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, \(0 < a \leq n\). Here \(n\) is a positive integer.
Solution
Goal. Show \(\int_0^\infty \frac{x^{a-1}}{1+x^n}\,dx = \frac{\pi}{n\sin(a\pi/n)}\) for \(0 < a \le n\).
1. Use the keyhole contour (or a sector contour) around the branch cut of \(z^{a-1}\).
the integrand \(z^{a-1}/(1+z^n)\) has a branch point at \(0\) and poles at the \(n\)-th roots of \(-1\).
2. The poles of \(1/(1+z^n)\) are at \(z = e^{i(2k+1)\pi/n}\), \(k = 0, \dots, n-1\).
\(z^n = -1\) at these points.
3. The residue at \(z_0 = e^{i\pi/n}\) (the pole in the upper half-plane nearest the positive real axis) is \(\frac{z_0^{a-1}}{n z_0^{n-1}} = \frac{z_0^a}{n z_0^n} = -\frac{z_0^a}{n}\).
\(\operatorname{Res}(z^{a-1}/(1+z^n), z_0) = z_0^{a-1}/(n z_0^{n-1}) = z_0^a/(n z_0^n) = -z_0^a/n\) (since \(z_0^n = -1\)).
4. Sum the residues over the poles in the upper half-plane and apply the residue theorem to the sector contour.
4.1. The sector contour (angle \(2\pi/n\)) encloses exactly one pole, \(z_0 = e^{i\pi/n}\).
the sector \(0 \le \arg z \le 2\pi/n\) contains the single pole \(e^{i\pi/n}\).
4.2. The integral over the sector gives \((1 - e^{2\pi i a/n})\int_0^\infty \frac{x^{a-1}}{1+x^n}\,dx = 2\pi i \operatorname{Res}(f, z_0)\).
the two radial segments differ by the factor \(e^{2\pi i a/n}\) (from the branch of \(z^{a-1}\)), and the arc contribution vanishes as the radius \(\to \infty\).
4.3. Hence \(\int_0^\infty \frac{x^{a-1}}{1+x^n}\,dx = \frac{2\pi i \operatorname{Res}(f, z_0)}{1 - e^{2\pi i a/n}}\).
solve for the integral.
5. Evaluate.
5.1. \(\operatorname{Res}(f, z_0) = -\frac{e^{i a\pi/n}}{n}\).
3 with \(z_0 = e^{i\pi/n}\).
5.2. \(\frac{2\pi i \cdot (-e^{ia\pi/n}/n)}{1 - e^{2\pi i a/n}} = \frac{-2\pi i e^{ia\pi/n}}{n(1 - e^{2\pi i a/n})}\).
substitute.
5.3. \(= \frac{\pi}{n \sin(a\pi/n)}\).
\(\frac{-2\pi i e^{ia\pi/n}}{1 - e^{2\pi i a/n}} = \frac{-2\pi i}{e^{-ia\pi/n} - e^{ia\pi/n}} = \frac{-2\pi i}{-2i\sin(a\pi/n)} = \frac{\pi}{\sin(a\pi/n)}\).
6. Q.E.D.
5.3 gives the result.