Methods of contour integration

Not to be confused with Line integral.

In the mathematical field of complex analysis, contour integration is a method of evaluating certain integrals along paths in the complex plane.^[1]^[2]^[3]

Contour integration is closely related to the calculus of residues,^[4] a methodology of complex analysis.

One use for contour integrals is the evaluation of integrals along the real line that are not readily found by using only real variable methods.^[5]

Contour integration methods include

direct integration of a complex-valued function along a curve in the complex plane (a contour)
application of the Cauchy integral formula
application of the residue theorem.

One method can be used, or a combination of these methods, or various limiting processes, for the purpose of finding these integrals or sums.

1 Direct methods
- 1.1 Example
2 Applications of integral theorems
3 See also
4 References and notes
5 Further reading
6 External links

Direct methods

Direct methods involve the calculation of the integral by means of methods similar to those in calculating line integrals in several-variable calculus. This means that we use the following method:

parametrizing the contour

The contour is parametrized by a differentiable complex-valued function of real variables, or the contour is broken up into pieces and parametrized separately

substitution of the parametrization into the integrand

Substituting the parametrization into the integrand transforms the integral into an integral of one real variable.

direct evaluation

The integral is evaluated in a method akin to a real-variable integral.

Example

A fundamental result in complex analysis is that the contour integral of z⁻¹ is 2πi, where the path of the contour is taken to be the unit circle traversed counterclockwise (or any Jordan curve about 0). In the case of the unit circle there is a direct method to evaluate the integral

$\oint_C {1 \over z}\,dz.$

In evaluating this integral, use the unit circle |z| = 1 as contour, parametrized by z(t) = e^it, with t ∈ [0, 2π], then dz / dt = i e^it and

$\begin{align} \oint_C {1 \over z}\,dz & {} = \int_0^{2\pi} {1 \over e^{it}} \, ie^{it}\,dt = i\int_0^{2\pi} 1 \,dt \\ & {} = \Big[t\Big]_0^{2\pi} i=(2\pi-0)i = 2\pi i, \end{align}$

which is the value of the integral.

Applications of integral theorems

Applications of integral theorems are also often used to evaluate the contour integral along a contour, which means that the real-valued integral is calculated simultaneously along with calculating the contour integral.

Integral theorems such as the Cauchy integral formula or residue theorem are generally used in the following method:

a specific contour is chosen:

The contour is chosen so that the contour follows the part of the complex plane that describes the real-valued integral, and also encloses singularities of the integrand so application of the Cauchy integral formula or residue theorem is possible

application of the Cauchy–Goursat theorem

The integral is reduced to only an integration around a small circle about each pole.

application of the Cauchy integral formula or residue theorem

Application of these integral formula gives us a value for the integral around the whole of the contour.

division of the contour into a contour along the real part and imaginary part

The whole of the contour can be divided into the contour that follows the part of the complex plane that describes the real-valued integral as chosen before (call it R), and the integral that crosses the complex plane (call it I). The integral over the whole of the contour is the sum of the integral over each of these contours.

demonstration that the integral that crosses the complex plane plays no part in the sum

If the integral I can be shown to be zero, or if the real-valued integral that is sought is improper, then if we demonstrate that the integral I as described above tends to 0, the integral along R will tend to the integral around the contour R + I.

conclusion

If we can show the above step, then we can directly calculate R, the real-valued integral.

Example (I)

Consider

$\int_{-\infty}^\infty {1 \over (x^2%2B1)^2} dx,$

To evaluate this integral, we look at the complex-valued function

$f(z)={1 \over (z^2%2B1)^2}$

which has singularities at i and −i. However, we will want to choose a contour that will enclose the real-valued integral, so we choose a semicircle like the one shown on the left, which we will let expand as to contain the whole real axis (a will tend to infinity). Call this contour C.

Now, there are two ways of proceeding, using the Cauchy integral formula or by the method of residues.

