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Principal Value Integrals

Principal Value Integrals

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

\[p.v.\int_a^b f(x)dx = \lim_{\epsilon\rightarrow 0} \big(\int_a^{c-\epsilon} f(x)dx + \int_{c+\epsilon}^b f(x)dx\big)\]

Example Consider \(x^{-1}\), which is not defined on \(x=0\), but

\[\begin{align*} p.v. \int_{-1}^1 x^{-1}dx &= \lim_{\epsilon\rightarrow 0} \big(\int_{-1}^{-\epsilon} x^{-1}dx + \int_{\epsilon}^1 x^{-1}dx\big) \\ &= \lim_{\epsilon\rightarrow 0} [\log(x)]^{-\epsilon}_{-1} + [\log(x)]^{1}_{\epsilon}\\ &= 0 \end{align*}\]

Note that \(p.v.\int_a^b f(x)dx = \int_a^b f(x)dx\) if the integral exists.
And the principal value integral at infinity is defined as

\[\begin{align*} p.v.\int_{-\infty}^\infty f(x)dx &= \lim_{R\rightarrow\infty}p.v.\int_{-R}^R f(x)dx\\ &=\lim_{R\rightarrow\infty} \lim_{\epsilon\rightarrow 0} \big(\int_{-R}^{c-\epsilon} f(x)dx + \int_{c+R}^\infty f(x)dx\big) \end{align*}\]

Solving Integrals with PVI

Then, for some real functions, we can extend to complex, and take a way around its undefined point.

Lemma 1

\(\lim_{z\rightarrow z_0}(z-z_0)f(z) = 0\implies \lim_{\epsilon\rightarrow 0^+} \int_{C(\epsilon, \theta_1, \theta_2)_{z_0} }f(z)dz = 0\).

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 \(\int_{C_1} = \int_{-C_1} = 0\)

Lemma 2

If \(f(z)\) has a pole of order 1 at \(z_0\) then

\[\lim_{\epsilon\rightarrow 0^+} \int_{C(\epsilon, \theta_1, \theta_2)_{z_0} }f(z)dz = i(\theta_2 - \theta_1)Res(f, z_0)\]

proof. Consider the Laurent expansion of \(f\),

\[\begin{align*} \lim_{\epsilon\rightarrow 0^+} \int_{C(\epsilon, \theta_1, \theta_2)_{z_0} }f(z)dz &= \lim_{\epsilon\rightarrow 0^+} \int_{C(\epsilon, \theta_1, \theta_2)_{z_0} }\sum_{n=-1}^\infty c_n (z-z_0)^ndz\\ &= \lim_{\epsilon\rightarrow 0^+} \int_{C(\epsilon, \theta_1, \theta_2)_{z_0} } c_{-1} (z-z_0)^{-1}dz&\text{lemma 1}\\ &= \lim_{\epsilon\rightarrow 0^+}i\int_{\theta_1}^{\theta_2} c_{-1} (z_0 + \epsilon e^{it} - z_0)^{-1} \epsilon e^{it}dt\\ &= i(\theta_2 - \theta_1) c_{-1} \end{align*}\]

Example 1

Consider \(p.v. \frac{\sin^{x} }{x}dx\) First note that

\[p.v. \int_{-\infty}^\infty \frac{e^{ix} }{x}dx = p.v. \int_{-\infty}^\infty \frac{\cos x}{x}dx + i \:\:p.v.\int_{-\infty}^\infty \frac{\sin x}{x}dx \]

Because $\cos x / x $ is odd, $ p.v. \int_{-\infty}^\infty \frac{\cos x}{x}dx = 0$, we are left with

\[ \:\:p.v.\int_{-\infty}^\infty \frac{\sin x}{x}dx = i^{-1}p.v. \int_{-\infty}^\infty \frac{e^{ix} }{x}dx\]

Define \(f(z) = e^{iz}/z\) and consider the integrals

\[\begin{align*} I_C &= \lim_{R\rightarrow\infty}\lim_{\epsilon\rightarrow 0} \oint_{C_{R,\epsilon} } f(z)dz\\ &= \lim_{R\rightarrow\infty}\lim_{\epsilon\rightarrow 0} (\int_{C_{R, \epsilon}^{line} } f(z)dz + \int_{C(R,0,\pi)} f(z)dz - \int_{C(\epsilon,0,\pi)} f(z)dz )\\ &= \lim_{R\rightarrow\infty}\lim_{\epsilon\rightarrow 0} \int_{C_{R, \epsilon}^{line} } f(z)dz + \lim_{R\rightarrow\infty}\int_{C(R,0,\pi)} f(z)dz - \lim_{\epsilon\rightarrow 0}\int_{C(\epsilon,0,\pi)} f(z)dz \\ &= I_{line} + I_R - I_\epsilon \end{align*}\]

