If I would know the application and meaning of all variables, maybe I could help. Most likely you have a variable (y?) and some parameters (h1,h2, lambda, beta). Asymptotic expansions are used when we have either small (x1) parameter. Asymptotics can be in powers of this parameter, but also in other functions.

Your formulation of the problem is not rigorous. There is no final objective. Application of some method just to apply it is never a goal. It should have some advantage (precision, simplicity) over others to reach a certain objective.

Hi. Is it for lambda real going to +infinity ? Since exp(a lambda)

If I would know the application and meaning of all variables, maybe I could help. Most likely you have a variable (y?) and some parameters (h1,h2, lambda, beta). Asymptotic expansions are used when we have either small (x1) parameter. Asymptotics can be in powers of this parameter, but also in other functions.

Your formulation of the problem is not rigorous. There is no final objective. Application of some method just to apply it is never a goal. It should have some advantage (precision, simplicity) over others to reach a certain objective.

 Hi,  you could manage it using a computer algebra system, e.g. XCAS which is a freeware.

Just compute a 1st or 2nd order Taylor series with the Lambd parameter.

Thanks Miss and Mrs, i'll add some assumptions, h1, h2 and h3 are geometric parameters, they are given in order as follow h3>h1>h2, Lampda is the independant variable (it is the Fourrier variable since the Fourrier transform is used to arrive to this expression) the asymptot is needed for Lampda going to infinity, Beta is a physical parameter it is given by this expression: βh2=ln(μ3/μ1). Ln is the neperian logarithme, μ3 and μ1 are a physical properties, two cases can be considered which are  μ3>μ1 and μ3

Now that we know a lot about the relative magnitudes of your parameters, I think we can find your asymptotics without that many cases.

Do you want only the first dominant term in your asymptotics? I mean, if you had Lambda2 + Lambda + 1 for example, would you care only for the dominant term Lambda2, or would you want to know the second term Lambda as well, or even all of them in order? I'll assume you only want the first. But even in that case, we need to keep some second terms while working on the expression, because some dominant terms cancel each other out.

And is h2 > 0 by any chance? I guess that's something you meant when saying the h's were geometric parameters.

Now if I were you I would start with replacing all the

Sqrt(β2 + 4 Lambda2) = 2 Lambda Sqrt(1 + (β2/4)/Lambda2)

= 2 Lambda (1 + (1/2) (β2/4)/Lambda2 + ...)

with its equivalent form  2 Lambda + (β2/4) / Lambda + ...  

where ... stands for negligible terms compared to the previous term, when Lambda goes up to infinity. I've simply used the fact that Sqrt(1 + x) = 1 + (1/2) x + O(x2) when x goes towards 0. Note that we may need to further expand ... if all previous terms cancel out at some point during our simplifications, but you can keep it like that for now.

You'll then notice that several simplifications occur in your expression, because some 2 Lambda that were already there now cancel out with some new 2 Lambda term, leaving only some (β2/4) / Lambda + ...

Could you show us what is the resulting expression you get, so that we can keep working on it together?

Regards

all the physical and geomertic parameters are strictly >0, the goal is to find the domminant part when Lambda goes to infinity. In some Articles i am using they give up to the eigth term of the asymptot.

I'm working on it, i''l send you when if finish it. Thanks a lot

Dear Brick, I read your text. But I never studied such problems. I searched for articles relevant for your problem. Maybe like this: https://projecteuclid.org/download/pdf_1/euclid.jiea/1333560449 .

Claude gives you a correct suggestion how to present square root with function of lambda into power functions. Another idea is to replace exponents of (lambda y) with the sum of functions:  A (lambda) cosh (lambda y) +B(lambda) sinh (lambda y). Since y is very small, use Taylor expansion for them at y near 0. Then cosh x=1+x^2/2 +o(x^4), while sinh x= x +x^3/6 +o(x^5)), where x=lambda*y. See https://en.wikipedia.org/wiki/Hyperbolic_function . If lambda behaves like y^(-1/2), then x is like y^(1/2), i.e., small for y ->0. But this is just one of possible ideas. You will see when will start integrating those series. It is important that neglected terms are small comparing with those kept. Searching for asymptotic expansions that work for particular model is a bit of art, but general principles are like these.

Thanks Mr Yuri for your suggestions i'll take them along with Claude's suggestions to work on my problem, i'll keep you aware about the developements  later in a document as Claude suggested. I wish you a good weekend gentelmen

There is a great free course about getting asymptotics out of integrals on edx called "Complex Analysis with Physical Applications". It is of a very high level at some point but the first weeks are easier and wonderful. I have attached the link.

https://courses.edx.org/courses/course-v1:MISiSx+18.11x+1T2017/course/

The asymptotics of the expression k(y,lambda) is good to know for lambda going towards +infinity, but in order to estimate its integral over the positive reals, I think we need to find the lambda's for which k(y,lambda) is maximal (the dominant contribution to the integral). In order to find them, solving dk/dlambda = 0 for finite lambda's looks like a terrific task at first sight, but maybe it is manageable?

Hi All Followers,

I agree with most of the remarks above, in particular those  by  Yuri  and Claude: We need to know the physical meaning since the problem has a not sufficiently rigorous formulation.

My impression is that the formulation should contain  both variables  -  lambda   and  y.. Indeed, since the function k(y,lambda), for fixed  lmbda is a hiperbolic cosine or hyperbolic sine, one can guess that it describes a distribution of say an electric voltage or light in a line transporting energy in both opposite directions. In such cases the maximization should concern examplum gratie of  k(a,lambda)  or  k(b,lambda) at the ends  y=a  and/or  y=b  of the line.

Next: what kind of asymptotics is interesting?  If this is  the collection of  highest exponents as it is mentioned by Claude, then no expansion in powers of lambda is required.  For instance,  for given  y, the function is a sum  of  functions of the form

| Tj( lambda)  | := cj( lambda ) * exp( wj(  lambda ) ),

with coefficients   which have  asymptotics  of the form

cj(lambda)  ~  Cj * \lambda^(pj)   and    wj(lambda))  ~  Wj * \lambda^(qj)

each  with own  constan parameters   Cj, pj, Wj, qj

The terms Tj( lambda) determine the asymptotics of the function  k(y, lamda  )  according to the following ordering 

Tj 

Maybe look at "damped sine wave". Complex analysis yields the exact value of integrals such as from 0 to +oo of exp(-y*L) sin(a*L) dL. Maybe it would also work for integrals of stuffs like exp(-y*L) sin(a*L) / (1+exp(-b*L)) dL. When beta = 0 your expression is a finite sum of such stuffs. I'll have a go at it.

I wonder if you could use the section

Laplace Transforms with Small Parameters

in Chapter 2 Asymptotic Approximations

of the NIST Digital Library of Mathematical Functions

freely available online. I've attached a link. I agree it's hard to swallow at first sight.

http://dlmf.nist.gov/2.5#iii

I like Claude's  idea of using the LT or MT,  Mainly due to the possibility of changing the unreadable formula to a rational function (but then, considered near the origin of the domain of LT (or MT). However, the simplicity will not be evident untill the square-roots in the exponent are simplified by replacing them by linear functions of  lambda.

Best regards, Joachim