Editing Contour integration (section)

==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 [[Cauchy's integral 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 formulae 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 {{mvar|R}}), and the integral that crosses the complex plane (call it {{mvar|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 {{mvar|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 {{mvar|I}} as described above tends to 0, the integral along {{mvar|R}} will tend to the integral around the contour {{math|''R'' + ''I''}}.
* conclusion
*: If we can show the above step, then we can directly calculate {{mvar|R}}, the real-valued integral.

===Example 1===
Consider the integral
<math display=block>\int_{-\infty}^\infty \frac{1}{\left(x^2+1\right)^2}\,dx,</math>

To evaluate this integral, we look at the complex-valued function
<math display=block>f(z)=\frac{1}{\left(z^2+1\right)^2}</math>

which has [[Mathematical singularity|singularities]] at {{mvar|i}} and {{math|−''i''}}. We choose a contour that will enclose the real-valued integral, here a semicircle with boundary diameter<!-- What is a "boundary diameter"? I am guessing you want a semicircle centered at 0 and having that diameter. Something should say that here. --> on the real line (going from, say, {{math|−''a''}} to {{mvar|a}}) will be convenient. Call this contour {{mvar|C}}.

There are two ways of proceeding, using the [[Cauchy integral formula]] or by the method of residues:

====Using the Cauchy integral formula ====
Note that:
<math display=block>\oint_C f(z)\,dz = \int_{-a}^a f(z)\,dz + \int_\text{Arc} f(z)\,dz </math>
thus
<math display=block>\int_{-a}^a f(z)\,dz = \oint_C f(z)\,dz - \int_\text{Arc} f(z)\,dz </math>

Furthermore, observe that
<math display=block>f(z)=\frac{1}{\left(z^2+1\right)^2}=\frac{1}{(z+i)^2(z-i)^2}.</math>

Since the only singularity in the contour is the one at&nbsp;{{mvar|i}}, then we can write
<math display=block>f(z)=\frac{\frac{1}{(z+i)^2}}{(z-i)^2},</math>

which puts the function in the form for direct application of the formula. Then, by using Cauchy's integral formula,
<math display=block>\oint_C f(z)\,dz = \oint_C \frac{\frac{1}{(z+i)^2}}{(z-i)^2}\,dz = 2\pi i \, \left.\frac{d}{dz} \frac{1}{(z+i)^2}\right|_{z=i} =2 \pi i \left[\frac{-2}{(z+i)^3}\right]_{z = i} =\frac{\pi}{2}</math>

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

We need to show that the integral over the arc of the semicircle tends to zero as {{math|''a'' → ∞}}, using the [[estimation lemma]]
<math display=block>\left|\int_\text{Arc} f(z)\,dz\right| \le ML</math>

where {{mvar|M}} is an upper bound on {{math|{{abs|''f''(''z'')}}}} along the arc and {{mvar|L}} the length of the arc. Now,
<math display=block>\left|\int_\text{Arc} f(z)\,dz\right|\le \frac{a\pi}{\left(a^2-1\right)^2} \to 0 \text{ as } a \to \infty.</math>
So
<math display=block>\int_{-\infty}^\infty \frac{1}{\left(x^2+1\right)^2}\,dx = \int_{-\infty}^\infty f(z)\,dz = \lim_{a \to +\infty} \int_{-a}^a f(z)\,dz = \frac{\pi}2.\quad\square</math>

====Using the method of residues====
Consider the [[Laurent series]] of {{math|''f''(''z'')}} about {{mvar|i}}, the only singularity we need to consider. We then have
<math display=block>f(z) = \frac{-1}{4(z-i)^2} + \frac{-i}{4(z-i)} + \frac{3}{16} + \frac{i}{8}(z-i) + \frac{-5}{64}(z-i)^2 + \cdots</math>

(See the sample Laurent calculation from [[Laurent series]] for the derivation of this series.)