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,

\[\lim_{\epsilon\rightarrow 0}\int_{C(\epsilon,0,\pi)} f(z)dz = i(\pi - 0) Res(f, 0) = i\pi (\lim_{z\rightarrow 0} z \frac{e^{iz} }{z}) = i\pi\]

Therefore,

\[p.v. \frac{\sin x }{x}dx = i^{-1}I_{line} = \pi\]

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,

\[I_{\epsilon, 1} = \pi i Res(f, 1) = \lim_{z\rightarrow 1}(z+1)^{-1} = \pi i /2\]
\[I_{\epsilon, -1} = \pi i Res(f, -1) = \lim_{z\rightarrow -1}(z-1)^{-1} = -\pi i/2\]
\[ \int_{-\infty}^\infty (x^2-1)^{-1}dx = \pi i /2 - \pi i /2 = 0\]

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

\[f_+(r) = \lim_{\theta\rightarrow 0^+} f(re^{i\theta}), f_-(r) = \lim_{\theta\rightarrow 2\pi^-} f(re^{i\theta})\]

Example 1

Consider the sqrt function \(f(re^{i\theta}) = r^{1/2} e^{i\theta/2}\), then

\[f_+(r) = r^{1/2}, f_-(r) = -r^{1/2}\]

Example 2

evaluate \(\int_0^\infty \frac{\sqrt x}{x^3+1}dx\)

First, note that

\[\lim_{R\rightarrow\infty}\lim_{\epsilon\rightarrow 0}\lim_{\theta\rightarrow0^+}I_C = \lim_{R\rightarrow\infty}I_{C_R} - \lim_{\epsilon\rightarrow 0}I_{C_\epsilon} + \lim_{R\rightarrow\infty}\lim_{\epsilon\rightarrow 0}\lim_{\theta\rightarrow0^+} I_{C^{+}_{R,\epsilon}} + \lim_{R\rightarrow\infty}\lim_{\epsilon\rightarrow 0}\lim_{\theta\rightarrow0^+} I_{C^{-}_{R,\epsilon}}\]

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

\[\lim_{R\rightarrow\infty}I_{C_R} = 0, \lim_{\epsilon\rightarrow 0}I_{C_\epsilon} = 0\]

Then, consider the left two terms

\[\lim_{R\rightarrow\infty}\lim_{\epsilon\rightarrow 0}\lim_{\theta\rightarrow0^+} I_{C^{+}_{R,\epsilon}} = \lim_{R\rightarrow\infty}\lim_{\epsilon\rightarrow 0}\int_{\epsilon}^R f_+(x)dx = \int_{0}^\infty f_+(x)dx\]
\[\lim_{R\rightarrow\infty}\lim_{\epsilon\rightarrow 0}\lim_{\theta\rightarrow0^+} I_{C^{-}_{R,\epsilon}} = \lim_{R\rightarrow\infty}\lim_{\epsilon\rightarrow 0}\int_{R}^{\epsilon} f_-(z)dz = -\int_{0}^\infty f_-(z)dz\]

via branch cut of square root functions, we have \(-f_-(z) = f_+(z) = f(x)\) so that the whole integral

\[\lim_{R\rightarrow\infty}\lim_{\epsilon\rightarrow 0}\lim_{\theta\rightarrow0^+}I_C = \int_{0}^\infty f_+(x) -\int_{0}^\infty f_-(x) = 2 \int_0^\infty \frac{\sqrt x }{x^3+1} dx\]

Now, using residue theorem,

\[I_C = \int_{C} f(z)dz = 2\pi i (Res(f, -1) + Res(f, e^{i\pi/3}) + Res(f, e^{-i\pi/3}))\]

so that the original integral

\[\int_0^\infty \frac{\sqrt x }{x^3+1} dx = \pi i (Res(f, -1) + Res(f, e^{i\pi/3}) + Res(f, e^{-i\pi/3}))\]