Residue Theorem

The Residue Theorem says that a contour integral of an analytic function f over a closed curve \gamma (loop) is equal to the sum of residues \res_{z_k}f(z) of the function at all singularities z_k inside the loop:

\int_\gamma f(z) \d z = 2\pi i \sum_{z_k} \res_{z=z_k} f(z)

Residue \res_{z_0}f(z) is defined as the contour integral around z_0 divided by 2\pi i:

\res_{z=z_0} f(z) = {1\over 2\pi i}\int_{|z - z_0| = \epsilon} f(z) \d z

and it is equal to the coefficient of 1\over z-z_0 in the Laurent series of f(z) around the point z_0, as can be easily calculated:

\res_{z=z_0} f(z) = {1\over 2\pi i}\int\limits_{|z - z_0| = \epsilon} f(z) \d z
= {1\over 2\pi i}\int\limits_{|z - z_0| = \epsilon} \sum_{n=-\infty}^\infty c_n
(z-z_0)^n \d z

= \sum_{n=-\infty}^\infty c_n {1\over 2\pi i}\int\limits_{|z - z_0| = \epsilon}
(z-z_0)^n \d z
= \sum_{n=-\infty}^\infty c_n \delta_{n, -1} = c_{-1}

where we used the result of the following integral (we integrate over the curve z = z_0 + \epsilon e^{i\varphi}, 0\le\varphi<2\pi, so \d z = i\epsilon

{1\over 2\pi i}\int\limits_{|z - z_0| = \epsilon} (z-z_0)^n \d z
{1\over 2\pi i}\int\limits_0^{2\pi} (z_0+\epsilon e^{i\varphi}-z_0)^n
i\epsilon e^{i\varphi}\d\varphi
{\epsilon^{n+1}\over 2\pi}\int\limits_0^{2\pi} e^{i\varphi (n + 1)}

\epsilon^{n+1}\over 2\pi} \left[ {e^{i\varphi (n + 1)}\over i(n+1)}
\right]_0^{2\pi}=0\quad\text{for $n\neq-1$}\cr
{1\over 2\pi}\int\limits_0^{2\pi} \d\varphi=1\quad\text{for $n=-1$}\cr
=\delta_{n, -1}

Computation of Residues

One has to calculate the c_{-1} coefficient in the Laurent series. One way to do that is to write f(z) as:

f(z) = {H(z)\over (z-z_0)^m}

where H(z) is analytic in the vicinity of z_0, e.g. f(z) has a pole of order m at z_0. Then:

\res_{z=z_0} f(z) = c_{-1} = {1\over(m-1)!}
\left.{\d^m H(z)\over\d z^m}\right|_{z = z_0}

in particular for m=1:

\res_{z=z_0} f(z) = H(z_0) = \lim_{z\to z_0}(z-z_0) f(z)

for m=2:

\res_{z=z_0} f(z) = H'(z_0) = \lim_{z\to z_0}{\d\over\d z}[(z-z_0)^2 f(z)]

f has a pole of order 1 at z_0, g is analytic at z_0:

\res_{z=z_0} f(z)g(z) = \lim_{z\to z_0}(z-z_0) f(z)g(z)
= g(z_0)\lim_{z\to z_0}(z-z_0) f(z) = g(z_0)\res_{z=z_0}f(z)

f(z_0) = 0, but f'(z_0) \neq 0 and g is analytic at z_0:

\res_{z=z_0} {g(z)\over f(z)} = g(z_0)\lim_{z\to z_0}{z-z_0\over f(z)}
= g(z_0)\lim_{z\to z_0}{z-z_0\over f(z)-f(z_0)}
= {g(z_0)\over f'(z_0)}

Useful Formulas

Jordan’s Lemma

For estimating integrals over semicircles \Omega (z = Re^{i\varphi}, 0\le\varphi\le\pi), we can use the following estimates:

\left|\int_\Omega g(z) \d z \right| \le \pi R \max_\Omega |g(z)|

\left|\int_\Omega e^{i\alpha z}g(z) \d z \right| \le {\pi\over\alpha}
\max_\Omega |g(z)|\quad\text{for $\alpha>0$}

(In the first case the integration path can be extended to the full circle if needed (0\le\varphi\le2\pi), in the second case the semicircle is the maximum path. Also if \alpha<0, we need to integrate over the lower semicircle.) These formulas can be used to make sure the integral over the semicircle goes to zero as R\to\infty. Intuitively speaking, in the first case g(z) must vanish faster than 1\over R (e.g. 1\over R^2 is ok), in the second case it’s enough if g(z) just goes to 0 (no matter how fast).

