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    I was messing around with WolframAlpha and came up with the following question:

    Show that

     \displaystyle \int^{\infty}_{0} x\cos(x^3) \ dx = \dfrac{\Gamma (\frac{2}{3})}{6}

    I can think of two possible (incomplete) approaches:

    One idea is to use the substitution  u=x^3 and then use the identity  \cos (x) = \dfrac{e^{ix}+e^{-ix}}{2}

    Which transforms the integral into  \displaystyle \dfrac{1}{6} \int^{\infty}_{0} u^{-\frac{1}{3}}\left(e^{iu} + e^{-iu}\right) \ du, which begins to resemble the definition of the Gamma function, but I'm not quite sure where to go from here. I also think that there are issues with using a substitution such as  t=-iu .

    Another idea I had was to use series:

    I've shown that
     \displaystyle \int x\cos(x^3) \ dx = \sum^{\infty}_{n=0} \dfrac{(-1)^n x^{6n+2}}{(2n)!(6n+2)} + C

    but I'm new to using infinite series in this way so again I'm not sure how to proceed.
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    (Original post by ThatPerson)
    I was messing around with WolframAlpha and came up with the following question:

    Show that

     \displaystyle \int^{\infty}_{0} x\cos(x^3) \ dx = \dfrac{\Gamma (\frac{2}{3})}{6}

    I can think of two possible (incomplete) approaches:

    One idea is to use the substitution  u=x^3 and then use the identity  \cos (x) = \dfrac{e^{ix}+e^{-ix}}{2}

    Which transforms the integral into  \displaystyle \dfrac{1}{6} \int^{\infty}_{0} u^{-\frac{1}{3}}\left(e^{iu} + e^{-iu}\right) \ du, which begins to resemble the definition of the Gamma function, but I'm not quite sure where to go from here. I also think that there are issues with using a substitution such as  t=-iu .

    Another idea I had was to use series:

    I've shown that
     \displaystyle \int x\cos(x^3) \ dx = \sum^{\infty}_{n=0} \dfrac{(-1)^n x^{6n+2}}{(2n)!(6n+2)} + C

    but I'm new to using infinite series in this way so again I'm not sure how to proceed.
    after the substitution you leave it as cosu/(3u).
    this is known as the cosine integral if limits are 0 to infinity.

    The techniques from that point are not elementary to change it into Gamma.
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    (Original post by TeeEm)
    after the substitution you leave it as cosu/(3u).
    this is known as the cosine integral if limits are 0 to infinity.

    The techniques from that point are not elementary to change it into Gamma.
    Isn't it u^(1/3) on the denominator?
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    (Original post by ThatPerson)
    Isn't it u^(1/3) on the denominator?
    correct (typo)
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    (Original post by TeeEm)
    correct (typo)
    I just quote

    ∫cosu/ut du =π/[2Gamma(t)cos(πt/2)] with limits 0 to inf

    this sorts the answer I hope (I have not checked it)


    the proof (if you need to derive the main results along the way) is very long
    (it includes double integrals, more substitutions and finally complex residue integration)
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    (Original post by TeeEm)
    after the substitution you leave it as cosu/(3u).
    this is known as the cosine integral if limits are 0 to infinity.

    The techniques from that point are not elementary to change it into Gamma.

    I just quote

    ∫cosu/ut du =π/[2Gamma(t)cos(πt/2)] with limits 0 to inf

    this sorts the answer I hope (I have not checked it)


    the proof (if you need to derive the main results along the way) is very long
    (it includes double integrals, more substitutions and finally complex residue integration)
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    (Original post by ThatPerson)
    I was messing around with WolframAlpha and came up with the following question:

    Show that

     \displaystyle \int^{\infty}_{0} x\cos(x^3) \ dx = \dfrac{\Gamma (\frac{2}{3})}{6}

    I can think of two possible (incomplete) approaches:

    One idea is to use the substitution  u=x^3 and then use the identity  \cos (x) = \dfrac{e^{ix}+e^{-ix}}{2}

    Which transforms the integral into  \displaystyle \dfrac{1}{6} \int^{\infty}_{0} u^{-\frac{1}{3}}\left(e^{iu} + e^{-iu}\right) \ du, which begins to resemble the definition of the Gamma function, but I'm not quite sure where to go from here. I also think that there are issues with using a substitution such as  t=-iu .

    very bad day
    Another idea I had was to use series:

    I've shown that
     \displaystyle \int x\cos(x^3) \ dx = \sum^{\infty}_{n=0} \dfrac{(-1)^n x^{6n+2}}{(2n)!(6n+2)} + C

    but I'm new to using infinite series in this way so again I'm not sure how to proceed.
    very bad day ... twice I replied to myself...


    I just quote

    ∫cosu/ut du =π/[2Gamma(t)cos(πt/2)] with limits 0 to inf

    this sorts the answer I hope (I have not checked it)


    the proof (if you need to derive the main results along the way) is very long
    (it includes double integrals, more substitutions and finally complex residue integration)
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    (Original post by TeeEm)
    very bad day ... twice I replied to myself...


    I just quote

    ∫cosu/ut du =π/[2Gamma(t)cos(πt/2)] with limits 0 to inf

    this sorts the answer I hope (I have not checked it)


    the proof (if you need to derive the main results along the way) is very long
    (it includes double integrals, more substitutions and finally complex residue integration)
    Thank you, I got the answer into the required form after some rearranging. Do you have a link to the proof?

    Also, would the series approach not work in this case because  \displaystyle \sum^{\infty}_{n=0} \dfrac{(-1)^n}{(2n)!}\int^{\infty}_{0} x^{6n+1} \ dx doesn't converge?
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    (Original post by ThatPerson)
    Thank you, I got the answer into the required form after some rearranging. Do you have a link to the proof?

    Also, would the series approach not work in this case because  \displaystyle \sum^{\infty}_{n=0} \dfrac{(-1)^n}{(2n)!}\int^{\infty}_{0} x^{6n+1} \ dx doesn't converge?
    I am rather a pen, paper and book man

    The proof must exists in several textbooks.

    I have it in ADVANCED MATHEMATICS, SCHAUM OUTLINE, murray Spiegel 9.22
 
 
 
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