1
$\begingroup$

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 $G(s)$?

$\endgroup$
0

1 Answer 1

2
$\begingroup$

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 $I_2(x_0)$.

$\endgroup$
4
  • $\begingroup$ Thanks for your answer. I understand that the result is false for intermediate values of $x$ in the sense that the integrals are not equal. I was looking for a way to quantify the difference for general $G(s)$. If I assume that $G(s)$ is a strictly decreasing function (and under assumptions that it decays fast enough), can I quantify the difference between the integrals in terms of G(s)? For example, if G(s) decays very fast, the limits of $I_1$ can be changed to $0,\infty$ without much error and maybe it becomes easier to compare the integrals? $\endgroup$ Commented Dec 4, 2013 at 17:13
  • $\begingroup$ My second comment shows that for functions decaying extremely fast (compactly supported), the difference $|I_1(x)-I_2(x)|$ can be as large as $\sup_x I_2(x)$. (Note that the supremum makes sense since $I_2$ is a continuous function of $x$ decaying at infinity if $G \in L^1$.) $\endgroup$
    – dls
    Commented Dec 4, 2013 at 17:40
  • $\begingroup$ I guess my answer doesn't respect your monotonicity criterion. It may be helpful if you include a more specific comparison in the original question: a function $G$ and a table of values of $I_1$ and $I_2$ for various values of $x$ to demonstrate the behavior you're describing. $\endgroup$
    – dls
    Commented Dec 4, 2013 at 19:15
  • $\begingroup$ How about we assume $G(s)=s^{-.5}$, or $\exp[-s]$. Of course for these specific simple cases the integrals can be done analytically and we can easily quantify the difference between $I_1(x)$ and $I_2(x)$. I'm looking for ways to quantify the difference in a general fashion for any strictly monotonically decreasing real valued (and positive) $G(s)$. $\endgroup$ Commented Dec 6, 2013 at 0:19

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .