6.23 Principal values of integrals.

It was assumed in §6.22, §6.221, §6.222 that the function \(Q(x)\) had no poles on the real axis; if the function has a finite number of simple poles on the real axis, we can obtain theorems corresponding to those already obtained, except that the integrals are all principal values (§4.5) and \(\sum R\) has to be replaced by \(\sum R + \frac{1}{2} \sum R_{0}\), where \(\sum R_{0}\) means the sum of the residues at the poles on the real axis. To obtain this result we see that, instead of the former contour, we have to take as contour a circle of radius \(\rho\) and the portions of the real axis joining the points \[ -\rho, \ a-\delta_{1}, \ \ a+\delta_{1}, \ b-\delta_{2}, \ \ b+\delta_{2}, \ c-\delta_{3}, \; \ldots \] and small semicircles above the real axis of radii \(\delta_{1}\), \(\delta_{2},\ldots\) with centres \(a\), \(b\), \(c, \ldots\) where \(a\), \(b\), \(c, \ldots\) are the poles of \(Q(z)\) on the real axis; and then we have to make \(\delta_{1}, \delta_{2}, \ldots \rightarrow 0\); call these semicircles \(\gamma_{1}\), \(\gamma_{2},\ldots\). Then instead of the equation \[ \int_{-\rho}^{\rho}\! Q(z) \,dz + \int_{\Gamma} Q(z) \,dz = 2 \pi i \sum R, \] we get \[ P \int_{-\rho}^{\rho}\! Q(z) \,dz + \sum_{n} \lim_{\delta_{n} \rightarrow 0} \int_{\gamma_{n}}\! Q(z) \,dz + \int_{\Gamma} Q(z) \,dz = 2 \pi i \sum R. \]

Let \(a'\) be the residue of \(Q(z)\) at \(a\); then writing \(z = a + \delta_{1} e^{i\theta}\) on \(\gamma_{1}\) we get \[ \int_{\gamma_{1}}\! Q(z) \,dz = \int_{\pi}^{0}\! Q(a + \delta_{1} e^{i\theta})\, \delta_{1} e^{i\theta} i \,d\theta. \] But \(Q(a + \delta_{1} e^{i\theta}) \rightarrow a'\) uniformly as \(\delta_{1} \rightarrow 0\); and therefore \( \lim\limits_{\delta_{1} \rightarrow 0} \int_{\gamma_{1}} Q(z) \,dz = - \pi i a' \); we thus get \[ P \int_{-\rho}^{\rho}\! Q(z) \,dz + \int_{\Gamma} Q(z) \,dz = 2 \pi i \sum R + \pi i \sum R_{0}, \] and hence, using the arguments of §6.22, we get \[ P \int_{-\infty}^{\infty}\! Q(x) \,dx = 2 \pi i \left( \sum R + \frac{1}{2} \sum R_{0} \right). \]

The reader will see at once that the theorems of §6.221, §6.222 have precisely similar generalisations.

The process employed above of inserting arcs of small circles so as to diminish the area of the contour is called indenting the contour.

6.24 Evaluation of integrals of the form \(\int_{0}^{\infty} x^{a-1} Q(x) \,dx.\)

Let \(Q(x)\) be a rational function of \(x\) such that it has no poles on the positive part of the real axis and \(x^{a} Q(x) \rightarrow 0\) both when \(x \rightarrow 0\) and when \(x \rightarrow \infty\).

Consider \(\int (-z)^{a-1} Q(z) \,dz\) taken round the contour \(C\) shewn in figure 1, consisting of the arcs of circles of radii \(\rho\), \(\delta\) and the straight lines joining their end points; \((-z)^{a-1}\) is to be interpreted as \[ \exp \left\{ ( a-1) \log (- z) \right\} \] and \[ \log (-z) = \log \left|\,z\,\right| + \arg (-z), \] where \[ -\pi \leq \arg (-z) \leq \pi; %TODO:is overlap correct? \] with these conventions the integrand is one-valued and analytic on and within the contour save at the poles of \(Q(z)\).

Figure 1: Contour C for \(\int_C (-z)^{a-1} Q(z) \,dz\\).
Figure 1: Contour C for \(\int_C (-z)^{a-1} Q(z) \,dz\).

Hence, if \(\sum r\) denote the sum of the residues of \((-z)^{a-1} Q(z)\) at all its poles, \[ \int_{C} (-z)^{a-1} Q(z) \,dz = 2 \pi i \sum r. \]

On the small circle write \(-z = \delta e^{i\theta}\), and the integral along it becomes \(- \int_{\pi}^{-\pi} (-z)^{a} Q(z) i \,d\theta\), which tends to zero as \(\delta \rightarrow 0\).

On the large semicircle write \(-z = \rho e^{i\theta}\), and the integral along it becomes \(- \int_{-\pi}^{\pi} (-z)^{n} Q(z) i \,d\theta\), which tends to zero as \(\rho \rightarrow \infty\).

