Prove that \[ \int_0^\infty\frac{x^{\alpha-1}}{1+x}\,dx =\frac{\pi}{\sin(\pi\alpha)}. \] What restrictions must be placed on \(\alpha\)?
Solution
For \(x>0\), define \[ x^{\alpha-1} = \exp((\alpha-1)\log x), \] where \(\log x\) is the real logarithm.
1. The improper integral\[ I(\alpha) \coloneqq \int_0^\infty\frac{x^{\alpha-1}}{1+x}\,dx \]converges exactly when\[ \boxed{0<\Re\alpha<1}. \]
Near \(0\), \[ \frac{x^{\alpha-1}}{1+x} \sim x^{\alpha-1}, \] and \[ \int_0^1x^{\alpha-1}\,dx \] converges exactly when \(\Re\alpha>0\).
Near infinity, \[ \frac{x^{\alpha-1}}{1+x} \sim x^{\alpha-2}, \] and \[ \int_1^\infty x^{\alpha-2}\,dx \] converges exactly when \(\Re\alpha<1\). Thus both endpoints converge exactly in the displayed strip. In that strip the integral is in fact absolutely convergent.
2. Assume henceforth that \(0<\Re\alpha<1\). On\[ \CC\setminus[0,\infty) \]use the branch\[ \Log z=\log\abs z+i\arg z, \qquad 0<\arg z<2\pi, \]and set\[ F(z) \coloneqq \frac{e^{(\alpha-1)\Log z}}{1+z}. \]Then \(F\) has a single pole in this slit plane, at \(z=-1\), with\[ \operatorname{Res}_{z=-1}F = -e^{i\pi\alpha}. \]
The only zero of the denominator is \(-1\), whose argument in the chosen branch is \(\pi\). Hence \[ \operatorname{Res}_{z=-1}F = e^{(\alpha-1)i\pi} = -e^{i\pi\alpha}. \]
3. Integrate \(F\) around the positively oriented keyhole contourwith radii \(0<\epsilon<1<R\) about the positive real axis. The outerand inner circular contributions tend to \(0\) as\[ R\to\infty, \qquad \epsilon\to0. \]
On either circular arc, the factor \[ e^{-\Im(\alpha)\arg z} \] is bounded by a constant depending only on \(\alpha\).
On \(\abs z=R\), \[ \abs{F(z)} \leq C_\alpha \frac{R^{\Re\alpha-1}}{R-1}. \] Multiplying by the arc length \(2\pi R\) gives \[ O(R^{\Re\alpha-1})\longrightarrow0 \] because \(\Re\alpha<1\).
On \(\abs z=\epsilon\), \[ \abs{F(z)} \leq C_\alpha \frac{\epsilon^{\Re\alpha-1}}{1-\epsilon}. \] Multiplying by the arc length \(2\pi\epsilon\) gives \[ O(\epsilon^{\Re\alpha})\longrightarrow0 \] because \(\Re\alpha>0\).
4. In the same limit, the sum of the two straight portions of thekeyhole contour is\[ \left(1-e^{2\pi i\alpha}\right)I(\alpha). \]
On the upper side of the positive axis, \(\arg z\to0\), so the integrand tends to \[ \frac{x^{\alpha-1}}{1+x}, \] and that segment runs from \(\epsilon\) to \(R\).
On the lower side, \(\arg z\to2\pi\), so \[ e^{(\alpha-1)\Log z} = e^{2\pi i(\alpha-1)}x^{\alpha-1} = e^{2\pi i\alpha}x^{\alpha-1}. \] That segment is traversed from \(R\) back to \(\epsilon\). Its limiting contribution is therefore \[ -e^{2\pi i\alpha}I(\alpha). \] Adding the two contributions gives the claim.
5. The residue theorem gives\[ \left(1-e^{2\pi i\alpha}\right)I(\alpha) = -2\pi i e^{i\pi\alpha}. \]
By step 3, the circular integrals disappear in the limit. By step 4, the remaining contour integral tends to the left-hand side. Step 2 and the residue theorem give \[ 2\pi i\operatorname{Res}_{z=-1}F = -2\pi i e^{i\pi\alpha}. \]
6. For \(0<\Re\alpha<1\),\[ \boxed{ I(\alpha) = \frac{\pi}{\sin(\pi\alpha)}. } \]
Use \[ 1-e^{2\pi i\alpha} = -2i e^{i\pi\alpha}\sin(\pi\alpha) \] in step 5 and cancel the nonzero common factor \(-2i e^{i\pi\alpha}\).
7. If \(\alpha\) is required to be real, the restriction is exactly\[ \boxed{0<\alpha<1}. \]
This is step 1 specialized to real \(\alpha\).
8. Q.E.D.
Steps 1, 6, and 7 give the convergence restriction and the required value of the integral.