The Theory of Residues;
Application to the Evaluation of Definite Integrals

6.1 Residues.

If the function \(f(z)\) has a pole of order \(m\) at \(z=a\), then, by the definition of a pole, an equation of the form \[ f(z) = \frac{a_{-m}}{ (z-a)^{m} } + \frac{ a_{-m+1} }{ (z-a)^{m-1} } + \cdots + \frac{a_{-1}}{z-a} + \phi(z), \] where \(\phi(z)\) is analytic near and at \(a\), is true near \(a\).

The coefficient \(a_{-1}\) in this expansion is called the residue of the function \(f(z)\) relative to the pole \(a\).

[1]The existence of such a circle is implied in the definition of a pole as an isolated singularity. ↩

Consider now the value of the integral \(\int_{\alpha} f(z) \, d z\), where the path of integration is a circle \(\alpha\),[1] whose centre is the point \(a\) and whose radius \(\rho\) is so small that \(\phi(z)\) is analytic inside and on the circle.

We have \[ \int_{\alpha} f(z) \, d z = \sum_{r=1}^{m} a_{r} \int_{\alpha}\! \frac{\, d z}{ (z-a)^{r}} + \int_{\alpha}\! \phi(z) \, d z. \]

Now \(\displaystyle \int_{\alpha}\! \phi(z) \, d z = 0\) by §5.2; and (putting \(z-a = \rho e^{i\theta}\)) we have, if \(r \neq 1\), \[\begin{align*} \int_{\alpha}\! \frac{ \, d z}{ (z-a)^{r} } =& \int_{0}^{2\pi}\! \frac{\rho e^{i\theta} i \, d \theta}{ \rho^{r} e^{ri\theta}} = \rho^{-r+1} \int_{0}^{2\pi}\! e^{(1-r)i\theta} i \, d \theta =\\ \\ &\rho^{-r+1} \left[ \frac{e^{(1-r) i \theta}}{1-r} \right]_{0}^{2\pi} = 0. \end{align*}\]

But, when \(r = 1\), we have \[ \int_{\alpha} \frac{d z}{z - a} = \int_{0}^{2\pi} i \, d \theta = 2\pi i. \]

Hence finally \[ \int_{\alpha} f(z) \, d z = 2 \pi i a_{-1}. \]

Now let \(C\) be any contour, containing in the region interior to it a number of poles \(a\), \(b\), \(c,\ldots\) of a function \(f(z)\), with residues \(a_{-1}\), \( b_{-1}\), \( c_{-1},\ldots\) respectively; and suppose that the function \(f(z)\) is analytic throughout \(C\) and its interior, except at these poles.

Surround the points \(a\), \(b\), \(c,\ldots\) by circles \(\alpha\), \(\beta\), \(\gamma,\ldots\) so small that their respective centres are the only singularities inside or on each circle; then the function \(f(z)\) is analytic in the closed region bounded by \(C\), \(\alpha\), \(\beta\), \(\gamma,\ldots\).

Hence, by §5.2 corollary 3, \[\begin{align*} \int_{C} f(z) \, d z =& \int_{\alpha} f(z) \, d z + \int_{\beta} f(z) \, d z + \cdots \\ =& 2\pi i a_{-1} + 2\pi i b_{-1} + \cdots. \end{align*} \]

Thus we have the theorem of residues, namely that if \(f(z)\) be analytic throughout a contour \(C\) and its interior except at a number of poles inside the contour, then \[ \int_{C} f(z) \, d z = 2\pi i \sum R, \] where \(\sum R\) denotes the sum of the residues of the function \(f(z)\) at those of its poles which are situated within the contour \(C\).

This is an extension of the theorem of §5.21.

Note. If \(a\) is a simple pole of \(f(z)\) the residue of \(f(z)\) at that pole is \(\lim\limits_{z\rightarrow a} \, (z-a)\, f(z)\).

6.2 The evaluation of definite integrals.

We shall now apply the result of §6.1 to evaluating various classes of definite integrals; the methods to be employed in any particular case may usually be seen from the following typical examples.