Using the Cauchy integral formula

Note that:

$\oint_C f(z)\,dz = \int_{-a}^a f(z)\,dz %2B \int_\text{Arc} f(z)\,dz$

thus

$\int_{-a}^a f(z)\,dz = \oint_C f(z)\,dz - \int_\text{Arc} f(z)\,dz$

Furthermore observe that

$f(z)={1 \over (z^2%2B1)^2}={1 \over (z%2Bi)^2(z-i)^2}.$

Since the only singularity in the contour is the one at i, then we can write

$f(z)={{1 \over (z%2Bi)^2} \over (z-i)^2},$

which puts the function in the form for direct application of the formula. Then, by using Cauchy's integral formula,

$\begin{align} \oint_C f(z)\,dz & = \oint_C {1 \over (z^2%2B1)^2}\,dz = \oint_C {{1 \over (z%2Bi)^2} \over (z-i)^2}\,dz = 2\pi i \frac{d}{dz} \left(\left.{1 \over (z%2Bi)^2}\right)\right|_{z=i} \\ & =2 \pi i \left.\left({-2 \over (z%2Bi)^3}\right)\right|_{z = i} =2 \pi i (1/4i)={\pi\over 2} \end{align}$

(We take the first derivative, in the above steps, because the pole is a second-order pole. That is, (z − i ) is taken to the second power, so we employ the first derivative of ƒ(z). If it were (z − i ) taken to the third power, we would use the second derivative and divide by 2!, etc. The case of (z − i ) to the first power corresponds to a zero order derivative—just ƒ(x) itself.)

If we call the arc of the semicircle Arc, we need to show that the integral over Arc tends to zero as a tends to infinity — using the estimation lemma

$\left|\int_\text{Arc} f(z)\,dz\right| \le ML$

where M is an upper bound on |ƒ(z)| along the Arc and L the length of Arc. Now,

$\int_\text{Arc} f(z)\,dz \le {a\pi \over (a^2-1)^2} \rightarrow 0\ \mathrm{as}\ a \rightarrow \infty.$

$\int_{-\infty}^\infty {1 \over (x^2%2B1)^2}\,dx = \int_{-\infty}^\infty f(z)\,dz = \lim_{a \rightarrow %2B\infty} \int_{-a}^a f(z)\,dz = {\pi\over 2}.\quad\square$

Using the method of residues

Consider the Laurent series of f(z) about i, the only singularity we need to consider. We then have

$f(z) = {-1 \over 4(z-i)^2} %2B {-i \over 4(z-i)} %2B {3 \over 16} %2B {i \over 8}(z-i) %2B {-5 \over 64}(z-i)^2 %2B \cdots$

(See Sample Laurent Calculation from Laurent series for the derivation of this series.)

It is clear by inspection that the residue is −i/4 (to see this, imagine that the above equation were multiplied by z − i, then both sides integrated via the Cauchy integral formula—only the second term would integrate to a non-zero quantity), so, by the residue theorem, we have

$\oint_C f(z)\,dz = \oint_C {1 \over (z^2%2B1)^2}\,dz = 2 \pi i \,\mathrm{Res}_{z=i} f = 2 \pi i (-i/4)={\pi\over 2}\quad\square$

Thus we get the same result as before.

Contour note

As an aside, a question can arise whether we do not take the semicircle to include the other singularity, enclosing −i. To have the integral along the real axis moving in the correct direction, the contour must travel clockwise, i.e., in a negative direction, reversing the sign of the integral overall.

This does not affect the use of the method of residues by series.

Example (II) – Cauchy distribution

The integral

$\int_{-\infty}^\infty {e^{itx} \over x^2%2B1}\,dx$

(which arises in probability theory as (a scalar multiple of) the characteristic function of the Cauchy distribution) resists the techniques of elementary calculus. We will evaluate it by expressing it as a limit of contour integrals along the contour C that goes along the real line from −a to a and then counterclockwise along a semicircle centered at 0 from a to −a. Take a to be greater than 1, so that the imaginary unit i is enclosed within the curve. The contour integral is

$\int_C {e^{itz} \over z^2%2B1}\,dz.$

Since e^itz is an entire function (having no singularities at any point in the complex plane), this function has singularities only where the denominator z² + 1 is zero. Since z² + 1 = (z + i)(z − i), that happens only where z = i or z = −i. Only one of those points is in the region bounded by this contour. The residue of f(z) at z = i is

