Compute the following integrals. (i) \(\displaystyle \int_0^\infty \frac{x^{a-1}}{1 + x^n} \, dx\), \(0< a < n\) (ii) \(\displaystyle \int_0^\infty \frac{\log x}{(1 + x^2)^2}\, dx\)
Solution
Goal: Compute (i) \(\int_0^\infty \frac{x^{a-1}}{1 + x^n}\,dx\) for \(0 < a < n\), and (ii) \(\int_0^\infty \frac{\log x}{(1 + x^2)^2}\,dx\).
1. For (i), use the keyhole contour or the sector contour of angle \(2\pi/n\) around the pole \(e^{i\pi/n}\).
Integrate \(f(z) = z^{a-1}/(1 + z^n)\) over the sector \(\{0 \le \arg z \le 2\pi/n\}\) (with \(z^{a-1}\) the branch \(\abs{z}^{a-1}e^{i(a-1)\arg z}\)): boundary rays are the positive real axis and the ray at angle \(2\pi/n\); the only pole inside is \(z_0 = e^{i\pi/n}\).
Let \(I = \int_0^\infty x^{a-1}/(1+x^n)\,dx\). On the second ray, \(z = xe^{2\pi i/n}\), \(z^{a-1} = x^{a-1} e^{2\pi i(a-1)/n}\), \(dz = e^{2\pi i/n}dx\), so the integral there is \(e^{2\pi i a/n} I\). The small circle around 0 and the large arc vanish (\(a > 0\) handles the small circle; \(a < n\) handles the large arc). Hence \(I(1 - e^{2\pi i a/n}) = 2\pi i \Res_{z_0} f\).
2. Compute the residue.
\(\Res_{z_0} \frac{z^{a-1}}{1+z^n} = \frac{z_0^{a-1}}{n z_0^{n-1}} = \frac{1}{n} z_0^{a-n} = \frac{1}{n} e^{i\pi(a-n)/n} = -\frac{1}{n} e^{i\pi a/n}\) since \(e^{-i\pi} = -1\).
3. Solve for \(I\).
From <1>1–<1>2: \(I(1 - e^{2\pi i a/n}) = 2\pi i \cdot \qty(-\frac{1}{n} e^{i\pi a/n})\), so \(I = \frac{-2\pi i}{n} \frac{e^{i\pi a/n}}{1 - e^{2\pi i a/n}} = \frac{2\pi i}{n} \frac{e^{i\pi a/n}}{e^{2\pi i a/n} - 1} = \frac{2\pi i}{n} \cdot \frac{1}{e^{i\pi a/n} - e^{-i\pi a/n}} = \frac{2\pi i}{n} \cdot \frac{1}{2i\sin(\pi a/n)} = \frac{\pi}{n \sin(\pi a/n)}\).
4. For (ii), differentiate the \(a\)-family at the right point or integrate by parts: use \(F(a) = \int_0^\infty \frac{x^{a-1}}{1 + x^2}\,dx = \frac{\pi}{\sin(\pi a/2)}\) (case \(n = 2\) of (i), valid for \(0 < a < 2\)) and differentiate with respect to \(a\) at \(a = 1\)… but the integrand is \(\log x/(1+x^2)^2\), not \(\log x \cdot x^{a-1}/(1+x^2)\).Instead, differentiate \(G(a) = \int_0^\infty \frac{x^{a-1}}{1 + x^2}\,dx\): \(G'(a) = \int_0^\infty \frac{\log x \, x^{a-1}}{1+x^2}\,dx\).We need \(\int \frac{\log x}{(1+x^2)^2}\).Relate: \(\ddd{}{a} \int_0^\infty \frac{x^{a-1}}{(1+x^2)^2}\,dx = \int_0^\infty \frac{\log x \, x^{a-1}}{(1+x^2)^2}\,dx\); at \(a = 1\) this is exactly the desired integral.So compute \(H(a) = \int_0^\infty \frac{x^{a-1}}{(1+x^2)^2}\,dx\) by differentiating the standard \(\int_0^\infty \frac{x^{a-1}}{1 + x^2}\,dx = \frac{\pi}{\sin(\pi a/2)}\) with respect to the exponent of the denominator: alternatively use \(\frac{1}{(1+x^2)^2} = -\dd{}{\beta}\frac{1}{\beta + x^2}\big|_{\beta = 1}\), or directly: \(H(a) = \lim_{\beta \to 1} -\dd{}{\beta} \int_0^\infty \frac{x^{a-1}}{\beta + x^2}\,dx\).Using \(\int_0^\infty \frac{x^{a-1}}{\beta + x^2}\,dx = \beta^{a/2 - 1}\frac{\pi}{2\sin(\pi a/2)}\) (substitute \(x = \sqrt\beta t\)): \(H(a) = -\dd{}{\beta}\qty[\beta^{a/2 - 1}]\big|_{\beta=1} \frac{\pi}{2\sin(\pi a/2)} = \qty(1 - a/2)\frac{\pi}{2\sin(\pi a/2)}\).At \(a = 1\): \(H(1) = \frac{1}{2} \cdot \frac{\pi}{2\sin(\pi/2)} = \frac{\pi}{4}\).
5. Then \(\int_0^\infty \frac{\log x}{(1+x^2)^2}\,dx = H'(1)\).
\(H'(a) = \int_0^\infty \frac{\log x \, x^{a-1}}{(1+x^2)^2}\,dx\) by differentiation under the integral (justified by dominated convergence on \([\delta, R]\) plus uniform convergence of the derivative integrals away from \(0\) and \(\infty\)). So evaluate \(H'(1)\): with \(H(a) = \frac{\pi}{2}\qty(1 - a/2)\csc(\pi a/2)\), \(H'(a) = \frac{\pi}{2}\qty[-\frac12 \csc\qty(\frac{\pi a}{2}) + \qty(1 - \frac a2)\qty(-\frac{\pi}{2}\cot\qty(\frac{\pi a}{2})\csc\qty(\frac{\pi a}{2}))]\).
At \(a = 1\): \(\csc(\pi/2) = 1\), \(\cot(\pi/2) = 0\), so \(H'(1) = \frac{\pi}{2} \cdot \qty(-\frac12) = -\frac{\pi}{4}\).
6. Q.E.D.
3 gives (i): \(\int_0^\infty \frac{x^{a-1}}{1+x^n}\,dx = \frac{\pi}{n\sin(\pi a/n)}\).
5 gives (ii): \(\int_0^\infty \frac{\log x}{(1+x^2)^2}\,dx = -\frac{\pi}{4}\).