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Math Help - Integrals

  1. #1
    MHF Contributor chiph588@'s Avatar
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    Integrals

    Show  \displaystyle \int_0^\infty \cos(x^k)dx = \Gamma\left(\frac{k+1}k\right)\cos\left(\frac\pi{2  k}\right) and  \displaystyle \int_0^\infty \sin(x^k)dx = \Gamma\left(\frac{k+1}k\right)\sin\left(\frac\pi{2  k}\right) .
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  2. #2
    MHF Contributor chisigma's Avatar
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    Let's start from the general formula You can find in any 'Manual of Math' valid for n>1 ...

    \displaystyle \int_{0}^{\infty} \frac{dx}{x^{n} + a^{n}} = \pi\ \frac{a^{1-n}}{n\ \sin \frac{\pi}{n}} (1)

    Now we define...

    \displaystyle \chi (t) = \int_{0}^{\infty} \cos (x^{k}\ t)\ dx (2)

    ... and compute...

    \displaystyle \mathcal{L} \{\chi(t)\} = \int_{0}^{\infty} e^{-s\ t}\ \int_{0}^{\infty} \cos (x^{k}\ t)\ dx\ dt = \int_{0}^{\infty} dx\ \int_{0}^{\infty} e^{-s\ t} \cos (x^{k}\ t)\ dt =

    \displaystyle = \int_{0}^{\infty} \mathcal{L} \{\cos (x^{k}\ t)\}\ dx =\int_{0}^{\infty} \frac{s}{s^{2} + x^{2k}}\ dx (3)

    From (3) and (1), setting n=2\ k and a=s^{\frac{1}{k}} we obtain...

    \displaystyle \mathcal{L} \{\chi(t)\} = \pi\ \frac{s^{\frac{1}{k} -1}}{2\ k\ \sin \frac{\pi}{2k}} (4)

    ... from which we derive \chi(t) ...

    \displaystyle \chi(t) = \frac{\pi\ t^{-\frac{1}{k}}}{2\ k\ \Gamma(1-\frac{1}{k})\ \sin \frac{\pi}{2 k}} (5)

    ... and finally, setting t=1 in (5), we obtain...

    \displaystyle \int_{0}^{\infty} \cos x^{k}\ dx =  \frac{\pi}{2\ k\ \Gamma(1-\frac{1}{k})\ \sin \frac{\pi}{2 k}} (6)

    At first it seems that the result (6) is not correct ... but using the fact that...

    \displaystyle \frac{1}{\Gamma(1-\frac{1}{k})} = \frac{\sin \frac{\pi}{k}\ \Gamma(\frac{1}{k})}{\pi}

    \displaystyle \Gamma(\frac{1}{k}) = k\ {\Gamma (1+\frac{1}{k})

    \displaystyle \sin \frac{\pi}{k} = 2\  \sin \frac{\pi}{2k} \cos \frac{\pi}{2k}

    ... You find that (6) is equivalent to the expression given by chiph588 ...

    Kind regards

    \chi \sigma
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  3. #3
    MHF Contributor Bruno J.'s Avatar
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