# How is this asymptotic expansion of an integral calculated?

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.

• Unfortunately, 13.0 fails with AsymptoticIntegrate[t^-t*t^x, {t, 0, Infinity}, {x, Infinity, 1}, Assumptions -> x > 0]. Dec 16, 2021 at 18:57
• Fails because for integral: Integrate[t^-t, {t, 0, \[Infinity]}] closed form solution Not exist yet. Dec 16, 2021 at 19:14
• @MariuszIwaniuk: Did you carefully read my question before having posted your comment? Version 13.0 fails with Integrate[(t^10 + 3) Exp[I*\[Lambda] (t^5 + t + 1)], {t, -2, 2}, Assumptions -> \[Lambda] > 0], but AsymptoticIntegrate cracks it as $\lambda \to \infty$. Dec 16, 2021 at 19:20
• @MariuszIwaniuk: The integral $$\int_0^\infty t^{-t}t^x\,dt$$ is known to be a test example for Laplace's method. Don't hesitate to ask for further explanation in need. Dec 16, 2021 at 19:27
• @MariuszIwaniuk: AsymptoticIntegrate[ Sin[t]/t* Exp[I \[Lambda] (t^5 + t + 1)], {t, -2, 2}, {\[Lambda], Infinity, 1}] fails in a long time, though Integrate[Sin[t]/t,{t,-2,2}] is expressed through special functions. Dec 16, 2021 at 20:00

Too long for a comment. My play with modifications of the example under consideration, e.g.

AsymptoticIntegrate[(t^2 + 2)*Sinh[t]*Cos[2 t]*
Exp[I \[Lambda] (t^12 + 2*t + 1)], {t, -2, Infinity}, {\[Lambda],Infinity, 1}]


-(((2 + 1/6^(2/11)) E^((I \[Pi])/4 + I (1 - 2^(10/11)/3^(1/11) + 1/(6 6^(1/11))) \[Lambda]) Sqrt[\[Pi]/ 11] Cos[2^(10/11)/3^(1/11)] Sinh[1/6^(1/11)])/( 6^(1/22) Sqrt[\[Lambda]]))

and

AsymptoticIntegrate[(t^2 + 2)*Sinh[t]*Cos[2 t]*
Exp[I \[Lambda] (Pi*t^12 + 2*t + 1)], {t, -2, 5}, {\[Lambda],Infinity, 3}]


which output is too long to be quoted here,

suggests that the asymptitic expansion for the integral $$\int\limits_a^b Q(t)e^{i\lambda P(t)}\,dt$$ as $$\lambda \to \infty$$, where $$Q(t)$$ is a quasi-polynomial over the reals (see Russian edition of Wiki which is better on this topic than English one) and $$P(t)$$ is a polynomial over the reals and the range of the integration may be infinite, is currently implemented in Mathematica. This is a special case of the method of stationary phase.