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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.

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  • $\begingroup$ Is there a typo in your integral? $\endgroup$ – Yagna Patel Jan 21 '16 at 2:16
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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:

$$=\boxed{\frac{1}{\pi n^{2}}\cdot \Gamma(1/n)\sin\left(\frac{\pi}{2n}\right)}$$

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