Dogbone

The function $\sqrt{z-1}$ is analytic on the cut plane $\mathbb{C} \backslash(-\infty,1]$ , while the function $\sqrt{z+1}$ is analytic on the cut plane $\mathbb{C} \backslash (-\infty,-1]$ . Putting these functions together, we can define $\sqrt{z^2-1}$ on the domain $\mathbb{C} \backslash [-1,1]$ .

For $R > 1$ , let $\gamma_R$ be the positively oriented contour $|z| = R$ . For $0 < \epsilon < \tfrac12$ , let $D_\epsilon$ be the ``dogbone" contour consisting of:

• the semicircular arc $z = 1 + \epsilon e^{i\theta}$ for $-\pi \le \theta \le \pi$ ,
• the straight line segment from $1-\epsilon$ to $-1 +\epsilon$ , running just above the cut,
• the semicircular arc $z = -1 + \epsilon e^{i\theta}$ for $0 \le \theta \le 2\pi$ ,
• the straight line segment from $-1+\epsilon$ to $1 - \epsilon$ , running just below the cut.

For any $n \in \mathbb{N}$ with $n \ge 2$ , let $S_n$ be the roots of the equation $z^{2n} + 1 \,=\,0$ . For sufficiently small $\epsilon > 0$ , we have $\int_{\gamma_R} \frac{\sqrt{z^2-1}}{z^{2n}+1}\,dz - \int_{D_\epsilon} \frac{\sqrt{z^2-1}}{z^{2n}+1}\,dz \; = \; 2\pi i \sum_{w \in S_n} \, \mathrm{Res}_{z = w} \frac{\sqrt{z^2-1}}{z^{2n}+1}$ Since $\int_{\gamma_R} \frac{\sqrt{z^2-1}}{z^{2n}+1}\,dz \; = \; O\big(R^{2-2n}\big) \qquad \qquad R \to \infty$ and since $\lim_{\epsilon \to 0} \int_{D_\epsilon} \frac{\sqrt{z^2-1}}{z^{2n}+1}\,dz \; = \; -2i\int_{-1}^1 \frac{\sqrt{1-x^2}}{x^{2n}+1}\,dx \; = \; -4i\int_0^1 \frac{\sqrt{1-x^2}}{x^{2n}+1}\,dx$ we deduce that $\int_0^1 \frac{\sqrt{1-x^2}}{x^{2n}+1}\,dx \; = \; \tfrac12\pi \sum_{w \in S_n} \, \mathrm{Res}_{z=w} \frac{\sqrt{z^2-1}}{z^{2n}+1}\; = \; -\tfrac{\pi}{4n} \sum_{w \in S_n} w\sqrt{w^2-1}$ If $w = e^{i\theta}$ where $|\theta| < \pi$ , then $\begin{array}{rcl} w + 1 & = & 2\cos\tfrac12\theta \,e^{\frac12i\theta} \\ w - 1 & = & 2i\sin\tfrac12\theta \,e^{\frac12i\theta} \\ \sqrt{w+1} & = & \sqrt{2\cos\tfrac12\theta}\,e^{\frac14i\theta} \\ \sqrt{w-1} & = & \sqrt{2|\sin\tfrac12\theta|}\,e^{\frac14i(\theta + \mathrm{sgn}(\theta)\pi)} \\ \sqrt{w^2-1} & = & \sqrt{2|\sin\theta|}\,e^{\frac14i(2\theta + \mathrm{sgn}(\theta)\pi)} \\ w\sqrt{w^2-1} & = & \sqrt{2|\sin\theta|}\,e^{\frac14i(6\theta + \mathrm{sgn}(\theta)\pi)} \end{array}$ Thus we deduce, by letting $R \to \infty$ and $\epsilon \to 0$ that $\int_0^1 \frac{\sqrt{1-x^2}}{x^{2n}+1}\,dx \; = \; -\tfrac{\pi}{2n}\sum_{k=0}^{n-1} \sqrt{2\sin\left(\tfrac{(2k+1)\pi}{2n}\right)}\cos\left(\tfrac{(6k+n+3)\pi}{4n}\right)$ and so, putting $n=4$ , $\begin{array}{rcl} \displaystyle \int_0^1 \frac{\sqrt{1-x^2}}{x^8+1}\,dx & = & -\frac{\pi}{8}\left( \begin{array}{l} \sqrt{2\sin\tfrac{\pi}{8}}\cos\frac{7\pi}{16} + \sqrt{2\sin\tfrac{7\pi}{8}}\cos\tfrac{25\pi}{16} \\ + \sqrt{2\sin\tfrac{3\pi}{8}}\cos\tfrac{13\pi}{16} + \sqrt{2\sin\tfrac{5\pi}{8}}\cos\tfrac{19\pi}{16} \end{array} \right) \\[5ex] & = & -\frac{\pi}{4}\left(\sqrt{2\sin\tfrac{\pi}{8}}\cos\tfrac{7\pi}{16} - \sqrt{2\sin\tfrac{3\pi}{8}} \cos\tfrac{3\pi}{16}\right) \\[2ex] & = & \frac{\pi}{\sqrt{8}}\left(\sqrt{\cos\tfrac{\pi}{8}} \cos\tfrac{3\pi}{16} - \sqrt{\sin\tfrac{\pi}{8}}\sin\tfrac{\pi}{16}\right) \end{array}$ making the answer $1+8+8+3+16+1 = \boxed{37}$ .

