Show that \[ \int_{0}^{\infty} \frac{x^{a-1}}{1+x^{n}} d x=\frac{\pi}{n \sin \frac{a \pi}{n}} \] using complex analysis, \(0<a<n\). Here \(n\) is a positive integer.
Solution
Goal: Show \(\int_0^\infty \frac{x^{a-1}}{1+x^n}\, dx = \frac{\pi}{n\sin \frac{a\pi}{n}}\) for \(0 < a < n\) by complex analysis.
1. Set up the integrand and the contour: \(f(z) = \frac{z^{a-1}}{1 + z^n}\) with \(z^{a-1}\) the branch \(\abs{z}^{a-1} e^{i(a-1)\arg z}\), \(0 \leq \arg z \leq \frac{2\pi}{n}\); integrate over the sector \(\Gamma_R\): the segment \([0, R]\), the arc \(\abs{z} = R\) with \(0 \leq \arg z \leq \frac{2\pi}{n}\), and the ray \(\arg z = \frac{2\pi}{n}\) back to \(0\).
On the sector the branch of \(z^{a-1}\) is single-valued and holomorphic, and the only pole of \(f\) inside is at \(z = e^{i\pi/n}\) (where \(z^n = -1\)), since \(0 < \pi/n < 2\pi/n\).
2. The integral along the ray \(\arg z = \frac{2\pi}{n}\) equals \(-e^{2\pi i a/n} \int_0^R \frac{x^{a-1}}{1 + x^n}\, dx\).
On that ray \(z = re^{2\pi i/n}\), \(dz = e^{2\pi i/n} dr\), \(z^{a-1} = r^{a-1} e^{2\pi i (a-1)/n}\), and \(z^n = r^n e^{2\pi i} = r^n\); traversed from \(R\) down to \(0\), the integral is \(-\int_0^R \frac{r^{a-1} e^{2\pi i(a-1)/n}}{1 + r^n} e^{2\pi i/n}\, dr = -e^{2\pi i a/n} \int_0^R \frac{r^{a-1}}{1+r^n}\, dr\).
3. The arc integral tends to \(0\) as \(R \to \infty\).
On the arc, \(\abs{f(z)} \leq \frac{R^{a-1}}{R^n - 1}\) (since \(\abs{1 + z^n} \geq \abs{z}^n - 1\)), and the arc has length \(\frac{2\pi R}{n}\); hence the arc integral is bounded by \(\frac{2\pi}{n}\frac{R^a}{R^n - 1} \to 0\) as \(R \to \infty\) because \(a < n\).
4. The residue of \(f\) at \(z = e^{i\pi/n}\) is \(\frac{1}{n} e^{i(a-n)\pi/n}\).
At a simple pole, \(\Res_{z = \zeta} \frac{z^{a-1}}{1+z^n} = \frac{\zeta^{a-1}}{n\zeta^{n-1}} = \frac{1}{n} \zeta^{a-n} = \frac{1}{n} e^{i(a-n)\pi/n}\), using \(\zeta = e^{i\pi/n}\).
5. The sector integral equals \(2\pi i\) times the residue.
By the residue theorem applied on \(\Gamma_R\) (with the arc contribution handled in <1>3), \(\qty(1 - e^{2\pi i a/n}) \int_0^R \frac{x^{a-1}}{1+x^n}\, dx + o(1) = 2\pi i \cdot \frac{1}{n} e^{i(a-n)\pi/n}\); here 1, 2 and 3 give the left-hand side, and 4 the right-hand side.
6. Solve for the integral.
From 5, \(\int_0^\infty \frac{x^{a-1}}{1+x^n}\, dx = \frac{2\pi i}{n} \frac{e^{i(a-n)\pi/n}}{1 - e^{2\pi i a/n}}\).
Since \(1 - e^{2\pi i a/n} = -2i e^{\pi i a/n} \sin\frac{a\pi}{n}\) and \(e^{i(a-n)\pi/n} e^{-i a\pi/n} = e^{-i\pi} = -1\), this simplifies to \(\frac{2\pi i}{n} \cdot \frac{-e^{i a\pi/n}}{-2i e^{i a\pi/n} \sin\frac{a\pi}{n}} = \frac{\pi}{n \sin\frac{a\pi}{n}}\).
7. Q.E.D.
6 establishes the claimed formula for all \(0 < a < n\).