"Asymptotic behavior of int_inf^-inf (x^2 exp( (x−a)^2))/(1+Aa^(−1) exp(−x^2/(1+a^2))) dx as a->0 Let a>0 and A>0, I am looking for the decay rate of the integralint_inf^-inf (x^2 exp( (x−a)^2))/(1+Aa^(−1) exp(−x^2/(1+a^2))) as a->0. Do we have some literature discussing this kind of issue?

s2vunov 2022-09-03 Answered
Let a>0 and A>0, I am looking for the decay rate of the integral
M + M x 2 exp ( ( x a ) 2 ) 1 + A a 1 exp ( x 2 / ( 1 + a 2 ) ) d x  as  a 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
M + M x 2 exp ( ( x a ) 2 ) 1 + A a 1 exp ( x 2 / ( 1 + a 2 ) ) d x  as  a 0
be expanding the denominator in power series.
But I do not know to deal with the integral in [−M,M].
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Answers (1)

Mario Monroe
Answered 2022-09-04 Author has 12 answers
Let us assume a 1. Approximating the function (or simply plotting it), we can see that the maximum of the integrand is at | x | 1 (in fact it is | x | = 1 for A a and becomes larger when increasing A).
Because if this we can approximate 1 + a 2 1 and ( x a ) 2 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 ) I 0 ( a , A ) = x 2 exp ( x 2 ) 1 + A a 1 exp ( x 2 ) d x = π a 2 A Li 3 / 2 ( A / a ) .
Here, Lis is the polylogarithm function. In fact, I0 is an excellent approximation to I for a 0.1
The polylogarithm function has a known asymptotic expansions in terms of logx. In particular, we have x 1
Li s ( x ) log s ( x ) Γ ( s ) ( 1 + O ( log 2 ( x ) ) ) .
As a result, we obtain
I ( a , A ) I 0 ( a , A ) 2 a 3 A log 3 / 2 ( A / a ) .
Note that this is not a rigorous mathematical answer. What remains is to show that I I 0 is small for a 0. Numerically, it seems to hold that I I 0 = o ( a 2 )
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