Do These Integrals Always Exist Under Certain Conditions?

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    Existence Integrals
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SUMMARY

The integrals discussed, specifically \(\int_{-\infty}^{\infty} dx \cos(uf(x))\) and \(\int_{-\infty}^{\infty} dxe^{-ag(x)}\), exist under the conditions where \(a > 0\) and \(g(x) > 0\) for all real \(x\). As \(a\) and \(u\) approach infinity, both integrals tend to zero. The discussion highlights the relationship between these integrals and the WKB approach in physics, particularly in defining the wave function \(\psi(x) = A \cos(S(x)/\hbar)\) in the semi-classical limit. The transformation of \(g(x)\) into its inverse \(h(x)\) illustrates the Laplace transform behavior, reinforcing the conclusion that the integrals converge to zero for large parameters.

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lokofer
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Let be the integrals:

[tex]\int_{-\infty}^{\infty}dx Cos(uf(x))[/tex] (or the same but a sine) and

[tex]\int_{-\infty}^{\infty}dxe^{-ag(x)}[/tex]

Where "a" is a a>0 positive constant, u can be either positive or negative.. and g(x)>0 for every real x.. my question is will these integrals "always2 exist under these conditions?..what would happen if we take the limit a-->oo and u-->oo ? are in this case equal to 0?
 
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For the second case, set g(x)=1 (for all x). What do you see?
 
Oh..then sorry "Arildno".. then for similar case with f(x) and g(x) different from f(x)=C (C a real constant) and g(x)>0 i think the integrals should tend to 0 for big u and a, and that they exist..for example for the WKB approach in Physics if you definte the "Action" S of the system the approximate wave function can be written as [tex]\psi(x)= ACos(S(x)/\hbar)[/tex] so in the "Semi-classical " limit (h-->0 ) the Wave function is 0 , A is a normalization constant.

- for the case of real exponential if you take g(x)=h(x) so h(x) is the "inverse" of g(x) the function becomes a "Laplace transform" of h'(x) that tends to 0 for big a
 

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