Consider the following two integrals:
$I_1=\int\limits_{0}^{1/x}G(s)\ ds$
$I_2=\Big|\int\limits_0^\infty G(s)\exp[-ixs]\ ds\Big|$
where $G(s)$ is a monotonically decreasing positive function, $i=\sqrt{-1}$ and $\Big|\cdot\Big|$ denotes absolute value. I have observed from playing around in Mathematica that for functions such as $G(s)=s^{-0.5}$ or $G(s)=\exp[-s]$ which satisfy the conditions placed on $G(s)$ the two integrals are very close in value, i. e., $I_1\approx I_2$. I would like to come up with some quantitative measure of the difference between the functions. Can this be done in a general fashion without a specific form for $G(s)$?
Assume $G$ is contained in $L^1(\mathbb{R})$. Then $\lim_{x\rightarrow\infty} I_2(x) = 0$ by the Riemann-Lebesgue lemma. The first integral vanishes for large $x$ since the domain of integration shrinks; apply dominated convergence theorem to $G\cdot\chi_{[0,1/n]}$.
For intermediate values of $x$, the result is false. Fix a compactly supported function $G$ and fix a value of $x_0>0$ so that $I_2(x_0)>0$. Via translation, you can place the support of $G$ in the complement of $(-\infty,1/x_0]$ so that $I_1(x_0)=0$. This will not change the modulus of the Fourier transform, leaving $I_2(x_0)$ unaffected. Moreover, by multiplying $G$ by a scalar we can attain any desired positive value for $I_2(x_0)$.