Integrate and find closed form for $\int_0^\infty\frac{\sin x^n}{n\pi}dx$ I’m working on a practice exam for my analysis class, and I was asked to find a general form for
$$\int_0^\infty\frac{\sin x^n}{n\pi}dx$$
When I first looked at this, my mind instantly went to the Dirchlet Integral. Its been some time and I haven’t been able to see the relation.
 A: This integral is kind of famous. It can be done with residues or real methods.
It is probably on the site somewhere. It is more like what is known as a Fresnel integral. 
Write $$\frac{1}{\pi n}\int_{0}^{\infty}\sin(x^{n})dx$$
Let $t=x^{n}, \;\ t^{1/n}=x, \;\ dx=\frac{1}{n}t^{1/n-1}dt$
$$\frac{1}{n}\cdot \frac{1}{\pi n}\int_{0}^{\infty}t^{1/n-1}\sin(t)dt$$
By the Gamma function, we can write a double integral:
$$\frac{1}{\pi n^{2}}\cdot \frac{1}{\Gamma(1-\frac{1}{n})}\int_{0}^{\infty}\int_{0}^{\infty}u^{-1/n}e^{-ut}\sin(t)dudt$$
switch integrals and integrate w.r.t t:
$$\frac{1}{\pi n^{2}\Gamma(1-1/n)}\int_{0}^{\infty}\frac{1}{u^{1/n}(u^{2}+1)}du$$
Among other means, we can use the Beta function: $\beta(p,q)=\frac{\Gamma(p)\Gamma(q)}{\Gamma(p+q)}=\int_{0}^{\infty}\frac{y^{p-1}}{(1+y)^{p+q}}dy$
Thus, by using the sub $y=u^{2}$,  we get:
$$\frac{1}{2\pi n^{2}\Gamma(1-1/n)}\Gamma\left(\frac{1-n}{2}\right)\cdot \Gamma\left(\frac{n+1}{2}\right)$$
Using the reduction formulas for Gamma:
$\Gamma(1/2-\frac{1}{2n})\Gamma(1/2+\frac{1}{2n})=\pi \sec(\frac{\pi}{2n})\to \frac{\pi \sec\left(\frac{\pi}{2n}\right)}{\Gamma(1-1/n)}=2\sin\left(\frac{\pi}{2n}\right)\Gamma(1/n)$,
we can write:
$$=\boxed{\frac{1}{\pi n^{2}}\cdot \Gamma(1/n)\sin\left(\frac{\pi}{2n}\right)}$$
