Sure way to convert variable integral to infinite integral?

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  • #1
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Sometimes I would like to transform an integral ##F(x) = \int_{a}^{x}f(s)ds## into an infinite integral of the form ##F(x) = \int_{0}^{\infty}f(g(u),x)du##. Is there some kind of change of variables that can guarantee this conversion on the boundaries and still give me a function of ##x##, at least with some assumptions like differentiable or invertibility? Or are there maybe certain classes of functions for which this would work like it exists on ##(0,\infty)##?
 

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  • #2
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 ?
 
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  • #3
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That looks interesting, but is there a method to deduce ##\theta##?
 
  • #5
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Wait, are you saying that you just multiply the function by 0 over a certain bound and take it to infinity? I don't know if that quite works because I am looking for is a change of variables and it doesn't quite make sense to change a general integrable function to a constant. But for instance, it is reasonable to make a change of variables that might change an integral to something like the integral for the gamma function.
 
  • #6
anuttarasammyak
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I am pretty sure it works or not depends on what you want to do after conversion. Tell me your investigation on your problem if you please.
 
  • #7
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I am pretty sure it works or not depends on what you want to do after conversion. Tell me your investigation on your problem if you please.
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.
 
  • #8
anuttarasammyak
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[tex]y(x)=\int_0^x f(x,s)ds[/tex]
[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.
 
  • #9
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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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  • #10
pasmith
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[itex][0, \infty) \to [0, 1) : x \mapsto \tanh(x)[/itex] is the usual choice for turning an infinite reange into a finite range.
 
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  • #11
pasmith
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For completeness: [tex]
\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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