Right contour for integrating goniometric function with the $x^n$ as an argument

Question

How would you integrate:

$\int_0^\infty \sin (x^n) \,dx$ $\;$for $n \gt 1$

• I mean the result is via gamma function and there exists a formula for that gamma function but I struggled with the rooting of the complex number

• Can I integrate this function using some convenient contour (complex)? Different way that using gamma function?No Idea how it would look like

The solution according to my textbook is $\ \frac {1}{n} \Gamma (\frac {1}{n}) \sin (\frac {\pi}{2n})$

My attempt:

$\int_0^\infty \sin (x^n) \,dx = \int_0^\infty \sin (z^n) \,dz = Im \int_0^\infty \exp (iz^n) \,dz$

• where $z$ is a complex number, $x$ is real and I don't know the Laplace transform so please lets avoid it.

Answer 1

$$\int_0^\infty \sin(x^n)dx=-\Im\int_0^\infty e^{-ix^n}dx$$ $$=-\Im \lim_{a\to0}\int_0^\infty e^{-(a+i)x^n}dx=-\Im\lim_{a\to0}\frac{1}{n}\int_0^\infty x^{\frac{1}{n}-1}e^{-(a+i)x}dx$$ $$=-\Im\lim_{a\to 0}\Gamma(\frac{1}{n})\frac{(a+i)^{-\frac{1}{n}}}{n}=\frac{\Gamma(\frac{1}{n})}{n}(-\Im(e^{-\frac{\pi i}{2n}}))=\frac{\Gamma(\frac{1}{n})}{n}\sin(\frac{\pi}{2n}).$$

No need to use countour integration all you need is the definition of the gamma function and Euler's Formula.

Answer 2

$$I=\int_0^\infty\sin(x^n)=\Im\left(\int_0^\infty \cos(x^n)+i\sin(x^n)dx\right)=\Im\left(\int_0^\infty e^{ix^n}dx\right)$$ Now you could try and use a summation to solve this: $$I=\Im\left(\int_0^\infty \sum_{r=0}^\infty \frac{(ix^n)^r}{r!}dx\right)=\Im\left(\sum_{r=0}^\infty \frac{i^r}{r!}\int_0^\infty x^{nr}dx\right)$$ $$=\Im\left(\sum_{r=0}^\infty \frac{i^r}{r!}\left[\frac{x^{nr+1}}{nr}\right]_0^\infty\right)=\lim_{x\to\infty}\Im\left(\sum_{r=0}^\infty\frac{i^rx^{nr+1}}{nr.r!}\right),$$ although I don't think this appears to lead anywhere.

coming back to a previous step: $$I(n)=\Im\left(\int_0^\infty e^{ix^n}dx\right)$$ $$I'(n)=\Im\left(i\int_0^\infty\frac{\partial}{\partial_x}e^{ie^{n\ln(x)}}dx\right)$$ $$=\Im\left(i\int_0^\infty ix^n\ln(x)e^{ix^n}dx\right) =-\Im\left(\int_0^\infty x^n\ln(x)e^{ix^n}dx\right)$$ this may lead somewhere