Sure way to convert variable integral to infinite integral?

[LieToMe]

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 $(0,\infty)$?

 

[anuttarasammyak]

How about
$$\int_0^\infty f(u+a)[1-\theta(u-x+a)]du$$
where $\theta(x)$ = 0 for x<0, 1 for x>0 ?

 

[LieToMe]

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

 

[anuttarasammyak]

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

 

[LieToMe]

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.

 

[anuttarasammyak]

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.

 

[LieToMe]

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.

 

[anuttarasammyak]

$$y(x)=\int_0^x f(x,s)ds$$
$$u= tan \frac{\pi s}{2x}$$
$$s=\frac{2x}{\pi} tan^{-1}u$$
$$y(x)=\frac{2x}{\pi} \int_0^\infty f(x,\frac{2x}{\pi} tan^{-1}u)\frac{d\tan^{-1}u}{du}du$$
Though divergence at u=0 should be investigated, it might be a kind of your line of transformation.

 

[LieToMe]

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

 

[pasmith]

$[0, \infty) \to [0, 1) : x \mapsto \tanh(x)$ is the usual choice for turning an infinite reange into a finite range.

 

[pasmith]

For completeness: $$\int_a^x f(s)\,ds = \int_0^\infty f\left(a + (x - a)\tanh t\right)\frac{(x - a)}{\cosh^2 t}\,dt.$$