Problem 8

Exercise 3.8 (A sector-based contour integral)   Evaluate for $p\geq 1$

$$\int_{\mathbb{R}}\frac{dx}{1+x^{2p}}$$ (3.57)

Proof. Consider the function $f : \mathbb{C} \to \mathbb{C}$ defined by

$$f(z) = \frac{1}{1+z^{2p}}$$ (3.58)

Capture the pole with least argument $x_0=\exp(i\pi/2p)$ in the sector $S_R = \{re^{i\theta} \;\vert\; 0\leq r\leq R, 0\leq\theta\leq \pi/p\}$. Compute the residue

$$\operatorname{Res}(f,x_0) = \lim_{z\to x_0} (z-x_0)f(z) = \lim_{z... ...-x_0}{1+z^{2p}} = \lim_{z\to x_0} \frac{1}{2pz^{2p-1}} = \frac{1}{2px_0^{2p-1}}$$ (3.59)

The residue theorem states

$$\int_{\partial S_R}f(z)dz = \left(\int_0^R + \int_{\text{arc}} + ... ...e}}\right)f(z)dz = 2\pi i \operatorname{Res}(f,z_0) = \frac{\pi i}{px_0^{2p-1}}$$ (3.60)

The line integral can be found by parametrizing $\gamma : [0,R] \to\mathbb{C}$ by $\gamma(t) = (R-t)e^{i\pi/p}$. Then

$$\int_{\text{line}}f(z)dz = \int_0^R \frac{1}{1 + (R-t)^{2p}}(-e^{i\pi/p})dt$$ (3.61)

$$= -e^{i\pi/p}\int_0^R f(z)dz$$ (3.62)

$$= -x_0^2\int_0^R f(z)dz$$ (3.63)

The arc integral vanishes as $R\to\infty$. Set $\rho(t) = Re^{it}$ for $t\in[0,\pi/p]$. Then $\rho'(t)=Rie^{it}$ and the arc integral equals

$$\int_{\text{arc}}f(z)dz = \int_0^{\pi/p} f(\rho(t))\rho'(t)dt = \int_0^{\pi/p} \frac{Rie^{it}}{1+R^{2p}e^{2ipt}}$$ (3.64)

Taking absolute values, we see

$$\left\vert\int_{\text{arc}}f(z)dz\right\vert \leq \int_0^{\pi/p} ... ..._0^{\pi/p} \frac{R}{R^{2p}-1}dt \leq \frac{\pi}{p}\cdot\frac{R}{R^{2p}-1} \to 0$$ (3.65)

Therefore, we can take limits and rearrange the integral-residue equation to find

$$(1-x_0^2)\int_0^\infty f(z)dz = \frac{\pi i}{px_0^{2p-1}} = \frac{\pi i}{px_0^{2p}/x_0} = \frac{\pi i}{p(-1)/x_0} = -\frac{\pi i x_0}{p}$$ (3.66)

Therefore,

$$\int_{\mathbb{R}}\frac{dx}{1+x^{2p}} = \frac{2\pi i x_0}{p(x_0^2-... ...x_0 - x_0^{-1})} = \frac{2\pi i}{p(2i\sin(\pi/2p))} = \frac{\pi}{p\sin(\pi/2p)}$$ (3.67)

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