6.21 The evaluation of the integrals of certain periodic functions taken between the limits \(0\) and \(2\pi\).

An integral of the type \[ \int_{0}^{2\pi} R(\cos \theta, \sin \theta) \, d \theta, \] where the integrand is a rational function of \(\cos\theta\) and \(\sin\theta\) finite on the range of integration, can be evaluated by writing \(e^{i\theta}=z\); since \[ \cos\theta = \frac{1}{2}(z+z^{-1}), \quad \sin\theta = \frac{1}{2i} (z-z^{-1}), \] the integral takes the form \(\int_{C} S(z) \, d z\), where \(S(z)\) is a rational function of \(z\) finite on the path of integration \(C\), the circle of radius unity whose centre is the origin.

Therefore, by §6.1, the integral is equal to \(2\pi i\) times the sum of the residues of \(S(z)\) at those of its poles which are inside that circle.

Example 1. If \(0 < p < 1\), \[ \int_{0}^{2\pi} \!\! \frac{\, d \theta}{ 1 - 2p\cos\theta + p^{2}} = \int_{C} \frac{ \, d z}{ i\, (1-pz) (z-p) }. \] The only pole of the integrand inside the circle is a simple pole at \(p\); and the residue there is \[ \lim_{z \rightarrow p} \frac{z-p}{ i\, (1-pz)(z-p)} = \frac{1}{i\, (1-p^{2})}. \] Hence \[ \int_{0}^{2\pi} \!\!\frac{\, d\theta}{1 - 2p\cos\theta + p^{2}} = \frac{2\pi}{1-p^{2}}. \]

Example 2. If \(0 < p < 1\), \[\begin{align*} & \int_{0}^{2\pi}\! \frac{ \cos^{2} 3\theta}{1-2p\cos 2\theta + p^{2}} \, d \theta =\\ \\ & \qquad \qquad\int_{C} \frac{\, d z}{iz} \left(\frac{1}{2} z^{3} + \frac{1}{2} z^{-3}\right)^{2} \frac{1}{ (1-pz^{2})(1-pz^{-2})} \end{align*} \] where \(\sum R\) denotes the sum of the residues of \( \dfrac{ (z^{6} + 1)^{2} }{ 4z^{5} (1-pz^{2})(z^{2}-p)} \) at its poles inside \(C\); these poles are \(0\), \(-p^{\frac{1}{2}}\), \(p^{\frac{1}{2}}\); and the residues at them are \(\, -\dfrac{1+p^{2}+p^{4}}{4p^{3}}\), \(\,\dfrac{(p^{3}+1)^{2}}{ 8p^{3}(1-p^{2})}\), \(\,\dfrac{(p^{3}+1)^{2}}{ 8p^{3}(1-p^{2})}\); and hence the integral is equal to \[ \frac{\pi (1-p+p^{2})}{1-p}. \]

Example 3. If \(n\) be a positive integer, \[\begin{align*} \int_{0}^{2\pi}\! e^{\cos\theta} \cos(n\theta - \sin\theta) \, d \theta = \frac{2\pi}{n!}, \quad \int_{0}^{2\pi}\! e^{\cos\theta} \sin(n\theta - \sin\theta) \, d \theta = 0. \end{align*}\]

Example 4. If \(a>b>0\), \[ \int_{0}^{2\pi}\! \frac{\, d \theta}{ (a+b\cos\theta)^{2}} = \frac{2\pi a}{ (a^{2}-b^{2})^{3/2} }, \quad \int_{0}^{2\pi}\! \frac{\, d\theta}{ (a+b\cos^{2}\theta)^{2}} = \frac{\pi(2a+b)}{ a^{3/2} (a+b)^{3/2}}. \]

6.22 The evaluation of certain types of integrals taken between the limits \(-\infty\) and \(+\infty\).

We shall now evaluate \(\int_{-\infty}^{\infty} Q(x) \, d x\), where \(Q(z)\) is a function such that

  1. It is analytic when the imaginary part of \(z\) is positive or zero (except at a finite number of poles).
  2. It has no poles on the real axis and
  3. As \(\left|\, z \,\right|\rightarrow\infty\), \(zQ(z) \rightarrow 0\) uniformly for all values of \(\arg z\) such that \(0 \leq \arg z \leq \pi\); provided that
  4. When \(x\) is real, \(x Q(x) \rightarrow 0\), as \(x \rightarrow \pm\infty\), in such a way that \(\int_{0}^{\infty} Q(x) \, d x\) and \(\int_{-\infty}^{0} Q(x) \, d x\) both converge.[2]
[2]The condition \(x Q(x)\rightarrow 0\) is not in itself sufficient to secure the convergence of \(\int^{\infty} Q(x) \, d x\); consider \(Q(x) = (x \log x)^{-1}\). ↩

Given \(\epsilon\), we can choose \(\rho_{0}\) (independent of \(\arg z\)) such that \(\left|\, z Q(z) \,\right| < \left.\epsilon\middle/\pi\right.\) whenever \(\left|\, z \,\right| > \rho_{0}\) and \(0 \leq \arg z \leq \pi\).

