Integral Theory?

Question

Let R+ = {x | x in R, x > 0}, where R is the field of real numbers.
Suppose A is a subset of R+, and define 1/A = {1/y | y is in A}

Note that 1/A is also a subset of R+

Further, let C be the field of complex numbers and suppose
f: A --> C
is such that f is integrable on A, and further, f is integrable on 1/A (integrable in this context could be restricted to general Riemann integration, but Lebesgue integration may be more useful)

Question:
Is there a way to evaluate the following expression:

∫ (A) f(x) f(1/x) dx

I know it should be defined, because, letting g(x) = f(1/x), as f is integrable on 1/A, then g is integrable on A, so the product fg is integrable on A...

I'm wondering if there's a way to evaluate it.

As a particular case, consider the integral:
∫ (pi to 3pi) e^((e^(ix))/2) e^((e^(-ix))/2) dx

(yes, this is ∫ (pi to 3pi) e^(cos(x)) dx, but I wanted to see if there was a trick to the more general case, as I'm guessing this integral is done by contour integration)

JCS · Accepted Answer

EDIT: how did you obtain the expression

-∫[A]x^2*f(x)dg(x) ?

If you define g(x) = f(1/x) then, assuming differentiability a.e. for f(x), you'll have 

dg(x) = −(1/x^2)f'(1/x)dx

not

dg(x) = −(1/x^2)f(1/x)dx 

And substituting the former would not result in your original expression, but in

∫ (A) f(x) f'(1/x) dx

-----------------------------------------------------------------------------------

If you just want to compute the integral, then my bet would also be countour integration.

If you want to look at integrals of this type from the theoretical side, then what you have is a Lebesgue - Stieltjes integral (or Riemann - Stieltjes, depending on how you choose to define the base integral). In a nutshell, these are Lebesgue (or Riemann) integrals of the form:

∫[A]f(x)dg(x)

Where g(x) is a sufficiently regular function that defines a measure on A; in your particular case, you have an integral over (R+,μ), where μ is the measure defined by g(x) = f(1 / x).

Note that these integrals are mainly theoretical tools, it's unlikely that they will be of some help in evaluating particular cases.

? · Answer

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匿名 · Answer

Interesting question. I would remark that if we define g(t)=f(exp(t)) and ln(A)={ln s;s in A}, then your integral equals ∫ (ln A) g(x)g(-x)exp(x) dx

Integral Theory?

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