$\lim_{z\to i}(z-i)f(z)=\lim_{z\to i}(z-i){e^{itz} \over z^2%2B1}=\lim_{z\to i}(z-i){e^{itz} \over (z-i)(z%2Bi)}$

$=\lim_{z\to i}{e^{itz} \over z%2Bi}={e^{iti} \over i%2Bi}={e^{-t}\over 2i}.$

According to the residue theorem, then, we have

$\int_C f(z)\,dz=2\pi i\cdot\operatorname{Res}_{z=i}f(z)=2\pi i{e^{-t} \over 2i}=\pi e^{-t}.$

The contour C may be split into a "straight" part and a curved arc, so that

$\int_{\mbox{straight}}%2B\int_{\mbox{arc}}=\pi e^{-t},$

and thus

$\int_{-a}^a =\pi e^{-t}-\int_{\mbox{arc}}.$

It can be shown that if t > 0 then

$\int_{\mbox{arc}}{e^{itz} \over z^2%2B1}\,dz \rightarrow 0\ \mbox{as}\ a\rightarrow\infty.$

Therefore if t > 0 then

$\int_{-\infty}^\infty{e^{itx} \over x^2%2B1}\,dx=\pi e^{-t}.$

A similar argument with an arc that winds around −i rather than i shows that if t < 0 then

$\int_{-\infty}^\infty{e^{itx} \over x^2%2B1}\,dx=\pi e^t,$

and finally we have this:

$\int_{-\infty}^\infty{e^{itx} \over x^2%2B1}\,dx=\pi e^{-\left|t\right|}.\quad\square$

(If t = 0 then the integral yields immediately to real-valued calculus methods and its value is π.)

Example (III) – trigonometric integrals

Certain substitutions can be made to integrals involving trigonometric functions, so the integral is transformed into a rational function of a complex variable and then the above methods can be used in order to evaluate the integral.

As an example, consider

$\int_{-\pi}^\pi {1 \over 1 %2B 3 (\cos{t})^2} \,dt.$

We seek to make a substitution of z = e^it.

Now, recall

$\cos{t} = {1 \over 2} \left(e^{it}%2Be^{-it}\right) = {1 \over 2} \left(z%2B{1 \over z}\right)$

and

${dz \over dt} = iz,\ dt = {dz \over iz}.$

Taking C to be the unit circle, we substitute to get:

$\begin{align} & {} \quad \oint_C {1 \over 1 %2B 3 ({1 \over 2} (z%2B{1 \over z}))^2} \,{dz\over iz} \\[6pt] & = \oint_C {1 \over 1 %2B {3 \over 4} (z%2B{1 \over z})^2}{1 \over iz} \,dz = \oint_C {-i \over z%2B{3\over 4}z(z%2B{1\over z})^2}\,dz \\[6pt] & = -i \oint_C { 1 \over z%2B{3\over 4}z(z^2%2B2%2B{1\over z^2})} \,dz \\[6pt] & = -i \oint_C {1\over z%2B{3\over 4}(z^3%2B2z%2B{1 \over z})} \,dz = -i \oint_C {1 \over {3\over 4 }z^3%2B{5 \over 2}z%2B{3 \over 4z}} \,dz \\[6pt] & = -i \oint_C {4 \over 3z^3%2B10z%2B{3\over z}}\,dz = -4i \oint_C {1 \over 3z^3%2B10z%2B{3\over z}}\,dz \\[6pt] & = -4i \oint_C { z \over 3z^4%2B10z^2%2B3 } \,dz. \end{align}$

We use the Cauchy integral formula. Factorize the denominator:

$\begin{align} & = -4i \oint_C { z \over 3z^4%2B10z^2%2B3 } \,dz = -4i \oint_C {z \over 3(z^2%2B3)(z^2%2B1/3)}\,dz \\[6pt] & = -4i \oint_C {z \over 3(z%2B\sqrt{3}i)\left(z-\sqrt{3}i\right)\left(z%2B\sqrt{1\over 3}i\right)\left(z-\sqrt{1\over 3}i\right)}\,dz \\[6pt] & = -{4\over 3}i \oint_C {z \over (z%2B\sqrt{3}i)(z-\sqrt{3}i)\left(z%2B\sqrt{1\over 3}i\right)\left(z-\sqrt{1\over 3}i\right)}\,dz. \end{align}$