Consider \(\int_{C} Q(z) \, d z\) taken round a contour \(C\) consisting of the part of the real axis joining the points \(\pm\rho\) (where \(\rho > \rho_{0}\)) and a semicircle \(\Gamma\), of radius \(\rho\), having its centre at the origin, above the real axis.

Then, by §6.1, \(\int_{C} Q(z) \, d z = 2 \pi i \sum R\), where \(\sum R\) denotes the sum of the residues of \(Q(z)\) at its poles above the real axis.[3]

[3]Assume \(\rho\) is sufficiently large that \(Q(z)\) has no poles above the real axis outside the contour. ↩

Therefore \[ \left|\, \int_{-\rho}^{\rho}\! Q(z) \, d z - 2 \pi i \sum R \,\right| = \left|\, \int_{\Gamma} Q(z) \, d z \,\right|. \]

In the last integral write \(z = \rho e^{i\theta}\), and then \[\begin{align*} \left|\, \int_{\Gamma} Q(z) \, d z \,\right| &= \left|\, \int_{0}^{\pi}\! Q(\rho e^{i\theta}) \rho e^{i\theta} i \, d\theta \,\right| \\ \\ &< \int_{0}^{\pi}\! (\left.\epsilon \middle/ \pi\right.) \, d \theta \\ \\ \\ &= \epsilon \end{align*} \] by §4.62.

Hence \[ \lim_{\rho\rightarrow\infty} \int_{-\rho}^{\rho}\! Q(z) \, d z = 2\pi i \sum R. \]

But the meaning of \(\displaystyle \int_{-\infty}^{\infty}\! Q(x) \, d x\); is \(\displaystyle \lim_{\rho,\,\sigma\rightarrow\infty}\int_{-\rho}^{\sigma}\! Q(x) \, d x\); and since \(\displaystyle \lim_{\sigma\rightarrow\infty}\int_{0}^{\sigma}\! Q(x) \, d x\) and \(\displaystyle \lim_{\rho\rightarrow\infty}\int_{-\rho}^{0}\! Q(x) \, d x\) both exist, this double limit is the same as \(\displaystyle \lim_{\rho\rightarrow\infty}\int_{-\rho}^{\rho}\! Q(x) \, d x\).

Hence we have proved that \[ \int_{-\infty}^{\infty}\! Q(x) \, d x = 2\pi i \sum R. \]

This theorem is particularly useful in the special case when \(Q(x)\) is a rational function.

Note. Even if condition (iv) is not satisfied, we still have \[ \int_{0}^{\infty}\! \left\{ Q(x) + Q(-x) \right\} \, d x = \lim_{\rho\rightarrow\infty} \int_{-\rho}^{\rho}\! Q(x) \, d x = 2 \pi i \sum R. \]

Example 1. The only pole of \((z^{2} + 1)^{-3}\) in the upper half plane is a pole at \(z=i\) with residue there \(-\frac{3}{16} i\). Therefore \[ \int_{-\infty}^{\infty}\! \frac{\, d x}{ (x^{2}+1)^{3} } = \frac{3}{8} \pi. \]

Example 2. If \(a > 0, b > 0\), shew that \[ \int_{-\infty}^{\infty}\! \frac{x^{4} \, d x}{ (a + bx^{2})^{4} } = \frac{\pi}{ 16 a^{3/2} b^{5/2}}. \]

Example 3. By integrating \(\int e^{-\lambda z^{2}} \, d z\) round a parallelogram whose corners are \(-R\), \(R\), \(R + ai\), \(-R + ai\) and making \(R \rightarrow \infty\), shew that, if \(\lambda > 0\), then \[ \int_{-\infty}^{\infty}\! e^{-\lambda x^{2}} \cos (2\lambda a x) \, d x = e^{-\lambda a^{2}}\int_{-\infty}^{\infty}\! e^{-\lambda x^{2}} \, d x = 2 \lambda^{-\frac{1}{2}} e^{-\lambda a^{2}}\int_{0}^{\infty}\! e^{-x^{2}} \, d x. \]
6.221 Certain infinite integrals involving sines and cosines.

