Chances are it will behave similarly to Integrate[Exp[-x t^3 ], {t, 0, π/2}]
. I show a numerical verification below. We'll use the dominant series term, which we can compute, for the above integral, and compare to numerical integrations of the one we cannot compute analytically.
ee1[x_] := Exp[-x*t^3]
ee2[x_] := Exp[-x*t^3*Cos[t]];
Let's get the candidate analytic term.
i1 = Integrate[ee1[n], {t, 0, π/2}]
s1 = Expand[Normal[Series[i1, {n, Infinity, 2}]]]
(* Out[158]= (3 Gamma[4/3] - Gamma[1/3, (n π^3)/8])/(3 n^(1/3))
Out[159]= (64 E^(-((n π^3)/8)))/(9 n^2 π^5) - (
4 E^(-((n π^3)/8)))/(3 n π^2) + (1/n)^(1/3) Gamma[4/3] *)
The dominant term is (1/n)^(1/3) Gamma[4/3]
. Let's test that against the integral of actual interest.
t1 = Gamma[4/3.]*Table[(1/n)^(1/3), {n, 10., 1000., 10}];
t2 = Table[NIntegrate[ee2[n], {t, 0, π/2}], {n, 10, 1000, 10}];
t1/t2
(* Out[178]= {0.889613957994, 0.93398701513, 0.950173592407, \
0.959053918538, 0.964797571829, 0.968871281155, 0.971937888145, \
0.974344883597, 0.976293611756, 0.977909513436, 0.979275174531, \
0.980447382775, 0.98146658481, 0.982362438315, 0.983157228034, \
0.983868051412, 0.984508264217, 0.985088464385, 0.98561717834, \
0.986101350477, 0.986546698826, 0.986957978112, 0.987339177279, \
0.9876936699, 0.988024330159, 0.988333623268, 0.98862367681, \
0.988896337531, 0.989153216798, 0.989395727565, 0.989625114446, \
0.989842478448, 0.990048797416, 0.990244943055, 0.990431695169, \
0.990609753656, 0.990779748639, 0.990942249097, 0.991097770216, \
0.991246779697, 0.991389703197, 0.991526929023, 0.991658812204, \
0.991785678043, 0.991907825207, 0.992025528451, 0.992139041267, \
0.992248596638, 0.9923544116, 0.992456686183, 0.99704297563, \
0.997081574489, 0.997118917853, 0.997155073418, 0.997190103398, \
0.997224065159, 0.997257011748, 0.997288992353, 0.997320052698, \
0.997350235384, 0.997379580188, 0.997408124317, 0.997435902639, \
0.997462947887, 0.997489290828, 0.997514960431, 0.997539984002, \
0.997564387337, 0.997588194736, 0.997611429257, 0.997634112709, \
0.99765626576, 0.997677908018, 0.997699058098, 0.997719733686, \
0.997739951586, 0.997759727841, 0.997779077673, 0.99779801561, \
0.997816555502, 0.997834710569, 0.99785249343, 0.997869916144, \
0.997886990235, 0.997903726729, 0.997920136175, 0.997936228675, \
0.997952013907, 0.997967501146, 0.997982699287, 0.997997616863, \
0.998012262067, 0.998026642763, 0.998040766509, 0.998054640568, \
0.998068271924, 0.998081667296, 0.998094833147, 0.998107775704, \
0.998120500959} *)
Getting pretty close to 1.
A rigorous proof would take more work though. I guess a possible approach would be to show that as x
gets large, the dominant part of the integral is on a segment that shrinks toward the origin.
AsymptoticIntegrate[]
function doesn't know how to handle this. $\endgroup$