The singularities then to be considered are at 3^−1/2i, −3^−1/2i. We can now reduce the integral:

$= -{4\over 3}i \oint_{C_1} {\left( {z \over (z%2B\sqrt{3}i)(z-\sqrt{3}i)\left(z%2B\sqrt{1\over 3}i\right)} \right) \over z-\sqrt{1\over 3}i }\,dz %2B -{4\over 3}i \oint_{C_2} {\left( {z \over (z%2B\sqrt{3}i)(z-\sqrt{3}i)\left(z-\sqrt{1\over 3}i\right)}\right) \over z%2B\sqrt{1\over 3}i }$

where C₁ is a small circle about 3^−1/2i, and C₂ is a small circle about −3^−1/2i. We can now apply the formula:

$= -{4\over 3}i \left[ 2\pi i \left.\left({z \over (z%2B\sqrt{3}i)(z-\sqrt{3}i)(z%2B\sqrt{1\over 3}i)}\right)\right|_{z=\sqrt{1 \over 3}i} \right.$

$\left. {} %2B 2\pi i \left.\left({z \over (z%2B\sqrt{3}i)(z-\sqrt{3}i)(z-\sqrt{1\over 3}i)}\right)\right|_{z=-\sqrt{1 \over 3}i} \right]$

$= -{4\over 3}i \left[ 2\pi i \left( { \sqrt{1\over 3}i \over (\sqrt{1\over 3}i%2B\sqrt{3}i)(\sqrt{1\over 3}i-\sqrt{3}i)(\sqrt{1\over 3}i%2B\sqrt{1\over 3}i)} \right) \right.$

$\left. {} %2B2\pi i \left( { -\sqrt{1\over 3}i \over (-\sqrt{1\over 3}i%2B\sqrt{3}i)(-\sqrt{1\over 3}i-\sqrt{3}i)(-\sqrt{1\over 3}i-\sqrt{1\over 3}i } \right)\right]$

$= -{4\over 3}i \left[ 2\pi i \left( {\sqrt{1\over 3}i \over ({4 \over \sqrt{3}}i)(-{2 \over \sqrt{3}}i)({2 \over \sqrt{3}}i)} \right) %2B 2 \pi i \left( {-\sqrt{1\over 3}i \over ({2 \over \sqrt{3}}i)(-{4 \over \sqrt{3}}i)(-{2 \over \sqrt{3}}i)} \right) \right]$

$= -{4\over 3}i \left[ 2\pi i \left( {\sqrt{1\over 3}i \over i({4 \over \sqrt{3}})({2 \over \sqrt{3}})({2 \over \sqrt{3}})} \right) %2B 2\pi i \left( {-\sqrt{1\over 3}i \over -i({2 \over \sqrt{3}})({4 \over \sqrt{3}})({2 \over \sqrt{3}})} \right) \right]$

$= -{4\over 3}i \left[ 2\pi i \left( {\sqrt{1\over 3} \over ({4 \over \sqrt{3}})({2 \over \sqrt{3}})({2 \over \sqrt{3}})} \right) %2B 2\pi i \left( {\sqrt{1\over 3} \over ({2 \over \sqrt{3}})({4 \over \sqrt{3}})({2 \over \sqrt{3}})} \right) \right]$

$= -{4\over 3}i \left[ 2\pi i \left( { \,\sqrt{1\over 3} \,\over {16 \over 3\sqrt{3}} } \right) %2B 2\pi i \left( {\, \sqrt{1\over 3} \,\over {16 \over 3\sqrt{3}} } \right) \right]$

$= -{4\over 3}i \left[2 \pi i \left({3\over 16}\right) %2B 2 \pi i \left({3\over 16}\right)\right] = \pi.\quad\square$

Example (IIIa) trigonometric integrals, the general procedure

The above method may be applied to all integrals of the type

$\int_0^{2\pi} \frac{P(\sin(t),\sin(2t),\ldots,\cos(t),\cos(2t),\ldots)} {Q(\sin(t),\sin(2t),\ldots,\cos(t),\cos(2t),\ldots)}\, dt$

where P and Q are polynomials, i.e. a rational function in trigonometric terms is being integrated.