If \(Q(z)\) satisfies the conditions (i), (ii) and (iii) of §6.22, and \(m > 0\), then \(Q(z) e^{miz}\) also satisfies those conditions.

Hence \( \int_{0}^{\infty} \left\{ Q(x) e^{mix} + Q(-x) e^{-mix} \right\} \, d x \) is equal to \(2\pi i \sum R'\), where \(\sum R'\) means the sum of the residues of \(Q(z) e^{mix}\) at its poles in the upper half plane; and so

  1. If \(Q(x)\) is an even function, i.e. if \(Q (- x) = Q (x)\), \[ \int_{0}^{\infty} Q(x) \cos(mx) \, d x = \pi i \sum R'. \]

  2. If \(Q(x)\) is an odd function,\[ \int_{0}^{\infty} Q(x) \sin(mx) \, d x = \pi \sum R'. \]

6.222 Jordan’s lemma.[4]
[4]Jordan, Cours d’Analyse, ii. (1894), pp. 285, 286. ↩

The results of §6.221 are true if \(Q(z)\) be subject to the less stringent condition \(Q(z) \rightarrow 0\) uniformly when \(O \leq \arg z \leq \pi\) as \(\left|\, z \,\right| \rightarrow \infty\) in place of the condition \(z Q(z) \rightarrow 0\) uniformly.

To prove this we require a theorem known as Jordan’s lemma, viz.

If \(Q(z) \rightarrow 0\) uniformly with regard to \(\arg z\) as \(\left|\, z \,\right| \rightarrow \infty\) when \(0 \leq \arg z \leq \pi\), and if \(Q(z)\) is analytic when both \(\left|\, z \,\right| > c\) (a constant) and \(0 \leq \arg z \leq \pi\), then \[ \lim_{\rho\rightarrow\infty} \left( \int_{\Gamma} e^{miz} Q(z) \, d z \right) = 0, \] where \(\Gamma\) is a semicircle of radius \(\rho\) above the real axis with centre at the origin.

Given \(\epsilon\), choose \(\rho_{0}\) so that \(\left|\, Q(z) \,\right| < \left.\epsilon\middle/\pi\right.\) when \(\left|\, z \,\right| > \rho_{0}\) and \(0 \leq \arg z \leq \pi\); then, if \(\rho > \rho_{0}\), \[ \left|\, \int_{\Gamma} e^{miz} Q(z) \, d z \,\right| = \left|\, \int_{0}^{\pi}\! e^{mi (\rho \cos\theta + i\rho \sin\theta)} Q(\rho e^{i\theta}) \rho e^{i \theta} i \, d \theta \,\right|. \] But \(\left|\, e^{mi\rho\cos\theta} \,\right| = 1\), and so \[\begin{align*} \left|\, \int_{\Gamma} e^{miz} Q(z) \, d z \,\right| <& \int_{0}^{\pi}\! (\left.\epsilon\middle/\pi\right.) \rho e^{-m\rho\sin\theta} \, d\theta \\ =& (\left.2\epsilon\middle/\pi\right.) \int_{0}^{\frac{1}{2}\pi}\! \rho e^{-m\rho\sin\theta} \, d\theta. \end{align*} \]

[5]This inequality appears obvious when we draw the graphs \(y = \sin x\), \(y = \left.2x\middle/\pi\right.\); it may be proved by shewing that \((\left.\sin\theta)\middle/\theta\right.\) decreases as \(\theta\) increases from \(0\) to \(\frac{1}{2}\pi\). ↩

Now \(\sin\theta \geq \left.2\theta\middle/\pi\right.\), when \(0 \leq \theta \leq \frac{1}{2}\pi\),[5] and so \[\begin{align*} \left|\, \int_{\Gamma} e^{miz} Q(z) \, d z \,\right| <& (\left.2\epsilon\middle/\pi\right.) \int_{0}^{\frac{1}{2}\pi}\! \rho e^{\left.-2m\rho\theta\middle/\pi\right.} \, d\theta \\ =& (\left.2\epsilon\middle/\pi\right.) \cdot (\left.\pi\middle/2m\right.) \left[ -e^{\left.-2m\rho\theta\middle/\pi\right.} \vphantom{\sum}\right]_{0}^{\frac{1}{2}\pi} \\ \\ <& \left.\epsilon\middle/m\right. . \end{align*} \]

Hence \[ \lim_{\rho\rightarrow\infty} \int_{\Gamma} e^{miz} Q(z) \, d z = 0. \]

This result is Jordan’s lemma.

