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How should I proceed to work out following convolution integral:

$\int_0^y x^{-\alpha} (y-x)^{-\alpha} dx$

for real $\alpha$ > 0.

It is the convolution of a powerlaw decaying impulse response with itself. My goal is to find the decay exponent of the autocorrelation function.

user3817704
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  • You mean $0<\alpha<1$, since for $\alpha>1$ there is no solution. – Marcel Mar 16 '16 at 13:58
  • Thank you, very relevant to my cause ! Could you explain how you concluded this? – user3817704 Mar 16 '16 at 14:04
  • $1/x^\alpha$ is not integrable in the vicinity of zero in that case, since its integral would be $x^{1-\alpha}$ which is infinite if $\alpha>1$. Analogous reason in the vicinity of $x=y$. – Marcel Mar 16 '16 at 14:06
  • @Marcel: Dear Marcel, how would the conclusion "no solution for $\alpha > 1$" change if the limits were not [0, $y$] but rather [c > 0, $y$] ? The impulse response originates from a Pareto distribution which I forgot is defined only above a minimum value. – user3817704 Mar 18 '16 at 15:53
  • Conclusion still holds, because of the upper limit.You can have $\alpha>1$ if you integrate in $[c,d]$ with $c>0$ and $d<y$. – Marcel Mar 18 '16 at 16:53
  • @Marcel: Does that mean, for $\alpha>1$, that the $1-2\alpha$ decay from answer below, is at least approximately correct for $t_{min}/t_{max}$ very small ? Thank you. – user3817704 Mar 18 '16 at 19:32

2 Answers2

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Rescale with $x=yt$, then

$$\int_0^y x^{-\alpha} (y-x)^{-\alpha} dx=\int_0^1 (yt)^{-\alpha} (y-yt)^{-\alpha} y\,dt=y^{1-2\alpha}\int_0^1 t^{-\alpha} (1-t)^{-\alpha}\,dt\\ =y^{1-2\alpha}B(1-\alpha,1-\alpha).$$

  • Thank you! Please would you confirm that the solution thus decays as ~ $y^{1-2 \alpha}$? Beta function is just acting as a constant right? – user3817704 Mar 16 '16 at 14:29
  • @user3817704: in terms of $y$ or in terms of $\alpha$ ? –  Mar 16 '16 at 14:46
  • In terms of $y$. Impulse response was defined IRF($y$) = $y^{-\alpha}$ and I need to express the powerlaw decay of autocorrelation in terms of $\alpha$ parameter. – user3817704 Mar 16 '16 at 14:50
  • @user3817704: the answer should be obvious to you. –  Mar 16 '16 at 14:51
  • Dear Yves, I think I understand it correctly. Just that my question serves a project at work and I want explicit confirmations that what I do is correct. Anyway, I thank you sincerely for you answer, best regards. – user3817704 Mar 16 '16 at 15:09
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Hint write it as $(xy-x^2)^{-\alpha}dx$ now y is a constant as variable is x which is easy to integrate.

Archis Welankar
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