The trick is to use the substitution $z = \exp(i t)$ where $dz = i \exp(i t) \,dt$ and hence

$\frac{1}{iz} \,dz = dt.$

This substitution maps the interval $[0, 2\pi]$ to the unit circle. Furthermore,

$\sin(k t) = \frac{\exp(i k t) - \exp(- i k t)}{2 i} = \frac{z^k - z^{-k}}{2i}$

and

$\cos(k t) = \frac{\exp(i k t) %2B \exp(- i k t)}{2} = \frac{z^k %2B z^{-k}}{2}$

so that a rational function ƒ(z) in z results from the substitution, and the integral becomes

$\oint_{|z|=1} f(z) \frac{1}{iz}\, dz$

which is in turn computed by summing the residues of $f(z)\frac{1}{iz}$ inside the unit circle.

The image at right illustrates this for

$I = \int_0^\frac{\pi}{2} \frac{1}{1 %2B \sin(t)^2}\, dt,$

which we now compute. The first step is to recognize that

$I = \frac{1}{4} \int_0^{2\pi} \frac{1}{1 %2B \sin(t)^2} \,dt.$

The substitution yields

$\frac{1}{4} \oint_{|z|=1} \frac{4 i z}{z^4 - 6z^2 %2B 1}\, dz = \oint_{|z|=1} \frac{i z}{z^4 - 6z^2 %2B 1}\, dz .$

The poles of this function are at 1 ± √2 and −1 ± √2. Of these, 1 + √2 and −1 −√2 are outside the unit circle (shown in red, not to scale), whereas 1 − √2 and −1 + √2 are inside the unit circle (shown in blue). The corresponding residues are both equal to −i√2/16, so that the value of the integral is

$I = 2 \pi i \; 2 \left( - \frac{\sqrt{2}}{16} i \right) = \pi \frac{\sqrt{2}}{4}.$

Example (IV) – branch cuts

Consider

$\int_0^\infty {\sqrt{x} \over x^2%2B6x%2B8}\,dx.$

We can begin by formulating the complex integral

$\int_C {\sqrt{z} \over z^2%2B6z%2B8}\,dz=I.$

We can use the Cauchy integral formula or residue theorem again to obtain the relevant residues. However, the important thing to note is that z^1/2 = e^1/2·Log(z), so z^1/2 has a branch cut. This affects our choice of the contour C. Normally the logarithm branch cut is defined as the negative real axis, however, this makes the calculation of the integral slightly more complex, so we define it to be the positive real axis.

Then, we use the so-called keyhole contour, which consists of a small circle about the origin of radius ε say, extending to a line segment parallel and close to the positive real axis but not touching it, to an almost full circle, returning to a line segment parallel, close, and below the positive real axis in the negative sense, returning to the small circle in the middle.

Note that z = −2 and z = −4 are inside the big circle. These are the two remaining poles, derivable by factoring the denominator of the integrand. The branch point at z = 0 was avoided by detouring around the origin.

Let γ be the small circle of radius ε, Γ the larger, with radius r, then

$\begin{align} \int_C {\sqrt{z} \over z^2%2B6z%2B8}\,dz & = \int_\varepsilon^R {\sqrt{z} \over z^2%2B6z%2B8}\,dz \\ & %2B \int_\Gamma {\sqrt{z} \over z^2%2B6z%2B8}\,dz \\ & %2B \int_R^\varepsilon {\sqrt{z} \over z^2%2B6z%2B8}\,dz \\ & %2B \int_\gamma {\sqrt{z} \over z^2%2B6z%2B8}\,dz. \end{align}$

Since z^1/2 = e^1/2 Log(z), on the contour outside the branch cut, we have gained 2π in argument along γ (by Euler's Identity, $e^{i \pi} \,$ represents the unit vector, which therefore has ${i \pi} \,$ as its log. This ${ \pi} \,\!$ is what is meant by the argument of z. The coefficient of 1/2 forces us to use 2 times $\pi \,$ ), so

