This is from a paper I am reading.
It is given that $g(x)$ is a $C^1$ function on $(1,\infty)$.
The authors consider a cosine-type transform $\psi[g]$ of $g$ as $$ \psi(\xi) = \frac{2}{\pi} \int_{1}^{\infty}g(x) \cos(x \xi) \; dx $$ (the lower limit is 1, not 0)
Next, they let the integral $$ h(s) = -\int_{0}^{\infty} \frac{1}{\xi} \; \psi(\xi) \cos(\xi \,s)\; d\xi $$
It is given that $s > 1$. I am trying to understanding the restrictions that must apply to the original function $g$ so that $h$ exists.
For instance, it is is easy to verify that there are $C^1$ functions like $g(x) = \exp{(-x)}$ for which the second integral diverges. In this case (ignoring a constant coefficient)
$$ \psi(\xi) \sim \frac{\cos{\xi}-\xi \sin{\xi}}{1 + \xi^2} $$
On substituting this, the leading produces a divergent integral, i.e.
$$ \int_{0}^{\infty} \frac{1}{\xi} \; \frac{\cos{\xi}}{1 + \xi^2} \cos(\xi \,s)\; d\xi $$ does not exist, even as a Cauchy principal value, so I assume there must be some other conditions on $g(x)$.
Alternatively, is there some way to regularize such integrals or interpret them in the sense of generalized functions?
I must add that the authors of the paper have nothing at all to say about this issue, not even an aside, and carry on with their formal manipulations from that point on.
Thanks.
Let $g$ continuous and $f(x) = g(|x|)1_{|x| > 1} \in L^1$. Then
$$\hat{f}(\xi) = \int_{-\infty}^\infty f(x) e^{-2 i\pi \xi x}dx = 2 \int_1^\infty g(x) \cos(2 \pi \xi x)dx = 2 \psi(\xi)$$
And so the Fourier inversion theorem gives if $\hat{f} \in L^1$ : $$g(|x|)1_{|x| > 1} = f(x) = \int_{-\infty}^\infty \hat{f}(\xi) e^{2 i\pi \xi x}d\xi = 4\int_0 \psi(\xi) \cos(2 \pi \xi x)d\xi$$ (except the special case $|x|=1$ giving $\frac{g(1)}{2}$)
Also if it converges then $$h(x ) = \int_{-\infty}^\infty \frac{\hat{f}(\xi)}{2i \pi \xi} e^{2 i\pi \xi x}d\xi \quad \implies \quad h'(x) = g(|x|)1_{|x| > 1}, \quad h(x) = C+\int_0^x g(|y|)1_{|y| > 1} dy$$ And we can show $C =0$ since $\frac{\hat{f}(\xi)}{2i \pi \xi}$ is odd.
And when it doesn't converge, the result stays the same when interpreted in principal value, or in the sense of distributions.