# Wrong condition for the convergence of the integral $\int_0^\infty t^a \exp(-i t^b) dt$

The integral $\int_0^\infty t^a \exp(-i t^b) dt$ converges for $-1 < a < b - 1$ or $b - 1 < a < -1$.

However,

Assuming[Element[a, Reals] && Element[b, Reals],Integrate[t^a Exp[I t^(-b)], {t, 0, Infinity}]]


gives

$\text{ConditionalExpression}\left[-\frac{e^{-\frac{i \pi (a+1)}{2 b}} \Gamma \left(-\frac{a+1}{b}\right)}{b},a>-1 ~\&~ a+b<-1\right]$

and

Assuming[a < -1 && Element[b, Reals], Integrate[t^a Exp[I t^(-b)], {t, 0, Infinity}]]


gives

Integrate::idiv: "Integral of E^(I\t^-b)\ t^a does not converge on {0,\[Infinity]}.


So Mathematica gives wrong results on the convergence of the integral. But for specific choice of a=3/2 and b=3 for which Mathematica says the integral diverges,

Integrate[t^(-3/2) Exp[I t^(-3)], {t, 0, Infinity}]


gives correct result $\frac{(1+i) \left(\sqrt{3}-i\right) \Gamma \left(\frac{7}{6}\right)}{\sqrt{2}}$. Is it merely a bug or do I miss something?

Ok, there are lots of typos in my original question. Here is the polished version.

$\int t^a e^{i t^b} ~dt$ converges for $-1 < a < b-1$ or $b-1 < a < -1$.

Let's see the two cases seprately.

Assuming[-1 < a < b - 1, Integrate[t^a Exp[I t^(b)], {t, 0, Infinity}]]
(* (E^((I (1 + a) \[Pi])/(2 b)) Gamma[(1 + a)/b])/b *)


This is a correct result.

Assuming[b - 1 > a < -1, Integrate[t^a Exp[I t^(b)], {t, 0, Infinity}]]
(* Integrate::idiv: "Integral of E^(I\t^b)\ t^a does not converge on {0,\[Infinity]}. " *)


This is definitely wrong. However,

Integrate[t^(-3/2) Exp[I t^(-3)], {t, 0, Infinity}]
(* ((1 + I) (-I + Sqrt[3]) Gamma[7/6])/Sqrt[2] *)


This is contradict to the previous result while correct.

Sorry again there is a typo in the code but I cannot edit or make a comment. The second command should be

Assuming[b - 1 < a < -1, Integrate[t^a Exp[I t^(b)], {t, 0, Infinity}]]


I'm sorry for typos.

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The integral in your Mathematica code is not the same integral as in your image.

Integrate[t^a Exp[-I t^b], {t, 0, Infinity}]

(* ConditionalExpression[Gamma[(1 + a)/b]/ (E^((I*(1 + a)*Pi)/(2*b))*b),
Re[a] > -1 && Re[b] > 1 + Re[a] &&  Im[b] == 0] *)


Using these stated conditions

Assuming[{Re[a] > -1, Re[b] > 1 + Re[a],
Im[b] == 0}, Integrate[t^a Exp[-I t^b], {t, 0, Infinity}]]

(* Gamma[(1 + a)/b]/(E^((I*(1 + a)*Pi)/(2*b))*b) *)


However, since using a variable in an inequality implies that it is real, these assumptions can be simplified to

Assuming[{b - 1 > a > -1}, Integrate[t^a Exp[-I t^b], {t, 0, Infinity}]]

(* Gamma[(1 + a)/b]/(E^((I*(1 + a)*Pi)/(2*b))*b) *)

% == %%

(* True *)