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Thread: Invalid integral solution?

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

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


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

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

    let $\displaystyle u = ix $

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

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

    $\displaystyle = -\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...

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

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

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

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
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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; Jun 15th 2011 at 08:10 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...

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

    ... valid for $\displaystyle -1 < a < 0$?...

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

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

    Now we define...

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

    ... and then compute...

    $\displaystyle \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 = $

    $\displaystyle = \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...

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

    ... and from (5)...

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

    ... so that is...

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

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

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
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