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Math Help - Invalid integral solution?

  1. #1
    Super Member Random Variable's Avatar
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    Invalid integral solution?

     \int^{\infty}_{0} \frac{\sin x}{x^{1-a}} \ dx converges for  -1<a<1. But is the following solution only valid for  0 < a <1 ? Is it valid at all?


     \int_{0}^{\infty} \frac{\sin x}{x^{1-a}} \ dx = -\Im \int^{\infty}_{0} e^{-ix} x^{a-1} \ dx

    The issue is that  \int^{\infty}_{0} \frac{\cos x}{x^{1-a}} \ dx does not converge for  a \le 0 .

    let  u = ix

     = -\Im \int^{\infty}_{0} e^{-u} \big(\frac{u}{i}\big)^{a-1} \ \frac{du}{i}

     = -\Im \ (-i)^{a} \int_{0}^{\infty} u^{a-1} e^{-u} \ du

     = -\Im \ e^{\frac{-i a \pi}{2}} \ \Gamma(a) = \sin \Big( \frac{a \pi}{2} \Big) \Gamma(a)
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  2. #2
    MHF Contributor chisigma's Avatar
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    Re: Invalid integral solution?

    The fact is that the definition of the Gamma function...

    \Gamma(z)= \int_{0}^{\infty} t^{z-1}\ e^{-t}\ dt (1)

    ... is valid only if \Re (z) >0, that is necessary condition for convergence of the integral in (1). For \Re (z)<0 the Gamma function can be computed using the relation...

    \Gamma(z)\ \Gamma(-z) = -\frac{\pi}{z\ \sin \pi z} (2)

    Kind regards

    \chi \sigma
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  3. #3
    Super Member Random Variable's Avatar
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    Re: Invalid integral solution?

    Can I invoke the gamma reflection formula for the other values of a?
    Last edited by Random Variable; June 15th 2011 at 09:10 PM.
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  4. #4
    MHF Contributor chisigma's Avatar
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    Re: Invalid integral solution?

    What Random Variable is asking for is probably: is the formula...

    \int_{0}^{\infty} \frac{\sin x}{x^{1-a}}\ dx = \sin \frac{a\ \pi}{2}\ \Gamma (a) (1)

    ... valid for -1 < a < 0?...

    All right!... a simple way is try to compute the (1) setting [for example...] a=-\frac{1}{2} in some way and verify. If (1) is valid it should be...

    \int_{0}^{\infty} \frac{\sin x}{x^{\frac{3}{2}}}\ dx = -\sin \frac{\pi}{4}\ \Gamma (-\frac{1}{2})= \sqrt{2 \pi} (2)

    Now we define...

    \varphi(t)= \int_{0}^{\infty} \frac{\sin t x}{x^{\frac{3}{2}}}\ dx (3)

    ... and then compute...

    \mathcal{L} \{\varphi(t)\} = \int_{0}^{\infty} e^{-s t}\ \int_{0}^{\infty} \frac{\sin t x}{x^{\frac{3}{2}}}\ dx\ dt = \int_{0}^{\infty} \frac{\mathcal{L} \{\sin t x \}}{x^{\frac{3}{2}}}\ dx =

    = \int_{0}^{\infty} \frac{dx}{\sqrt{x}\ (s^{2}+x^{2})} (4)

    The integral in (4) can be attacked with the usual complex analysis and we obtain...

    \mathcal {L} \{\varphi (t)\} = \frac{\pi}{\sqrt{2}\ s^{\frac{3}{2}}} (5)

    ... and from (5)...

    \varphi(t) = \frac{\pi}{\sqrt{2}}\ \frac{\sqrt{t}}{\Gamma (\frac{3}{2})} = \sqrt{2 \pi t} (6)

    ... so that is...

    \int_{0}^{\infty} \frac{\sin x}{x^{\frac{3}{2}}}\ dx = \varphi (1) = \sqrt{2 \pi} (7)

    ... and the (1) is verified for a=- \frac{1}{2}. Following the same procedure it is [probably...] possible to demonstrate that the same is for -1<a<0...

    Kind regards

    \chi \sigma
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