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Integral Manipulation

  1. Feb 6, 2012 #1
    If [itex]q>p>-1[/itex] and [itex]w=\cosh(x)[/itex] then how do I get smoothly from: [tex]\displaystyle \int^{\infty}_1 \sinh^{p-1}(x) w^{-q} \;dw[/tex] to [tex]\displaystyle \int^{\infty}_1 (w^2-1)^{\frac{p-1}{2}} w^{-q}\;dw[/tex] and if [itex]t=w^{-2}[/itex] how do I get smoothly from this to: [tex]\displaystyle \frac{1}{2} \int^{\infty}_0 t^{\frac{p+1}{2}-1} (1-t)^{\frac{q-p}{2}-1}\;dt[/tex]
     
  2. jcsd
  3. Feb 6, 2012 #2

    tiny-tim

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    (you know cosh2 = sinh2 + 1 ?)

    looks straightforward to me

    where exactly are you having difficulty?
     
  4. Feb 6, 2012 #3
    Ah, that identity gives me the first one.

    Now if [itex]t=w^{-2}[/itex] then [itex]\displaystyle dw = -\frac{dt}{2w^{-3}}[/itex] so: [tex]\displaystyle \int^{\infty}_1 (w^2-1)^{\frac{p-1}{2}} w^{-q}\;dw = -\frac{1}{2} \int^{\infty}_0 (t^{-1} -1)^{\frac{p-1}{2}} t^{\frac{q}{2}}\;\frac{dt}{w^{-3}}[/tex] and since [itex]w^{-3} = t^{\frac{3}{2}}[/itex] this equals [tex]\displaystyle -\frac{1}{2} \int^{\infty}_0 (t^{-1} -1)^{\frac{p-1}{2}} t^{\frac{1}{2}(q-3)}\;dt[/tex]

    How do I get from this to [tex]\displaystyle \frac{1}{2} \int^{\infty}_0 t^{\frac{p+1}{2}-1} (1-t)^{\frac{q-p}{2}-1}\;dt[/tex]
     
    Last edited: Feb 6, 2012
  5. Feb 7, 2012 #4

    tiny-tim

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    Hi Ted123! :wink:

    (just got up :zzz: …)

    You're right :smile:, it doesn't work …

    I think the question has a misprint, it should be …
    [tex]\displaystyle \int^{\infty}_1 \sinh^{q-p}(x) w^{-q} \;dw[/tex]
     
  6. Feb 7, 2012 #5
    I think I've got it to work. When we let [itex]t=w^{-2}[/itex] the limits of the integral change from [itex]\int^{\infty}_1[/itex] to [itex]\int^0_1 = -\int^1_0[/itex].

    So from [tex]\displaystyle \frac{1}{2} \int^1_0 (t^{-1} -1)^{\frac{p-1}{2}} t^{\frac{1}{2}(q-3)}\;dt[/tex]

    we have that [itex]t^{-1} - 1 = \frac{1-t}{t}[/itex]. Hence we have [tex]\displaystyle \frac{1}{2} \int^1_0 \left( \frac{1-t}{t} \right) ^{\frac{p-1}{2}} t^{\frac{1}{2}(q-3)}\;dt[/tex]

    [tex]= \frac{1}{2} \int^1_0 (1-t)^{\frac{p-1}{2}} \frac{ t^{\frac{1}{2} (q-3)} }{t^{ \frac{p-1}{2} }} \;dt[/tex]

    [tex]= \frac{1}{2} \int^1_0 (1-t)^{\frac{p-1}{2}} t^{\frac{q-p}{2}-1} \;dt = \frac{1}{2} \int^1_0 (1-t)^{\frac{p+1}{2} - 1} t^{\frac{q-p}{2}-1} \;dt[/tex] This is the beta function [itex]\displaystyle \frac{1}{2} B \left (\frac{q-p}{2} , \frac{p+1}{2} \right)[/itex] and the Beta function is symmetric so this equals [itex]\displaystyle \frac{1}{2} B \left (\frac{p+1}{2} , \frac{q-p}{2} \right)[/itex] which gives us [tex]\frac{1}{2} \int^1_0 (1-t)^{\frac{q-p}{2}-1} t^{\frac{p+1}{2} - 1} \;dt[/tex]
     
    Last edited: Feb 7, 2012
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