Now \[ \int_{0}^{\rho}\! \left\{ e^{mix} Q(x) + e^{-mix} Q(-x) \right\} \, d x = 2 \pi i \sum R' - \int_{\Gamma} e^{miz} Q(z) \, d z, \] and, making \(\rho\rightarrow\infty\), we see at once that \[ \int_{0}^{\infty}\! \left\{ e^{mix} Q(x) + e^{-mix} Q(-x) \right\} \, d x = 2 \pi i \sum R', \] which is the result corresponding to the result of §6.221.

Example 1. Shew that, if \(a > 0\), then \[ \int_{0}^{\infty} \frac{ \cos x}{x^{2} + a^{2}} \, d x = \frac{\pi}{2a} e^{-a}. \]

Example 2. Shew that, if \(a \geq 0\), \(b \geq 0\), then\[\int_{0}^{\infty}\frac{ \cos 2ax - \cos 2bx}{x^{2}}\, d x=\pi (b-a)\] (Take a contour consisting of a large semicircle of radius \(\rho\), a small semicircle of radius \(\delta\), both having their centres at the origin, and the parts of the real axis joining their ends; then make \(\rho \rightarrow \infty\), \(\delta \rightarrow 0\).)

Example 3. Shew that, if \(b > 0\), \(m \geq 0\), then \[ \int_{0}^{\infty} \frac{3x^{2} - a^{2}}{ (x^{2} + b^{2})^{2} } \cos mx \, d x = \frac{\pi e^{-mb}}{4b^{3}} \left\{ 3b^{2} - a^{2} - mb(3b^{2}+a^{2}) \right\}. \]

Example 4. Shew that, if \(k > 0\), \(a > 0\), then \[ \int_{0}^{\infty} \frac{ x \sin ax }{x^{2} + k^{2}} \, d x = \frac{1}{2} \pi e^{-ka}. \]

Example 5. Shew that, if \(m \geq 0\), \(a > 0\), then \[ \int_{0}^{\infty} \frac{ \sin mx }{ x (x^{2} + a^{2})^{2} } \, d x = \frac{\pi}{2a^{4}} - \frac{\pi e^{-ma}}{4a^{3}} \left( m + \frac{2}{a} \right). \] (Take the contour of example 2.)

Example 6. Shew that, if the real part of \(z\) be positive, \[ \int_{0}^{\infty}\! (e^{-t} - e^{-tz}) \frac{\, d t}{t} = \log z. \] [We have \[\begin{align*} \int_{0}^{\infty}\! (e^{-t} - e^{-tz}) \frac{ d t}{t} &= \phantom{.}\lim_{\delta\rightarrow 0, \,\rho\rightarrow\infty} \left\{ \int_{\delta}^{\rho} \!\frac{e^{-t}}{t} \, d t - \int_\delta^\rho \! \frac{e^{-tz}}{t} \, d t \right\} \\ \\ &= \phantom{.}\lim_{\delta\rightarrow 0,\, \rho\rightarrow\infty} \left\{ \int_{\delta}^{\rho}\! \frac{e^{-t}}{t} \, d t - \int_{\delta z}^{\rho z} \!\frac{e^{-u}}{u} \, d u \right\} \\ \\ &= \phantom{.}\lim_{\delta\rightarrow 0, \,\rho\rightarrow\infty} \left\{ \int_{\delta}^{\delta z}\! \frac{e^{-t}}{t} \, d t - \int_{\rho}^{\rho z}\! \frac{e^{-t}}{t} \, d t, \right\}, \end{align*} \] since \(t^{-1} e^{-t}\) is analytic inside the quadrilateral whose corners are \(\delta\), \(\delta z\), \(\rho z\), \(\rho\).

Now \(\int_{\rho}^{\rho z} t^{-1} e^{-t} \, d t \rightarrow 0\) when \(\mathfrak{Re} (z) > 0\); and \[ \int_{\delta}^{\delta z}\! t^{-1} e^{-t} \, d t = \log z - \int_{\delta}^{\delta z}\! t^{-1} (1 - e^{-t}) \, d t \rightarrow \log z, \] since \(t^{-1} (1 - e^{-t}) \rightarrow 1\) as \(t \rightarrow 0.\)]