I Assigning a value for integrating a divergent oscillatory function to infinity

Summary
Is it possible to assign a meaningful value to the integral (up to infinity) of oscillating divergent functions?
There are meaningful ways to assign values to things like
1 - 1 + 1 + ...
or
1 - 2 + 3 - 4 + ...
In a similar spirit, is it possible to assign a value to the integral of a function like this: ##f(x)=x*sin(x)##
or this one:
##g(x)=Re(x^{1+5i})##



1564329191332.png


(Integrals from some value, say zero, up to infinity)
 
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Perhaps "meaningful" is not the best term here. In the case of divergent sums, Hardy uses words like "reasonable" and "natural". So speaking very loosely, a procedure for which one can claim that it is the summation of a divergent series in some justifiable sense*, and which yields the correct answer for convergent series, and behaves like a conventional sum as much as possible.

So my question was, is there an analogous notion for integration that would extend the idea of a convergent integral and help to define the integral of a divergent, oscillating function?

* Hardy gives the hypothetical example of inventing a new sum and calling it a sum in the "Pickwickian" sense.
 
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If anything works for this kind of function at all, then its probably the residue theorem. But you need to turn it into a holomorphic function first (g(x) is not one).
 
If anything works for this kind of function at all, then its probably the residue theorem. But you need to turn it into a holomorphic function first (g(x) is not one).
I needed to look up the residue theorem, which I have done and got some very basic understanding.

I'm also trying to understand the properties of ##f(z)=z^{(a+ib)}##. For example, with a+ib = 1 + 5i there is a discontinuity along the negative real axis. Is this the reason that we say ##f(z)## is not holomorphic? Can we select a contour that avoids the break, such that it will help with the g(x) in my OP?

And BTW, is the discontinuity along the negative real axis caused by the argument (angle) of z going from ##\pi## to ##-\pi##?

1564709411455.png
 
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Summary: Is it possible to assign a meaningful value to the integral (up to infinity) of oscillating divergent functions?

[...] is it possible to assign a value to the integral of a function like this: f(x)=x∗sin(x) [...]
Sure! However, as you probably know, an antiderivative to ##f(x) = x\sin(x)## is ##F(x) = \sin(x) - x\cos(x)##. Thus, the original improper integral is
\begin{align}
I &= \int_0^\infty x\sin(x)\,dx \\ &= \lim_{a\rightarrow\infty}\Big[\sin(x) - x\cos(x)\big].
\end{align}
This limit doesn't exist and ##I## is accordingly divergent!

Now, consider the Laplace transform of our original function ##f(x)##. That is,
\begin{align}
\mathcal{L}[f](s) &= \int_0^\infty x\sin(x)e^{-sx}\,dx \\ &\ \ \vdots \nonumber\\ &= \frac{2s}{(1 + s^2)^2}.
\end{align}
This function is defined for ##\text{Re}(s)>0## in the complex plane. However we can use this result to regularize our original integral ##I## by noticing that
\begin{align}
I "=" \lim_{s\rightarrow 0} \mathcal{L}[f](s) = 0.
\end{align}
I've putted the equal sign in qutations, since this equality isn't strictly true in the conventional interpretation of this symbol.
 
Now, consider the Laplace transform of our original function f(x). ... We can use this result to regularize our original integral II by noticing that ##I "=" \lim_{s\rightarrow 0} \mathcal{L}[f](s) = 0##.
Is there any sort of equivalence between the Laplace transfrom method and the Cesaro method? Or can they sometimes give different answers?
 
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