$\begin{align} & {} \quad \int_R^\varepsilon {\sqrt{z} \over z^2%2B6z%2B8}\,dz=\int_R^\varepsilon {e^{{1\over 2} \mathrm{Log}(z)} \over z^2%2B6z%2B8}\,dz=\int_R^\varepsilon {e^{{1\over 2}(\log{|z|}%2Bi \arg{z})} \over z^2%2B6z%2B8}\,dz \\ & = \int_R^\varepsilon { e^{{1\over 2}\log{|z|}}e^{1/2(2\pi i)} \over z^2%2B6z%2B8}\,dz=\int_R^\varepsilon { e^{{1\over 2}\log{|z|}}e^{\pi i} \over z^2%2B6z%2B8}\,dz \\ & = \int_R^\varepsilon {-\sqrt{z} \over z^2%2B6z%2B8}\,dz=-\int_\varepsilon^R {-\sqrt{z} \over z^2%2B6z%2B8}\,dz \end{align}$

simplifying,

$=\int_\varepsilon^R {\sqrt{z} \over z^2%2B6z%2B8}\,dz,$

and then

$\begin{align} \int_C {\sqrt{z} \over z^2%2B6z%2B8}\,dz & = \int_\varepsilon^R {\sqrt{z} \over z^2%2B6z%2B8}\,dz \\ & %2B \int_\Gamma {\sqrt{z} \over z^2%2B6z%2B8}\,dz \\ & %2B \int_\varepsilon^R {\sqrt{z} \over z^2%2B6z%2B8}\,dz \\ & %2B \int_\gamma {\sqrt{z} \over z^2%2B6z%2B8}\,dz. \end{align}$

It can be shown that the integrals over Γ and γ both tend to zero as ε tends to zero and R tends to infinity, by an estimation argument above. Thus, then,

$\int_C {\sqrt{z} \over z^2%2B6z%2B8}\,dz=2\int_0^\infty {\sqrt{x} \over x^2%2B6x%2B8}\,dx.$

By using the residue theorem or the Cauchy integral formula (first employing the partial fractions method to derive a sum of two simple contour integrals) one obtains

$\pi i \left({i\over \sqrt{2}}-i\right)=\int_0^\infty {\sqrt{x} \over x^2%2B6x%2B8}\,dx = \pi\left(1-{1\over\sqrt{2}}\right).\quad\square$

Example (V) – the square of the logarithm

This section treats a type of integral of which

$\int_0^\infty \frac{\log x}{(1%2Bx^2)^2} \, dx$

is an example.

To calculate this integral, one uses the function

$f(z) = \frac{(\log z)^2}{(1%2Bz^2)^2}$

and the branch of the logarithm corresponding to $-\pi < \arg(z) \le \pi$ .

We will calculate the integral of ƒ(z) along the keyhole contour shown at right. As it turns out this integral is a multiple of the initial integral that we wish to calculate.

Let R be the radius of the large circle, and r the radius of the small one. We will denote the upper line by M, and the lower line by N. As before we take the limit when R goes to infinity and r to zero. The two contributions from the two circles vanish. For example, one has the following upper bound with the ML-lemma:

$\left| \int_R f(z) \, dz \right| \le 2 \pi R \frac{(\log R)^2 %2B \pi^2}{(R^2-1)^2} \rightarrow 0.$

By the Cauchy residue theorem we thus have

$\left( \int_R %2B \int_M %2B \int_N %2B \int_r \right) f(z) \, dz = 2 \pi i \left( \mathrm{Res}_{z=i} f(z) %2B \mathrm{Res}_{z=-i} f(z) \right)$

$2 \pi i \left( - \frac{\pi}{4} %2B \frac{1}{16} i \pi^2 - \frac{\pi}{4} - \frac{1}{16} i \pi^2 \right) = - i \pi^2.$

It remains to calculate the contributions of M and N. We set $z =-x %2B i\epsilon$ on M and $z = -x - i\epsilon$ on N, with $0 < x < \infty.$ This gives

$\left( \int_M %2B \int_N \right) f(z) \, dz = - \int_\infty^0 \frac{(\log(-x %2B i\epsilon))^2}{(1%2B(-x %2B i\epsilon)^2)^2} \, dx - \int_0^\infty \frac{(\log(-x - i\epsilon))^2}{(1%2B(-x - i\epsilon)^2)^2} \, dx$