The estimates can be proved easily:

\left|\int_\Omega g(z) \d z \right|
= \left|\int_0^\pi g(Re^{i\varphi})iRe^{i\varphi} \d\varphi \right|
\le \int_0^\pi \left|g(Re^{i\varphi})\right|R \d\varphi
\le R\max_\Omega |g(z)| \int_0^\pi \d\varphi
= \pi R \max_\Omega |g(z)|


\left|\int_\Omega e^{i\alpha z}g(z) \d z \right|
=\left|\int_0^\pi e^{i\alpha

\le\int_0^\pi e^{-\alpha R\sin\varphi}\left|g(Re^{i\varphi})\right| R \d\varphi
\le R \max_\Omega |g(z)| \int_0^\pi e^{-\alpha R\sin\varphi}\d\varphi

< {\pi\over\alpha} \max_\Omega |g(z)|

where we use the following useful estimate for the integral (valid for \alpha>0):

\int_0^\pi e^{-\alpha R\sin\varphi}\d\varphi
< 2\int_0^{\pi\over2} e^{-\alpha R{2\over\pi}\varphi}\d\varphi
= 2 \left[e^{-\alpha R{2\over\pi}\varphi}\over
    -\alpha R {2\over\pi}\right]_0^{\pi/2}

= {2\over-\alpha R {2\over\pi}}
    \left[e^{-\alpha R} - 1\right]
= {\pi\over\alpha R} (1-e^{-\alpha R}) < {\pi\over\alpha R}


Sometimes it is useful to integrate over the arc z = z_0 + \epsilon
e^{i\varphi}, \varphi_0 \le \varphi \le \varphi_0 + \alpha, and let \epsilon\to0 at the end. If the function is analytic, the result is 0. If the function has a pole of order n > 1, the result is infinity, unless it’s a full circle (in which case the result is 0). The remaining case is if the function has a pole of order one, e.g. it can be written (H(z) is analytic at z_0):

f(z) = {H(z)\over z - z_0}


\int_\Omega f(z) \d z =
\int_\Omega {H(z)\over z - z_0} \d z = \int_{\varphi_0}^{\varphi_0+\alpha}{H(z_0 +
\epsilon e^{i\varphi})\over z_0 + \epsilon e^{i\varphi} - z_0}\epsilon i
e^{i\varphi} \d \varphi

=\int_{\varphi_0}^{\varphi_0+\alpha}H(z_0 + \epsilon e^{i\varphi}) i \d \varphi
\to\int_{\varphi_0}^{\varphi_0+\alpha}H(z_0) i \d \varphi = i\alpha H(z_0)
= i\alpha \res_{z = z_0}f(z)

Complex Substitution

When substituting in integrals, as long as we just substitute for real functions, we use the regular substitution theorem, e.g. x = y+1 (f(x) can be a complex function):

\int_{-\infty}^\infty f(x) \d x = \int_{-\infty}^\infty f(y+1) \d y

if, on the other hand, we substitute for complex functions, e.g. x = iy:

\int_{-\infty}^\infty f(x) \d x = \int_{i\infty}^{-i\infty} f(iy) i\d y
\to \int_{\infty}^{-\infty} f(iy) i\d y

then the first two integrals in the left hand side are equal, however the integral on the right hand side is over a different integration path and we need to use the Residue Theorem to relate those integrals, e.g. in general the two integrals on the LHS and the integral on the RHS are not equal. However the idea is that the integral after the substitution (and changing the limits, e.g. the integration path) is easier to evaluate, so the substitution guides us which integration path to choose for the Residue Theorem.

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