On one of the lines we write \(-z = x e^{\pi i}\) on the other \(-z = x e^{-\pi i}\) and \((-z)^{a-1}\) becomes \(x^{a-1} e^{\pm (a-1) \;\!\pi i}\).

Hence \[ \lim_{\delta\rightarrow 0,\ \rho\rightarrow\infty} \int_{\delta}^{\rho}\! \left\{ x^{a-1} e^{-(a-1)\pi i} Q(x) - x^{a-1} e^{(a-1)\pi i} Q(x) \right\} \,dx = 2 \pi i \sum r; \] and therefore \[ \int_{0}^{\infty}\! x^{a-1} Q(x) \,dx = \pi \, \mathrm{cosec}(a\pi) \sum r. \]

Corollary. If \(Q(x)\) have a number of simple poles on the positive part of the real axis, it may be shewn by indenting the contour that \[ P \int_{0}^{\infty}\! x^{a-1} Q(x) \,dx = \pi \, \mathrm{cosec} (a\pi) \sum r - \pi \cot (1\pi) \sum r', \] where \(\sum r'\) is the sum of the residues of \(z^{a-1} Q(z)\) at these poles.

Example 1. If \(\,0 < a < 1\), \[ \int_{0}^{\infty}\! \frac{x^{a-1}}{1+x} \,dx = \pi \, \mathrm{cosec}\;\! a\pi, \quad P \int_{0}^{\infty}\! \frac{x^{a-1}}{1+x} \,dx = \pi \cot a\pi \]

Example 2. If \(\,0 < z < 1\) and \(-\pi < a < \pi\), \[ \int_{0}^{\infty}\! \frac{t^{\:\!z-1}}{ t+e^{ia} } \,dt = \frac{ \pi e^{i(z-1)\:\! a}}{\sin \pi z}. \] (Minding.)

Example 3. Shew that, if \(- 1 < z < 3\), then \[ \int_{0}^{\infty}\! \frac{ x^{z} }{ (1+x^{2})^{2} } \,dx = \frac{ \pi (1-z) }{ 4 \cos \frac{1}{2}\pi z}. \]

Example 4. Shew that, if \(-1 < p < 1\) and \(-\pi < \lambda < \pi\), then \[ \int_{0}^{\infty}\! \frac{ x^{-p} \,dx }{ 1 + 2x \cos\lambda + x^{2} } = \frac{\pi}{\sin p\pi} \frac{\sin p\lambda}{\sin \lambda}. \] (Euler.)

6.3 Cauchy’s integral.

We shall next discuss a class of contour-integrals which are sometimes found useful in analytical investigations.

Let \(C\) be a contour in the \(z\)-plane, and let \(f(z)\) be a function analytic inside and on \(C\). Let \(\phi(z)\) be another function which is analytic inside and on \(G\) except at a finite number of poles; let the zeros of \(\phi(z)\) in the interior[1] of \(C\) be \(a_{1}\), \(a_{2},\ldots\), and let their degrees of multiplicity be \(r_{1}\), \(r_{2},\ldots\); and let its poles in the interior of \(C\) be \(b_{1},b_{2},\ldots\), and let their degrees of multiplicity be \(s_{1}\), \(s_{2},\ldots\).

[1]\(\phi(z)\) must not have any zeros or poles on \(C\). ↩

Then, by the fundamental theorem of residues, \( \displaystyle \frac{1}{2\pi i} \int_{C} f(z) \frac{\phi'(z)}{\phi(z)} \,dz \) is equal to the sum of the residues of \(f(z) \dfrac{\phi'(z)}{\phi(z)}\) at its poles inside \(C\).

Now \(f(z) \dfrac{\phi'(z)}{\phi(z)}\) can have singularities only at the poles and zeros of \(\phi(z)\). Near one of the zeros, say \(a_{1}\), we have \[ \phi(z) = A (z-a_{1})^{r_{1}} + B (z-a_{1})^{r_{1}+1} + \cdots. \]

Therefore \[ \phi'(z) = A r_{1} (z-a_{1})^{r_{1}-1} + B (r_{1}+1) (z-a_{1})^{r_{1}} + \cdots, \] and \[ f(z) = f(a_{1}) + (z-a_{1}) f'(a_{1}) + \cdots. \]

Therefore \( \left\{f(z) \dfrac{\phi'(z)}{\phi(z)} - \dfrac{ r_{1} f(a_{1}) }{ z - a_{1} } \right\} \) is analytic at \(a_{1}\).

Thus the residue of \(f(z) \dfrac{\phi'(z)}{\phi(z)}\), at the point \(z=a_{1}\), is \(r_{1} f(a_{1})\).

Similarly the residue at \(z=b_{1}\) is \(-s_{1} f(b_{1})\); for near \(z=b_{1}\), we have \[ \phi(z) = C (z-b_{1})^{-s_{1}} + D (z-b_{1})^{-s_{1} + 1} + \cdots, \] and \[ f(z) = f(b_{1}) + (z - b_{1}) f'(b_{1}) + \cdots, \] so \( f(z) \dfrac{ \phi'(z) }{\phi(z)} + \dfrac{ s_{1} f(b_{1}) }{ z-b_{1} } \) is analytic at \(b_{1}\).

