calculus - integral of $int limits_{0}^{infty}frac {sin (x^n)} {x^n}dx$

what is the answer of $$\int \limits_{0}^{\infty}\frac {\sin (x^n)} {x^n}dx$$

From this A sine integral $\int_0^{\infty} \left(\frac{\sin x }{x }\right)^n\,\mathrm{d}x$ I saw the answer for $$\int \limits_{0}^{\infty}\left(\frac {\sin x} {x}\right)^ndx$$

but for my question i didn't see any answer

is there any help

thanks for all

Answer

$$n>1:$$

$$f(t)=\int_0^{\infty} \frac{\sin tx^n}{x^n}\,dx\Rightarrow f'(t)=\int_0^{\infty}\cos tx^n\,dx=\frac{1}{\sqrt[n]{t}}\int_0^{\infty}\cos x^n\,dx$$

This is the generalised Fresnel integral, which evaluates to:

$$ \int_0^{\infty} \cos x^n\,dx=\cos \left(\frac{\pi}{2n}\right) \Gamma \left(\frac{n+1}{n}\right)\;\;(\star)$$

Noting that $f(0)=0:$

$$f(1)=\cos \left(\frac{\pi}{2n}\right) \Gamma \left(\frac{n+1}{n}\right)\int_0^1\frac{1}{\sqrt[n]{t}}dt=\cos\left(\frac{\pi}{2n}\right)\Gamma \left(\frac{1}{n}\right)\frac{1}{n-1}$$

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As requested, here is a proof $(\star):$ consider the following paths:

$$\begin{aligned}\gamma(x)=x,\, 0 \leq x\leq r,\;\;\gamma'(t)=te^{i \frac{\pi}{2n}},\, 0\leq t\leq r,\;\;\mu(\theta)=re^{i \theta},\, 0\leq \theta \leq \tfrac{\pi}{2n}\end{aligned}$$

By Cauchy, $\displaystyle \int_{\gamma} e^{iz^n}\,dz+\int_{\mu} e^{iz^n}\,dz=\int_{\gamma'} e^{iz^n}\,dz$

On the RHS, as $r\to\infty:$

$$\begin{aligned}\displaystyle\int_{\gamma'} e^{iz^n}\,dz=e^{i\frac{\pi}{2n}}\int_0^{r} e^{-t^n}dt=\frac{e^{i\frac{\pi}{2n}}}{n}\int_0^{\sqrt[n]{r}} s^{\frac{1}{n}-1}e^{-s}\,ds\to \frac{e^{i\frac{\pi}{2n}}}{n}\, \Gamma \left(\frac{1}{n}\right)=e^{i \frac{\pi}{2n}}\Gamma\left(\frac{n+1}{n}\right)\end{aligned}$$

On the LHS, as $r\to\infty:$

$$ \left|\int_{\mu} e^{iz^n}\,dz\right|= \left|ir\int_0^{\frac{\pi}{2n}}e^{i\left(r^n e^{in\theta}+\theta\right)}\,d \theta\right| \leq r\int_0^{\frac{\pi}{2n}} e^{-r^n\sin n\theta}\,d\theta \to 0$$

and obviously $\displaystyle \int_{\gamma} e^{iz^n}\,dz=\int_0^{\infty} e^{ix^n}\,dx$ as $r\to\infty$

Thus, equating real & imaginary parts:

$$ \int_0^{\infty} \cos x^n\,dx=\cos \left(\frac{\pi}{2n}\right) \Gamma \left(\frac{n+1}{n}\right)$$

$$ \int_0^{\infty} \sin x^n\,dx=\sin \left(\frac{\pi}{2n}\right) \Gamma \left(\frac{n+1}{n}\right)$$