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anuttarasammyak

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How about

[tex]\int_0^\infty f(u+a)[1-\theta(u-x+a)]du[/tex]

where ##\theta(x)## = 0 for x<0, 1 for x>0 ?

[tex]\int_0^\infty f(u+a)[1-\theta(u-x+a)]du[/tex]

where ##\theta(x)## = 0 for x<0, 1 for x>0 ?

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That looks interesting, but is there a method to deduce ##\theta##?

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anuttarasammyak

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https://en.wikipedia.org/wiki/Sign_function shows some features of ##sgn(x) = 2\theta(x) -1##.

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anuttarasammyak

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I recall playing around with Beta-like integrals and there was a substitution I could make for ##y(x) = \int_{0}^{x}(f(x,s)ds); s \rightarrow \frac{u}{x}## that changed the bound from ##0 \rightarrow x## to ##0 \rightarrow 1## which then allowed me to satisfy the definition of the beta function. I feel like a similar thing should be possible but with an infinite bound, that there should be some guaranteed way to make a bound go from 0 to infinity.

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anuttarasammyak

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[tex]u= tan \frac{\pi s}{2x}[/tex]

[tex]s=\frac{2x}{\pi} tan^{-1}u[/tex]

[tex]y(x)=\frac{2x}{\pi} \int_0^\infty f(x,\frac{2x}{\pi} tan^{-1}u)\frac{d\tan^{-1}u}{du}du[/tex]

Though divergence at u=0 should be investigated, it might be a kind of your line of transformation.

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That seems like a good idea. So more generally, an invertible function which maps 0 to 0 and has a horizontal asymptote as ##x \rightarrow \infty##.

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pasmith

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pasmith

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\int_a^x f(s)\,ds = \int_0^\infty f\left(a + (x - a)\tanh t\right)\frac{(x - a)}{\cosh^2 t}\,dt.[/tex]

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