This a follow-up to my previous question.
I suspected that the following identity, $\int_0^{\infty} e^{-ax^n} \, \text{d}x = \frac1{n} \cdot a^{-\frac1{n}} \Gamma{\left(\frac1{n}\right)} = a^{-\frac1{n}} \Gamma{\left(1+\frac1{n}\right)}$, implies that $$\int_0^{\infty} e^{-ix^n} \, \text{d}x = i^{-\frac1{n}} \Gamma{\left(1+\frac1{n}\right)} $$
If you substitute $a = i$. I’m not sure if such a substitution is “legal,” but I’m going forward with this anyway for now.
Knowing that $i^{-\frac1{n}} = \cos{\left(\frac{\pi}{2n}(1+4k)\right)} - i\sin {\left(\frac{\pi}{2n}(1+4k)\right)}$, where $k \in \mathbb{N},$ and $0 < k \leq n$, this implies that
$$\int_0^{\infty} e^{-ix^n} \, \text{d}x = \cos{\left(\frac{\pi}{2n}(1+4k)\right)} \Gamma{\left(1+\frac1{n}\right)} - i\sin {\left(\frac{\pi}{2n}(1+4k)\right)}\Gamma{\left(1+\frac1{n}\right)} $$
It would then follow that $$ \int_0^{\infty} \cos(x^n)\, \text{d}x = \cos{\left(\frac{\pi}{2n}(1+4k)\right)} \Gamma{\left(1+\frac1{n}\right)} \\ \int_0^{\infty} \sin(x^n)\, \text{d}x = \sin{\left(\frac{\pi}{2n}(1+4k)\right)} \Gamma{\left(1+\frac1{n}\right)} $$
Through some testing, I found that this equation holds true for $n = 2$, but not for all values of $k$. My computer does not have the processing power to test for higher values of $n$.
Do these integral expressions hold true for all $n$? If so, for what $k$ are they true? And is my method for deriving these expressions valid?
P.S.: For integer $n$, I would assume that we could use induction to prove that this expression is true for $n \geq 2$; but I’m not sure how I could make a proof like that work.
