complex analysis integral with $z^8$ function

Question

can anyone help with this integral I'm trying to do. Will a semicircle contour work for this function?

$$\int_{0}^{\infty}\frac{1}{1+x^{8}} dx$$

Answer 1

Our analysis would be considerably streamlined by judiciously choosing the enclosing contour. Accordingly, we choose the contour $C_R$ that is comprised of the (1) line segment from $0$ to $R$, (ii) circular arc from $R$ to $Re^{i\pi/4}$, and (iii) the line segment from $Re^{i\pi/4}$ to $0$.

Note that for $R>1$, $C_R$ encloses only the pole at $z=e^{i\pi/8}$. Proceeding, we can write

$$\begin{align} \oint_{C_R}\frac{1}{1+z^8}\,dz&=\int_0^R \frac{1}{1+x^8}\,dx+\int_0^{\pi/4}\frac{1}{1+R^8e^{i8\phi}}\,iRe^{i\phi}\,d\phi+\int_R^0 \frac{1}{1+t^8}\,e^{i\pi/4}\,dt\\\\ &=(1-e^{i\pi/4})\int_0^R \frac{1}{1+x^8}\,dx+\int_0^{\pi/4}\frac{1}{1+R^8e^{i8\phi}}\,iRe^{i\phi}\,d\phi\\\\ &=2\pi i \text{Res}\left(\frac{1}{1+z^8},z=e^{i\pi/8}\right)\\\\ &=2\pi i \lim_{z\to e^{i\pi/8}}\frac{z-e^{i\pi/8}}{1+z^8}\\\\ &=2\pi i\frac{1}{8e^{i7\pi/8}} \end{align}$$

Letting $R\to \infty$ we find that

$$\int_0^\infty \frac{1}{1+x^8}\,dx=2\pi i\frac{1}{8e^{i7\pi/8}(1-e^{i\pi/4})}=\frac{\pi}{8\sin(\pi/8)}$$

Answer 2

Yes, it will work. Using that approach, you will get that the integral is eual to $2\pi i$ times the sum of the residues of $(1+z^8)^{-1}$ at $\exp\left(\frac{\pi i}8\right)$, $\exp\left(\frac{3\pi i}8\right)$, $\exp\left(\frac{5\pi i}8\right)$, and $\exp\left(\frac{7\pi i}8\right)$. And, if $\omega$ is one of these numbers, the ridues there is eual to $\dfrac1{8\omega^7}$.