$\int_0^\infty \frac{(\log x %2B i\pi)^2}{(1%2Bx^2)^2} \, dx - \int_0^\infty \frac{(\log x - i\pi)^2}{(1%2Bx^2)^2} \, dx.$

To conclude, one observes that

$(\log x %2B i\pi)^2 - (\log x - i\pi)^2 = 4 \pi i \log x\,$

and thus

$4 \pi i \int_0^\infty \frac{\log x}{(1%2Bx^2)^2} \, dx = -i \pi^2$

which gives

$\int_0^\infty \frac{\log x}{(1%2Bx^2)^2} \, dx = - \frac{\pi}{4}.$

Example (VI) – logarithms and the residue at infinity

We seek to evaluate

$I = \int_0^3 {\left(x^3 (3-x)\right)^{1/4} \over 5-x}\,dx.$

This requires a close study of

$f(z) = \left(z^3 (3-z)\right)^{1/4}.\,$

We will construct f(z) so that it has a branch cut on [0, 3], shown in red in the diagram. To do this, we choose two branches of the logarithm, setting

$(z^3)^{1/4} = z^{3/4} = \exp((3/4) \log(z)) \quad \mbox{where} \quad -\pi \le \arg(\log(z)) < \pi$

and

$(3-z)^{1/4} = \exp((1/4) \log(3-z)) \quad \mbox{where} \quad 0 \le \arg(\log(3-z)) < 2\pi.$

The cut of z^3/4 is therefore $(-\infty, 0]$ and the cut of $(3-z)^{1/4}$ is $(-\infty, 3]$ . It is easy to see that the cut of the product of the two, i.e. $f(z)$ , is $[0, 3]$ , because f(z) is actually continuous across $(-\infty, 0).$ This is because when $z = -r < 0$ and we approach the cut from above, f(z) has the value

$r^{3/4} \exp(3 \pi i / 4) (3%2Br)^{1/4} \exp(2 \pi i / 4) = r^{3/4} (3%2Br)^{1/4} \exp(5 \pi i / 4).\,$

When we approach from below, f(z) has the value

$r^{3/4} \exp(-3 \pi i / 4) (3%2Br)^{1/4} \exp(0 \pi i / 4) = r^{3/4} (3%2Br)^{1/4} \exp(-3 \pi i / 4).\,$

But $\exp(-3 \pi i / 4) = \exp(5 \pi i / 4),$ so that we have continuity across the cut. This is illustrated in the diagram, where the two black oriented circles are labelled with the corresponding value of the argument of the logarithm used in $z^{3/4}$ and $(3-z)^{1/4}.$

We will use the contour shown in green in the diagram. To do this we must compute the value of ƒ(z) along the line segments just above and just below the cut.

Let $z = r$ (in the limit, i.e. as the two green circles shrink to radius zero), where $0 \le r \le 3.$ Along the upper segment, we find that $f(z)$ has the value

$r^{3/4} \exp(0 \pi i / 4) (3-r)^{1/4} \exp(2 \pi i / 4) = i \, r^{3/4} (3-r)^{1/4}\,$

and along the lower segment,

$r^{3/4} \exp(3\, \cdot 0 \, \pi i / 4) (3-r)^{1/4} \exp(0 \pi i / 4) = r^{3/4} (3-r)^{1/4}.\,$

It follows that the integral of $f(z)/(5-z)\,$ along the upper segment is $- \, i \, I$ in the limit, and along the lower segment, $I.\,$

If we can show that the integrals along the two green circles vanish in the limit, then we also have the value of $I\,$ , by the Cauchy residue theorem. Let the radius of the green circles be $\rho$ , where $\rho < 1/1000$ and $\rho \rightarrow 0,$ and apply the ML-inequality. For the circle $C_L$ on the left, we find

$\left| \int_{C_L} \frac{f(z)}{5-z} dz \right| \le 2 \pi \rho \frac{\rho^{3/4} (3%2B1/1000)^{1/4}}{5-1/1000} \in \mathcal{O} \left( \rho^{7/4} \right) \rightarrow 0.$

