dx = \frac{1}{\alpha} (2)

    ... I write...

    \int_{0}^{\infty} e^{-i x}\ dx = \int_{0}^{\infty} \cos x\ dx - i\ \int_{0}^{\infty} \sin x\ dx = \frac{1}{i}= -i (3)

    ... and from (3)...

    \int_{0}^{\infty} \cos x\ dx =0\ ,\ \int_{0}^{\infty} \sin x\ dx =1 (4)

    But the (4) are a nonsense because none of the integrals exist...

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  10. #10
    Super Member Random Variable's Avatar
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    Re: integrallll

    I'm fairly certain that the fact that  \int_{-\infty}^{\infty} \cos x^{2} \ dx and  \int_{-\infty}^{\infty} \sin x^{2} \ dx converge doesn't by itself justify that  \int_{-\infty}^{\infty} e^{-ix^{2}} \ dx = \sqrt{\frac{\pi}{i}} .


    The justification is the uniqueness theorem from complex analysis.
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