Evaluate \[ \int_0^\infty \frac{x^{a-1}}{1+x}\,dx, \] where \(a\) is a complex number. What restrictions must be imposed on \(a\)?
Solution
Set \[ I(a)=\int_0^\infty\frac{x^{a-1}}{1+x}\,dx, \qquad \sigma=\operatorname{Re}a. \]
1. The improper integral \(I(a)\) converges exactly when\[ 0<\sigma<1. \]
Near \(0\), \[ \frac{\abs{x^{a-1}}}{1+x} = \frac{x^{\sigma-1}}{1+x}, \] so absolute convergence there holds when \(\sigma>0\). Near infinity, \[ \frac{\abs{x^{a-1}}}{1+x} = \frac{x^{\sigma-1}}{1+x}, \] which is comparable to \(x^{\sigma-2}\), so absolute convergence there holds when \(\sigma<1\).
These conditions are also necessary for ordinary improper convergence. As \(x\to0^+\), \[ \frac1{1+x}=1+O(x). \] If \(\sigma<0\), then for fixed sufficiently small \(\delta>0\), \[ \int_\varepsilon^\delta \frac{x^{a-1}}{1+x}\,dx = \frac{\delta^a-\varepsilon^a}{a} +o(\abs{\varepsilon^a}) \qquad (\varepsilon\to0^+). \] Since \(\abs{\varepsilon^a}=\varepsilon^\sigma\to\infty\), the partial integrals cannot converge. If \(\sigma=0\) and \(a\neq0\), the error term has a finite limit while \(\varepsilon^a=e^{a\log\varepsilon}\) has no limit; for \(a=0\) the leading integral is logarithmic. Thus convergence at \(0\) requires \(\sigma>0\).
Likewise, \[ \frac1{1+x} = x^{-1}(1+O(x^{-1})) \qquad (x\to\infty). \] If \(\sigma>1\), then \[ \int_\delta^R \frac{x^{a-1}}{1+x}\,dx = \frac{R^{a-1}-\delta^{a-1}}{a-1} +o(\abs{R^{a-1}}) \qquad (R\to\infty), \] so the partial integrals cannot converge. If \(\sigma=1\) and \(a\neq1\), the error term converges while \(R^{a-1}\) has no limit; for \(a=1\) the leading integral is logarithmic. Thus convergence at infinity requires \(\sigma<1\).
2. Assume henceforth that \(0<\sigma<1\), and define\[ F(z)=\frac{z^{a-1}}{1+z} \]using the branch\[ z^{a-1} = \exp\bigl((a-1)(\log\abs{z}+i\arg z)\bigr), \qquad 0<\arg z<2\pi. \]On a keyhole contour about the positive real axis, the circular arcintegrals tend to \(0\) as the outer radius tends to infinity and theinner radius tends to \(0\).
On the outer circle \(\abs{z}=R\), the factor coming from \(e^{-\operatorname{Im}(a)\arg z}\) is bounded uniformly for \(0\le\arg z\le2\pi\). Thus, for a constant \(C_a\) independent of \(R\), \[ \abs{F(z)} \le C_a\frac{R^{\sigma-1}}{R-1}. \] Multiplying by the arc length \(2\pi R\) gives a bound of order \(R^{\sigma-1}\), which tends to \(0\) because \(\sigma<1\).
On the inner circle \(\abs{z}=r\), \[ \abs{F(z)} \le C_a\frac{r^{\sigma-1}}{1-r}. \] Multiplication by the arc length \(2\pi r\) gives a bound of order \(r^\sigma\), which tends to \(0\) because \(\sigma>0\).
3. The limiting contour integral satisfies\[ (1-e^{2\pi ia})I(a) = -2\pi i e^{\pi ia}. \]
On the upper side of the positive real axis, \(\arg z=0\), so the contribution tends to \(I(a)\). On the lower side, \(\arg z=2\pi\) and the contour runs from infinity back to \(0\), so the contribution tends to \[ -e^{2\pi i(a-1)}I(a) = -e^{2\pi ia}I(a). \] By step 2, the circular contributions vanish in the limit.
The only pole inside the keyhole contour is \(z=-1\). Since \(\arg(-1)=\pi\) on the chosen branch, \[ \Res_{z=-1}F(z) = (-1)^{a-1} = e^{\pi i(a-1)} = -e^{\pi ia}. \] The residue theorem therefore gives the displayed identity.
4. For \(0<\operatorname{Re}a<1\),\[ \boxed{ I(a) = \frac{\pi}{\sin(\pi a)} }. \]
Using \[ 1-e^{2\pi ia} = -2i e^{\pi ia}\sin(\pi a), \] step 3 gives \[ I(a) = \frac{-2\pi i e^{\pi ia}} {-2i e^{\pi ia}\sin(\pi a)} = \frac{\pi}{\sin(\pi a)}. \]
5. Q.E.D.
Step 1 gives the exact restriction on \(a\), and step 4 evaluates the integral throughout that domain.