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If $f(x)=\text{arccot}(x)$ for non-negative $x$ and $0$ otherwise, how can I calculate

$$\int_{-\infty}^\infty f(x)f(y-x)\, \mathrm dx$$

for $y\in\mathbb{R}$?

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The question is when the convolution of two functions exist? –  Mhenni Benghorbal Jun 11 '13 at 23:36
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1 Answer

$f\left(x\right)$ is this plot but with $f$ zero for $x < 0$ and $f\left(y-x\right)$ is $f\left(x\right)$ flipped about $x = 0$ and then displaced in the positive $x$ direction by $y$. So, for:

  • $y < 0$, there is no overlap, and the integral is zero.
  • $y > 0$, the overlap is from 0 to $y$, so you should integrate $\cot^{-1}\left(x\right) \cot^{-1}\left(y-x\right)$ from 0 to $y$.
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This is just the question phrased differently, right? I am interested in integrating $\text{arccos}(x)\text{arccos}(y-x)\, dx$ i.e. what strategy do I use (forget the limits)? –  user39657 Sep 6 '12 at 22:55
    
Well, considering that $f\left(x\right) \ne \cot^{-1}\left(x\right)$, part of the solution consists of determining, for a given $y$ value, the values of $x$ for which $f\left(x\right)$ and $f\left(y-x\right)$ overlap. It seems that the actual integrating of $\cot^{-1}\left(x\right)\cot^{-1}\left(y-x\right)$ is nontrivial. –  Eric Angle Sep 6 '12 at 23:23
    
Hence my question on this site :) –  user39657 Sep 6 '12 at 23:32
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