Suppose \(f:[a, c)\cup (c,b]\rightarrow \mathbb C\) is integrable on \([a, c-\epsilon)\cup (c+\epsilon, b]\) for all \(\epsilon > 0\). Define the principal value integral as
proof. First, take \(\epsilon\) small enough s.t. \(z_0\) is the only singularity in \(C_\epsilon\). By residue theorem,
\[\int_{C_\epsilon} f(z)dz = 2\pi i Res(f, z_0) = 2\pi i \lim_{z\rightarrow z_0}(z-z_0)f(z) = 0\]
Then, let \(a, b\) be the two endpoints of \(C(\epsilon, \theta_1, \theta_2)_{z_0}\), so that \(C^{1} = C(\epsilon, \theta_1, \theta_2)_{z_0}\) is the upper arc of \(C_{\epsilon}\), then, let \(C_{2}\) be the lower arc from \(a\) to \(b\). Because \(\int_{C_\epsilon} = 0\), by deformation of curve, \(\int_{C_{1} } = \int_{C_{2} }\). Also, note that \(\int_{C_2} + (\int_{-C_1}) = 0\) so that \(\int_{C_1} = \int_{-C_1} = 0\)
Consider each integral, we have - By Residue Theorem, since \(I_C\) goes around the undefined point, \(I_C = 0\) - By definition, \(I_{line} = \lim_{R\rightarrow\infty}\lim_{\epsilon\rightarrow 0}p.v.\int_{-R}^R f(z)dz\) is exactly what we want. - By Jordon's Lemma, \(I_{R} = 0\). Thus, we are left with \(I_{line} = I_\epsilon = \lim_{\epsilon\rightarrow 0}\int_{C(\epsilon,0,\pi)} f(z)dz\) is what we want. By lemma 2,
\[p.v. \frac{\sin x }{x}dx = i^{-1}I_{line} = \pi\]
Example 2
Similarly, if the principal value integral has multiple undefined points, we can go around each of them. Consider \(p.v. \int_{-\infty}^\infty (x^2 - 1) dx\)
Similarly, \(I_C = I_{R} + I_{line} - I_{\epsilon, -1} - I_{\epsilon, 1}\) - \(I_C = 0\) by residue theorem - \(I_{line} = \lim_{R\rightarrow\infty}\lim_{\epsilon\rightarrow 0} \int_{C_{R, \epsilon}^{line} } (z^2 - 1)^{-1}dz = p.v. \int_{-\infty}^\infty (x^2-1)^{-1}dx\) is what we want - $ I_{\epsilon, -1} + I_{\epsilon, 1} = I_{line}$ is what we calculate. By lemma 2, since \(z=-1\) and \(z=1\) are both poles of order 1,
\[I_{\epsilon, 1} = \pi i Res(f, 1) = \lim_{z\rightarrow 1}(z+1)^{-1} = \pi i /2\]
\[ \int_{-\infty}^\infty (x^2-1)^{-1}dx = \pi i /2 - \pi i /2 = 0\]
Integrals with Branch Cuts
Consider functions on \(\mathbb C\) with branch points at \(0\), and we only defined the funciton on some domain \(D = \{re^i\theta: r> 0, \theta\in (0, 2\pi)\}\) with a branch cut.
For function \(f(z)\), define \(f_+, f_-: (0,\infty)\rightarrow\mathbb C\) as
Then, note that \(\lim_{z\rightarrow\infty} z f(z) = \lim_{z\rightarrow\infty}\frac{z^{3/2}}{z^3+1} = 0, \lim_{z\rightarrow 0} z f(z) = \frac{0^{3/2}}{0^3+1} = 0\), so that