# Asymptotic expansion of an Euler-type integral

I am looking for an asymptotic expansion for large $$x$$ for the following integral: $$\int_{0}^{1} \frac{t^{ia}(1-t/2)^{-ia}}{\sqrt{1-t}} e^{ixt} dt$$ $$a$$ and $$x$$ are real and positive. I tried using AsymptoticIntegrate but it just throws me the same expression back. Perhaps some clever series expansion could help? I have the first term from this expansion from a question here: https://math.stackexchange.com/questions/3876832/asymptotics-for-2-humbert-series-special-forms I am looking for more terms in order to approximate this function well for large enough $$x$$.

• This is rather math than Mathematica. That book by M.Fedoryuk (only in Russian) may be useful to you. Commented Feb 7, 2021 at 5:08

Try this:

ClearAll["*"]; Remove["*"];(*Clear All*)

INV = InverseLaplaceTransform[(E^(I t x) (Z - t/2)^(-I a) t^(I a))/Sqrt[1 - t], Z, s](*Where Z=1*)
INT = Integrate[INV, {t, 0, 1}, Assumptions -> {s > 0, a > 0, x > 0}] // FunctionExpand
SER = Series[INT, {x, Infinity, 2}, Assumptions -> {x > 0, s > 0}] // Normal // Expand(*Terms = 2 *)
f[x_, a_] = Limit[LaplaceTransform[SER, s, Z], Z -> 1, Assumptions -> {x > 0, a > 0}] // FunctionExpand // FullSimplify // Expand


$$\frac{1}{2} a^2 e^{-\frac{\pi a}{2}} x^{-2-i a} \Gamma (i a+1)-\frac{(1-i) \sqrt{\pi } 2^{-\frac{1}{2}+i a} a e^{i x}}{x^{3/2}}-i a e^{-\frac{\pi a}{2}} x^{-2-i a} \Gamma (i a+1)-\frac{1}{2} e^{-\frac{\pi a}{2}} x^{-2-i a} \Gamma (i a+1)+i e^{-\frac{\pi a}{2}} x^{-1-i a} \Gamma (i a+1)+\frac{(1-i) \sqrt{\pi } 2^{-\frac{1}{2}+i a} e^{i x}}{\sqrt{x}}$$

Check:

g[x_, a_] := NIntegrate[(t^(I a) (1 - t/2)^(-I a))/Sqrt[1 - t] Exp[I x t], {t, 0,1}, Method -> "LocalAdaptive"];
g[20000, 10] (*For: x=20000 and a=10*)
(*0.0110877 + 0.00582967 I*)
f[20000, 10] // N (*For: x=20000 and a=10*)
(*0.0110877 + 0.00582966 I*)

• The answer from the Maxim's comment in math.stackexchange.com/questions/3876832/… is h[x_, a_] = -Sqrt[Pi]*Exp[3*Pi*I/4]*2^(I*a) Exp[I*x]/Sqrt[x] and h[20000, 10] // N results in 0.0110933 + 0.00583258 I. Is the game worth the candle? Commented Feb 7, 2021 at 12:05
• @user64494. Yes, because now I have 2 candles, yours and Maxim's. Commented Feb 7, 2021 at 12:14