s2vunov

2022-09-03

Let a>0 and A>0, I am looking for the decay rate of the integral

$${\int}_{M}^{\mathrm{\infty}}+{\int}_{-\mathrm{\infty}}^{-M}\frac{{x}^{2}\mathrm{exp}(-(x-a{)}^{2})}{1+A{a}^{-1}\mathrm{exp}(-{x}^{2}/(1+{a}^{2}))}dx\phantom{\rule{1em}{0ex}}\text{as}a\to 0$$

There is no closed form answer for the integral. I have tried on Matlab that it should converge to zero much faster than power growth. I think the growth should be exponential types. Do we have some literature discussing this kind of issue? Thanks!

I have successfully obtained the growth rate of

$${\int}_{M}^{\mathrm{\infty}}+{\int}_{-\mathrm{\infty}}^{-M}\frac{{x}^{2}\mathrm{exp}(-(x-a{)}^{2})}{1+A{a}^{-1}\mathrm{exp}(-{x}^{2}/(1+{a}^{2}))}dx\phantom{\rule{1em}{0ex}}\text{as}a\to 0$$

be expanding the denominator in power series.

But I do not know to deal with the integral in [−M,M].

$${\int}_{M}^{\mathrm{\infty}}+{\int}_{-\mathrm{\infty}}^{-M}\frac{{x}^{2}\mathrm{exp}(-(x-a{)}^{2})}{1+A{a}^{-1}\mathrm{exp}(-{x}^{2}/(1+{a}^{2}))}dx\phantom{\rule{1em}{0ex}}\text{as}a\to 0$$

There is no closed form answer for the integral. I have tried on Matlab that it should converge to zero much faster than power growth. I think the growth should be exponential types. Do we have some literature discussing this kind of issue? Thanks!

I have successfully obtained the growth rate of

$${\int}_{M}^{\mathrm{\infty}}+{\int}_{-\mathrm{\infty}}^{-M}\frac{{x}^{2}\mathrm{exp}(-(x-a{)}^{2})}{1+A{a}^{-1}\mathrm{exp}(-{x}^{2}/(1+{a}^{2}))}dx\phantom{\rule{1em}{0ex}}\text{as}a\to 0$$

be expanding the denominator in power series.

But I do not know to deal with the integral in [−M,M].

Mario Monroe

Beginner2022-09-04Added 12 answers

Let us assume $a\ll 1$. Approximating the function (or simply plotting it), we can see that the maximum of the integrand is at $|x|\ge 1$ (in fact it is $|x|=1$ for $A\ll a$ and becomes larger when increasing A).

Because if this we can approximate $1+{a}^{2}\approx 1$ and $(x-a{)}^{2}\approx {x}^{2}$ in the exponents of the integrand to obtain the leading order behavior.

Denoting the integral by I(a,A), we obtain

$$I(a,A)\simeq {I}_{0}(a,A)={\int}_{-\mathrm{\infty}}^{\mathrm{\infty}}{\displaystyle \frac{{x}^{2}\mathrm{exp}(-{x}^{2})}{1+A{a}^{-1}\mathrm{exp}(-{x}^{2})}}dx=-\frac{\sqrt{\pi}a}{2A}{\mathrm{Li}}_{3/2}(-A/a)\phantom{\rule{thinmathspace}{0ex}}.$$

Here, Lis is the polylogarithm function. In fact, I0 is an excellent approximation to I for $a\lesssim 0.1$

The polylogarithm function has a known asymptotic expansions in terms of logx. In particular, we have $x\gg 1$

$${\mathrm{Li}}_{s}(-x)\sim -\frac{{\mathrm{log}}^{s}(x)}{\mathrm{\Gamma}(s)}{\textstyle (}1+O({\mathrm{log}}^{-2}(x)){\textstyle )}\phantom{\rule{thinmathspace}{0ex}}.$$

As a result, we obtain

$$I(a,A)\simeq {I}_{0}(a,A)\sim \frac{2a}{3A}{\mathrm{log}}^{3/2}(A/a)\phantom{\rule{thinmathspace}{0ex}}.$$

Note that this is not a rigorous mathematical answer. What remains is to show that $I-{I}_{0}$ is small for $a\to 0$. Numerically, it seems to hold that $I-{I}_{0}=o({a}^{2})$

Because if this we can approximate $1+{a}^{2}\approx 1$ and $(x-a{)}^{2}\approx {x}^{2}$ in the exponents of the integrand to obtain the leading order behavior.

Denoting the integral by I(a,A), we obtain

$$I(a,A)\simeq {I}_{0}(a,A)={\int}_{-\mathrm{\infty}}^{\mathrm{\infty}}{\displaystyle \frac{{x}^{2}\mathrm{exp}(-{x}^{2})}{1+A{a}^{-1}\mathrm{exp}(-{x}^{2})}}dx=-\frac{\sqrt{\pi}a}{2A}{\mathrm{Li}}_{3/2}(-A/a)\phantom{\rule{thinmathspace}{0ex}}.$$

Here, Lis is the polylogarithm function. In fact, I0 is an excellent approximation to I for $a\lesssim 0.1$

The polylogarithm function has a known asymptotic expansions in terms of logx. In particular, we have $x\gg 1$

$${\mathrm{Li}}_{s}(-x)\sim -\frac{{\mathrm{log}}^{s}(x)}{\mathrm{\Gamma}(s)}{\textstyle (}1+O({\mathrm{log}}^{-2}(x)){\textstyle )}\phantom{\rule{thinmathspace}{0ex}}.$$

As a result, we obtain

$$I(a,A)\simeq {I}_{0}(a,A)\sim \frac{2a}{3A}{\mathrm{log}}^{3/2}(A/a)\phantom{\rule{thinmathspace}{0ex}}.$$

Note that this is not a rigorous mathematical answer. What remains is to show that $I-{I}_{0}$ is small for $a\to 0$. Numerically, it seems to hold that $I-{I}_{0}=o({a}^{2})$

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