Show that \[ \displaystyle \int_0^\infty \frac{x^{a-1}} {1+x^n} \dx=\frac{\pi}{n\sin \frac{a\pi}{n}} .\] using complex analysis, \(0< a < n\). Here \(n\) is a positive integer.
Solution
1. Choose the integrand and contour:
1.1. Let \(f(z) = \frac{z^{a-1}}{1 + z^n}\), where \(z^{a-1} = \exp\bigl((a-1)\operatorname{Log}(z)\bigr)\) using the branch of the logarithm with argument \(\arg(z) \in [0, 2\pi)\).
definition of power function.
1.2. Let \(\Gamma = \Gamma_1 + \Gamma_R - \Gamma_2 - \Gamma_\varepsilon\) be the closed wedge contour in the complex plane:- \(\Gamma_1\): the line segment along the positive real axis from \(\varepsilon\) to \(R\).- \(\Gamma_R\): the circular arc \(z = R e^{i\theta}\) with \(\theta \in [0, 2\pi/n]\).- \(\Gamma_2\): the ray \(z = r e^{i 2\pi/n}\) with \(r \in [\varepsilon, R]\).- \(\Gamma_\varepsilon\): the circular arc \(z = \varepsilon e^{i\theta}\) with \(\theta \in [0, 2\pi/n]\).
standard sector contour of angle \(2\pi/n\).
2. Locate the poles of \(f(z)\) inside \(\Gamma\) and calculate the residue:
2.1. The poles of \(f(z)\) occur where \(1 + z^n = 0 \iff z^n = -1 = e^{i\pi}\).
roots of the denominator.
2.2. The only pole strictly inside the sector \(0 < \arg(z) < 2\pi/n\) is \(z_0 = e^{i\pi/n}\).
\(\arg(z_0) = \pi/n \in (0, 2\pi/n)\), while the next root has argument \(3\pi/n > 2\pi/n\).
2.3. \(z_0\) is a simple pole, and its residue is:\[ \operatorname{Res}(f, z_0) = \frac{z_0^{a-1}}{\left.\frac{d}{dz}(1+z^n)\right|_{z=z_0}} = \frac{z_0^{a-1}}{n z_0^{n-1}} = \frac{z_0^a}{n z_0^n} = \frac{e^{i a\pi/n}}{n (-1)} = -\frac{1}{n} e^{i a\pi/n}. \]
residue at a simple pole for a quotient \(g(z)/h(z)\) with \(h'(z_0) \neq 0\).
3. Evaluate the integrals along the boundary pieces:
3.1. Along \(\Gamma_1\):\[ \int_{\Gamma_1} f(z)\,dz = \int_\varepsilon^R \frac{x^{a-1}}{1 + x^n}\,dx \xrightarrow[\substack{\varepsilon \to 0^+ \\ R \to \infty}]{} I = \int_0^\infty \frac{x^{a-1}}{1 + x^n}\,dx. \]
definition of improper integral.
3.2. Along \(\Gamma_2\): substitute \(z = r e^{i 2\pi/n}\) so \(dz = e^{i 2\pi/n}\,dr\) and \(z^n = r^n e^{i 2\pi} = r^n\):\[ \int_{\Gamma_2} f(z)\,dz = \int_\varepsilon^R \frac{r^{a-1} e^{i(a-1)2\pi/n}}{1 + r^n} e^{i 2\pi/n}\,dr = e^{i 2\pi a/n} \int_\varepsilon^R \frac{r^{a-1}}{1 + r^n}\,dr \xrightarrow[\substack{\varepsilon \to 0^+ \\ R \to \infty}]{} e^{i 2\pi a/n} I. \]
parametrization of \(\Gamma_2\) and \(e^{i(a-1)2\pi/n} e^{i 2\pi/n} = e^{i 2\pi a/n}\).
3.3. Along the outer arc \(\Gamma_R\): for \(R > 1\), \(|1 + z^n| \ge R^n - 1\), so:\[ \left|\int_{\Gamma_R} f(z)\,dz\right| \le \frac{R^{a-1}}{R^n - 1} \cdot \frac{2\pi R}{n} = \frac{2\pi}{n} \frac{R^a}{R^n - 1} \xrightarrow[R \to \infty]{} 0 \quad (\text{since } a < n). \]
ML estimate on arc of length \(2\pi R / n\).
3.4. Along the inner arc \(\Gamma_\varepsilon\): for \(\varepsilon < 1\), \(|1 + z^n| \ge 1 - \varepsilon^n\), so:\[ \left|\int_{\Gamma_\varepsilon} f(z)\,dz\right| \le \frac{\varepsilon^{a-1}}{1 - \varepsilon^n} \cdot \frac{2\pi \varepsilon}{n} = \frac{2\pi}{n} \frac{\varepsilon^a}{1 - \varepsilon^n} \xrightarrow[\varepsilon \to 0^+]{} 0 \quad (\text{since } a > 0). \]
ML estimate on arc of length \(2\pi \varepsilon / n\).
4. Apply the Cauchy Residue Theorem and solve for \(I\):
4.1. By the Residue Theorem:\[ \oint_\Gamma f(z)\,dz = 2\pi i \operatorname{Res}(f, z_0) = -\frac{2\pi i}{n} e^{i a\pi/n}. \]
Cauchy Residue Theorem.
4.2. Taking the limit as \(\varepsilon \to 0^+\) and \(R \to \infty\):\[ (1 - e^{i 2\pi a/n}) I = -\frac{2\pi i}{n} e^{i a\pi/n}. \]
3 and 1.
4.3. Solving for \(I\):\[ I = \frac{-2\pi i e^{i a\pi/n}}{n(1 - e^{i 2\pi a/n})} = \frac{2\pi i}{n(e^{i a\pi/n} - e^{-i a\pi/n})} = \frac{\pi}{n \left(\frac{e^{i a\pi/n} - e^{-i a\pi/n}}{2i}\right)} = \frac{\pi}{n \sin\left(\frac{a\pi}{n}\right)}. \]
Euler’s formula \(\sin(\theta) = \frac{e^{i\theta} - e^{-i\theta}}{2i}\) with \(\theta = a\pi/n\).
5. Conclusion:\(\int_0^\infty \frac{x^{a-1}}{1 + x^n}\,dx = \frac{\pi}{n \sin(a\pi/n)}\). Q.E.D.
4.