TITLE: integral of $\int \limits_{0}^{\infty}\frac {\sin (x^n)} {x^n}dx$
QUESTION [17 upvotes]: 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

REPLY [6 votes]: Using the change of variables $x^n=t$ gives

$$ I = \int \limits_{0}^{\infty}\frac {\sin (x^n)} {x^n}dx=\frac{1}{n}\int _{0}^{\infty }\!\sin\left( t \right) {t}^{{\frac {1-2\,n}{n}}}
{dt}.$$

Now, recalling the Mellin transform of a function $f$

$$ F(s)=\int_{0}^{\infty}x^{s-1}f(x) dx. $$

One can right away evaluate $I$ by finding the Mellin transform of $\sin(x)$ which is
$$\Gamma  \left( s \right) \sin \left( \frac{\pi \,s}{2} \right), $$
and then substituting $s=\frac{1-n}{n}$ (since $s-1={\frac {1-2\,n}{n}}$ ) to get

$$I = \frac{1}{n} \Gamma\left( \frac{1-n}{n} \right) \sin\left(\frac{\pi}{2}\frac{1-n}{n} \right)\quad n\in \mathbb{N}. $$ 

Note: The following is a useful identity

$$ \Gamma(1-z) \Gamma(z) = {\pi \over \sin{(\pi z)}},  $$

which is known as the Euler's reflection formula.