$\displaystyle I=\int\limits_{0}^{1} \frac{\sqrt{1-x^2}}{x^8+1} dx = \mathrm{Re}\{\int\limits_{0}^{\infty} \frac{\sqrt{1-x^2}}{x^8+1} dx \}$

Use contour integration technique,(using a semicircle contour in the upper region of plane)

$f(z)=\frac{\sqrt{1-z^2}}{z^8+1}$

Evaluation of poles :

$z^8+1=0 , z = i^{1/4},i^{3/4},i^{5/4},i^{7/4}$

Using residue theorem :

$\displaystyle \int_Cf(z)dz=\int\limits_{-R}^{R} f(x)dx+\int_{\Gamma}f(z)dz$

$\int_{\Gamma}f(z)dz \to 0$ as $R \to \infty$

$\displaystyle \int_Cf(z)dz = 2\pi i \sum_{x \in poles} Res$

Calculation of residues :

$\displaystyle Resf(i^{7/4})= \frac{(z-i^{7/4})\sqrt{1-z^2}}{z^8+1}=-\frac18 e^{\frac{7i\pi}{8}}\sqrt{1+e^{\frac{3i\pi}{4}}}$

$\displaystyle = -\frac18 e^{\frac{7i\pi}{8}}\sqrt{1+\cos(\pi/4)+i\sin(\pi/4)}=-\frac18 e^{\frac{7i\pi}{8}}\sqrt{2\cos(\pi/8)}e^{i\pi/16}=-\frac{1}{4\sqrt2}\sqrt{\cos(3\pi/8)}e^{17\pi i/16}$

Similarly ,

$\displaystyle Resf(i^{1/4})=\frac{1}{4\sqrt2}\sqrt{\cos(3\pi/8)}e^{-17\pi i/16}$

$\displaystyle Resf(i^{3/4})=\frac{1}{4\sqrt2}\sqrt{\cos(\pi/8)}e^{-11\pi i/16}$

$\displaystyle Resf(i^{5/4})=-\frac{1}{4\sqrt2}\sqrt{\cos(\pi/8)}e^{11\pi i/16}$

Sum of residues:(after converting the angles into acute angles)

$\displaystyle \sum_{x \in poles} Resf = -\frac{1}{2\sqrt2}i\{\sqrt{\cos(\frac{\pi}{8})}\cos(\frac{3\pi}{16}) - \sqrt{\sin{\frac{\pi}{8}}}\sin(\frac{\pi}{16})\}$

$\displaystyle \int_C f(z)dz=2\pi i \times -\frac{1}{2\sqrt2}i\{\sqrt{\cos(\frac{\pi}{8})}\cos(\frac{3\pi}{16}) - \sqrt{\sin{\frac{\pi}{8}}}\sin(\frac{\pi}{16})\}$

$\displaystyle = \frac{\pi}{\sqrt2}\{\sqrt{\cos(\frac{\pi}{8})}\cos(\frac{3\pi}{16}) - \sqrt{\sin{\frac{\pi}{8}}}\sin(\frac{\pi}{16})\}$

This gives ,

$\displaystyle \int\limits_{-\infty}^{\infty}\frac{\sqrt{1-x^2}}{x^8+1} dx=\frac{\pi}{\sqrt2}\{\sqrt{\cos(\frac{\pi}{8})}\cos(\frac{3\pi}{16}) - \sqrt{\sin{\frac{\pi}{8}}}\sin(\frac{\pi}{16})\}$

$\displaystyle \boxed{\int\limits_{0}^{\infty}\frac{\sqrt{1-x^2}}{x^8+1} dx=\frac{\pi}{\sqrt8}\{\sqrt{\cos(\frac{\pi}{8})}\cos(\frac{3\pi}{16}) - \sqrt{\sin{\frac{\pi}{8}}}\sin(\frac{\pi}{16})\}}$