I am strongly impressed by this example from New in 13
as =
AsymptoticIntegrate[
(t^10 + 3) Exp[I λ (t^5 + t + 1)],
{t, -2, 2}, {λ, Infinity, 2}
]
(250400/(531441 λ^2) + (1027 I)/(81 λ)) Cos[33 λ] + (250400/(531441 λ^2) - (1027 I)/(81 λ)) Cos[35 λ] + (-((250400 I)/(531441 λ^2)) + 1027/(81 λ)) Sin[33 λ] + ((250400 I)/(531441 λ^2) + 1027/(81 λ)) Sin[35 λ]
Indeed,
N[as /. λ -> 20]
0.511368 + 1.14401 I
is in a good accordance with
NIntegrate[
(t^10 + 3) Exp[I*λ (t^5 + t + 1)] /. λ -> 20,
{t, -2, 2}
]
0.511551 + 1.14442 I
I am interested how AsymptoticIntegrate
derives it. Usually the command works as Series@Integrate
, but in the case under consideration
Integrate[
(t^10 + 3) Exp[I*λ (t^5 + t + 1)],
{t, -2, 2},
Assumptions -> λ > 0
]
fails and returns the input.
Is the Laplace method (with generalizations) really implemented in Mathematica? I clearly understand that this is proprietary knowledge of Wolfram Research Inc, but I hope to see a general answer; I do not expect implementation details.
AsymptoticIntegrate[t^-t*t^x, {t, 0, Infinity}, {x, Infinity, 1}, Assumptions -> x > 0]
. $\endgroup$Integrate[t^-t, {t, 0, \[Infinity]}]
closed form solution Not exist yet. $\endgroup$Integrate[(t^10 + 3) Exp[I*\[Lambda] (t^5 + t + 1)], {t, -2, 2}, Assumptions -> \[Lambda] > 0]
, butAsymptoticIntegrate
cracks it as $\lambda \to \infty$. $\endgroup$AsymptoticIntegrate[ Sin[t]/t* Exp[I \[Lambda] (t^5 + t + 1)], {t, -2, 2}, {\[Lambda], Infinity, 1}]
fails in a long time, thoughIntegrate[Sin[t]/t,{t,-2,2}]
is expressed through special functions. $\endgroup$