Similarly, for the circle C_R on the right, we have

$\left| \int_{C_R} \frac{f(z)}{5-z} dz \right| \le 2 \pi \rho \frac{(3%2B1/1000)^{3/4} \rho^{1/4}}{2-1/1000} \in \mathcal{O} \left( \rho^{5/4} \right) \rightarrow 0.$

Now using the Cauchy residue theorem, we have

$(-i %2B 1) I = - \, 2\pi i \left( \mathrm{Res}_{z=5} \frac{f(z)}{5-z} %2B \mathrm{Res}_{z=\infty} \frac{f(z)}{5-z} \right).$

where the minus sign is due to the clockwise direction around the residues. Using the branch of the logarithm from before, clearly

$\mathrm{Res}_{z=5} \frac{f(z)}{5-z} = - 5^{3/4} \exp(\log(-2)/4).$

The pole is shown in blue in the diagram. The value simplifies to

$- 5^{3/4} \exp((\log(2) %2B \pi i) / 4 ) = - \exp( \pi i / 4) 5^{3/4} 2^{1/4}.\,$

We use the following formula for the residue at infinity:

$\mathrm{Res}_{z=\infty} h(z) = \mathrm{Res}_{z=0} \Big[- \frac{1}{z^2} h\left(\frac{1}{z}\right)\Big].$

Substituting, we find

$1/(5-1/z) = -z \left(1 %2B 5z %2B 5^2 z^2 %2B 5^3 z^3 %2B \cdots\right)$

and

$\left((1/z^3) (3-1/z)\right)^{1/4} = (1/z) (3z-1)^{1/4} = (1/z) \exp(\pi i / 4) (1-3z)^{1/4},$

where we have used the fact that $-1 = \exp(\pi i)$ for the second branch of the logarithm. Next we apply the binomial expansion, obtaining

$(1/z) \exp(\pi i / 4) \left( 1 - {1/4 \choose 1} 3z %2B {1/4 \choose 2} 3^2 z^2 - {1/4 \choose 3} 3^3 z^3 %2B \cdots \right).$

The conclusion is that

$\mathrm{Res}_{z=\infty} \frac{f(z)}{5-z} = \exp(\pi i / 4) (5 - 3/4) = \exp(\pi i / 4) 17/4.$

Finally, it follows that the value of $I\,$ is

$I = 2 \pi i \, \frac{\exp(\pi i / 4)}{-1%2Bi} \left( 17/4 - 5^{3/4} 2^{1/4} \right) = 2 \pi \, 2^{-1/2} \left( 17/4 - 5^{3/4} 2^{1/4} \right)$

which yields

$I = \frac{\pi}{2\sqrt{2}} \left(17 - 5^{3/4} 2^{9/4} \right) = \frac{\pi}{2\sqrt{2}} \left(17 - 40^{3/4} \right).$

References and notes

^ John Stalker (1998). Complex Analysis: Fundamentals of the Classical Theory of Functions. Springer. p. 77. ISBN 081764038X. http://books.google.com/books?id=yl3GIXd3dFIC&pg=PP12&dq=%22calculus+of+residues%22#PPA77,M1.
^ Joseph Bak & Donald J. Newman (1997). Complex Analysis. Springer. Chapters 11 & 12, pp. 130–156. ISBN 0387947566. http://books.google.com/books?id=JX2YSgfZwbYC&pg=PA130&dq=%22contour+integral%22#PPA130,M1.
^ Steven George Krantz (1999). Handbook of Complex Variables. Springer. Chapter 2. ISBN 0817640118. http://books.google.com/books?id=aYU2AdF_0dIC&pg=PT13&dq=Calculus++Residues+inauthor:krantz#PPT47,M1.
^ Dragoslav S. Mitrinovic & Jovan D. Keckic (1984). The Cauchy Method of Residues: Theory and Applications. Springer. Chapter 2. ISBN 9027716234. http://books.google.com/books?id=-suKhxfPH5AC&printsec=frontcover&dq=%22calculus+of+residues%22#PPA5,M1.
^ Dragoslav S. Mitrinovic & Jovan D. Keckic (1984). Chapter 5. ISBN 9027716234. http://books.google.com/books?id=-suKhxfPH5AC&printsec=frontcover&dq=%22calculus+of+residues%22#PPA108,M1.