Laplace integrals - Watson's lemma

This notebook  applies Watson' s lemma to three cases of 
        Laplace - 
      type integrals 
        (in all cases in the limit of x -> ∞) :

\(TraditionalForm
    \`\(1.\ \ \ \ \ \ 
      \(∫\_0\%b\( f(t)\) \(e\^\(\(-\ x\)\ t\)\) 
          dt\)\ \)\)

As an example, let's derive a 6-term expansion in the case of f[t]=1/(1+t):

n = 6 ;  f[t_] := 1/(1 + t)

We Taylor expand f[t]:

ser = Series[f[t], {t, 0, n}]

and the asymptotic expansion through Watson's lemma becomes

expansion = 
      Sum[Coefficient[ser, t, k]*k!/x^(k + 1), {k, 0, n}]

\(TraditionalForm
    \`2.\ \ \ \ \ \ 
        \(∫\_a\%b\( f(t)\) \(e\^\(x\ \(ϕ(t)\)\)\) 
            dt\ \ \ \ with\ \ ϕ\_\(max\ \)\ taken\ at\ \ t
            \) = a; \ \ \(ϕ'\)[a] < 0\)

As an example, we consider

f[t_] := 1/(1 + t);   ϕ[t_] := -ArcTan[t];  a = 1; 
    b = Infinity;  n = 6;

The first step is to change variable -ϕ[t]=s. To express t as a function of s, we need first to expand ϕ[t] as a Taylor series in t:

ϕt = Series[ϕ[t], {t, a, n}]

We now get t as a function of s by series inversion (remembering the minus sign)

ts = InverseSeries[-ϕt, s]

We get  f expanded in this new variable s through  f[ts]. We also need to change dt into (dt/ds) ds

fsdtds = f[ts]  D[ts, s]

The  original  integral 
            (∫_a^bf)[t] (e^(x  ϕ[t])) 
                dt  has  now  been  transformed  over  to  (∫_(-ϕ (a))^□) fs  dt/ds 
             (e^(-x s)) ds.We  are  ready  to  apply 
             Watson' slemma.We need to extract the expansion 
            coefficients of  fsdtds.

It is slightly tricky to do this from a series when it is not expanded around the origin, but around a point ϕ[a]. The following gives the k:th coefficient (note the := symbol, which delays execution until the result is actually used)

ck := 
      Simplify[Coefficient[Normal[fsdtds]  /.   s -> τ - ϕ[a], 
          τ, k]]

All is now set for applying Watson's lemma

expansion = Sum[Exp[ϕ[a]]*ck*k!/x^(k + 1), {k, 0, n}]

Remove["Global`*"]

\(TraditionalForm
    \`3.\ \ \ \ \ \ 
        \(∫\_a\%b\( f(t)\) \(e\^\(x\ \(ϕ(t)\)\)\) 
            dt\ \ \ \ with\ \ ϕ\_\(max\ \)\ taken\ at\ \ t
            \) = c; \ \ \(\(ϕ'\)'\)[c] < 0\)

This example follows closely the structure of case 2 above. So we comment only on the mathematical concepts.

f[t_] := 1/t;   ϕ[t_] := -t + Log[t];  a = 0; b = Infinity; 
    n = 4;

We can use Mathematica to find  the location\(TraditionalForm\`	
    t = c\ \ \ at\ \ which\ \ ϕ\_max\ \ is\ taken.\ A\ plot\ of\ 
        ϕ[t]\ may\ give\ a\ good\ idea\ of\ where\ the\ max\ can\ be\ 
        found.\ In\ a\ simple\ case\ \((as\ this\ one)\), \ 
    it\ is\ enough\ to\ \ pick\ the\ first\ \((here\ only)\)\ zero\ of\ 
      \(ϕ'\)[t]\)

c = t  /.  Solve[ϕ'[t] == 0, t] [[1]]

We should now check that c is in the inside of [a,b]. To start with, we consider only the contribution from t>c. (The contribution from t<c turns out to be obtained with no additional effort), as commented on below. Turns out that we have to expand most series this time to 2n terms in order to get n terms in the filal result.

ϕt = Series[ϕ[t], {t, c, 2 n}]

ts = InverseSeries[-ϕt, s]

fdtds = Simplify[f[ts]]    D[ts, s]

At this point, we are ready to apply Watson's lemma. The contributions from the case t<c turn out to give rise to an identical expansion with the exception that the sign for all integer powers get reversed. So to get both contributions, we simply extract the fractional power terms, and double their coefficients.

\(ck := 
      Simplify[Coefficient[Normal[fdtds]\  /. \ \ s -> τ - ϕ[c], 
          τ, k]]; 

    expansion = 
      Exp[x*ϕ[c]]*
        Sum[2*ck*Gamma[k + 1]/x^\((k + 1)\), {k, \(-\(1\/2\)\), n - 1\/2}]\)