Hence \[ \frac{1}{2\pi i} \int_{C} f(z) \frac{ \phi'(z) }{\phi(z)} \,dz = \sum_m r_{m}\, f(a_{m}) - \sum_m s_{m}\, f(b_{m}), \] the summations being extended over all the zeros and poles of \(\phi(z)\).

6.31 The number of roots of an equation contained within a contour.

The result of the preceding paragraph can be at once applied to find how many roots of an equation \(\phi(z) = 0\) lie within a contour \(C\).

For, on putting \(f(z) = 1\) in the preceding result, we obtain the result that \[ \frac{1}{2\pi i} \int_{C} \frac{ \phi'(z) }{\phi(z)} \,dz \] is equal to the excess of the number of zeros over the number of poles of \(\phi(z)\) contained in the interior of \(C\), each pole and zero being reckoned according to its degree of multiplicity.

Example 1. Shew that a polynomial \(\phi(z)\) of degree \(m\) has \(m\) roots.

Let \[ \phi(z) = a_{0} z^{m} + a_{1} z^{m-1} + \cdots + a_{m}, \quad (a_{0} \neq 0). \] Then \[ \frac{ \phi'(z) }{ \phi(z) } = \frac{ m a_{m}z^{m-1} + \cdots + a_{m-1} }{ a_{0}z^{m} + \cdots a_{m}}. \] Consequently, for large values of \(\left|\,z\,\right|\), \[ \frac{ \phi'(z) }{ \phi(z) } = \frac{m}{z} + O\left( \frac{1}{z^{2}} \right) \] Thus, if \(C\) be a circle of radius \(\rho\) whose centre is at the origin, we have \[ \frac{1}{2 \pi i} \int_{C} \frac{\phi'(z)}{\phi(z)} \,dz = \frac{m}{2 \pi i} \int_{C} \frac{ \,dz }{z} + \frac{1}{2 \pi i} \int_{C} O\left( \frac{1}{z^{2}} \right) \,dz = m + \frac{1}{2 \pi i} \int_{C} O\left( \frac{1}{z^{2}} \right) \,dz. \]

But, as in §6.22, \[ \int_{C} O\left( \frac{1}{z^{2}} \right) \,dz \rightarrow 0 \] as \(\rho\rightarrow\infty\); and hence as \(\phi(z)\) has no poles in the interior of \(C\), the total number of zeros of \(\phi(z)\) is \[ \lim_{\rho\rightarrow\infty} \frac{1}{2\pi i} \int_{C} \frac{ \phi'(z) }{\phi(z)} \,dz = m. \]

Example 2. If at all points of a contour \(C\) the inequality \[ \left|\, a_{k} z^{k} \,\right| > \left|\, a_{0} + a_{1} z + \cdots + a_{k-1} z^{k-1} + a_{k+1} z^{k+1} + \cdots + a_{m} z^{m} \,\right| \] is satisfied, then the contour contains \(k\) roots of the equation \[ a_{m} z^{m} + a_{m-1} z^{m-1} + \cdots a_{1} z + a_{0} = 0. \]

For write \[ f(z) = a_{m} z^{m} + a_{m-1} z^{m-1} + \cdots a_{1} z + a_{0}. \] Then \[\begin{align*} f(z) &= a_{k} z^{k} \left( 1 + \frac{ a_{m}z^{m} + \cdots + a_{k+1}z^{k+1} + a_{k-1}z^{k-1} + \cdots + a_{0} }{ a_{k} z^{k} } \right) \\ &= a_{k} z^{k} (1 + U), \end{align*}\] where \(\left|\,U\,\right| \leq a \leq 1\) on the contour, \(a\) being independent of \(z\).[2]

Therefore the number of roots of \(f(z)\) contained in \(C\) \[ = \frac{1}{2 \pi i} \int_{C} \frac{f'(z)}{f(z)} \,dz = \frac{1}{2 \pi i} \int_{C} \left( \frac{k}{z} + \frac{1}{1+U} \frac{ d U }{ d z } \right) \,dz \]

But \(\int_{C} \dfrac{d z}{z} = 2 \pi i\); and, since \(\left|\,U\,\right| < 1\), we can expand \((1 + U)^{-1}\) in the uniformly convergent series \[ 1 - U + U^{2} - U^{3} + \cdots, \] so \[ \int_{C} \frac{1}{1+U} \frac{d U}{d z} \,dz = \left[ U - \frac{1}{2} U^{2} + \frac{1}{3} U^{3} - \cdots \right]_{C} \! = 0. \] Therefore the number of roots contained in \(C\) is equal to \(k\).

[2]\(\,\left|\,U\,\right|\) is a continuous function of \(z\) on \(C\), and so attains its upper bound (§3.62). Hence its upper bound \(a\) must be less than \(1\). ↩

Example 3. Find how many roots of the equation \[ z^{6} + 6z + 10 = 0 \] lie in each quadrant of the Argand diagram.\(\vphantom{\\ 3\\}\)
(Clare, 1900.)