It is clear by inspection that the residue is {{math|−{{sfrac|''i''|4}}}}, so, by the [[residue theorem]], we have
<math display=block>\oint_C f(z)\,dz = \oint_C \frac{1}{\left(z^2+1\right)^2}\,dz = 2 \pi i \,\operatorname{Res}_{z=i} f(z) = 2 \pi i \left(-\frac{i}{4}\right)=\frac{\pi}2 \quad\square</math>

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 {{math|−''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 2 – Cauchy distribution===
The integral
<math display=block>\int_{-\infty}^\infty \frac{e^{itx}}{x^2+1}\,dx</math>

[[Image:contourDiagram.png|right|250px|the contour]]

(which arises in [[probability theory]] as a scalar multiple of the [[characteristic function (probability theory)|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 {{mvar|C}} that goes along the [[real number|real]] line from {{math|−''a''}} to {{mvar|a}} and then counterclockwise along a semicircle centered at 0 from {{mvar|a}} to {{math|−''a''}}. Take {{mvar|a}} to be greater than 1, so that the [[imaginary number|imaginary]] unit {{mvar|i}} is enclosed within the curve. The contour integral is
<math display=block>\int_C \frac{e^{itz} }{ z^2+1}\,dz.</math>

Since {{math|''e''<sup>''itz''</sup>}} is an [[entire function]] (having no [[mathematical singularity|singularities]] at any point in the complex plane), this function has singularities only where the denominator {{math|''z''<sup>2</sup> + 1}} is zero. Since {{math|1=''z''<sup>2</sup> + 1 = (''z'' + ''i'')(''z'' − ''i'')}}, that happens only where {{math|1=''z'' = ''i''}} or {{math|1=''z'' = −''i''}}. Only one of those points is in the region bounded by this contour. The [[residue (complex analysis)|residue]] of {{math|''f''(''z'')}} at {{math|1=''z'' = ''i''}} is
<math display=block>\lim_{z\to i}(z-i)f(z) = \lim_{z\to i}(z-i)\frac{e^{itz} }{ z^2+1} = \lim_{z\to i}(z-i)\frac{e^{itz} }{ (z-i)(z+i)} = \lim_{z\to i}\frac{e^{itz} }{ z+i} = \frac{e^{-t}}{2i}.</math>

According to the [[residue theorem]], then, we have
<math display=block>\int_C f(z)\,dz=2\pi i \operatorname{Res}_{z=i}f(z)=2\pi i\frac{e^{-t} }{ 2i}=\pi e^{-t}.</math>

The contour {{mvar|C}} may be split into a "straight" part and a curved arc, so that
<math display=block>\int_\text{straight}+\int_\text{arc}=\pi e^{-t},</math>
and thus
<math display=block>\int_{-a}^a =\pi e^{-t}-\int_\text{arc}.</math>

According to [[Jordan's lemma]], '''if {{math|''t'' > 0}} then'''
<math display=block>\int_\text{arc}\frac{e^{itz} }{ z^2+1}\,dz \rightarrow 0 \mbox{ as } a\rightarrow\infty.</math>

Therefore, '''if {{math|''t'' > 0}} then'''
<math display=block>\int_{-\infty}^\infty \frac{e^{itx} }{ x^2+1}\,dx=\pi e^{-t}.</math>

A similar argument with an arc that winds around {{math|−''i''}} rather than {{mvar|i}} shows that '''if {{math|''t'' < 0}} then'''
<math display=block>\int_{-\infty}^\infty \frac{e^{itx} }{ x^2+1}\,dx=\pi e^t,</math>
and finally we have this:
<math display=block>\int_{-\infty}^\infty \frac{e^{itx} }{ x^2+1} \,dx=\pi e^{-|t|}.</math>

(If {{math|1=''t'' = 0}} then the integral yields immediately to real-valued calculus methods and its value is {{pi}}.)

===Example 3 – 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
<math display=block>\int_{-\pi}^\pi \frac{1 }{ 1 + 3 (\cos t)^2} \,dt.</math>

We seek to make a substitution of {{math|1=''z'' = ''e<sup>it</sup>''}}. Now, recall
<math display=block>\cos t = \frac12 \left(e^{it}+e^{-it}\right) = \frac12 \left(z+\frac{1}{z}\right)</math>
and
<math display=block>\frac{dz}{dt} = iz,\ dt = \frac{dz}{iz}.</math>

Taking {{mvar|C}} to be the unit circle, we substitute to get:

<math>\begin{align}
\oint_C \frac{1}{ 1 + 3 \left(\frac12 \left(z+\frac{1}{z}\right)\right)^2} \,\frac{dz}{iz} &= \oint_C \frac{1 }{ 1 + \frac34 \left(z+\frac{1}{z}\right)^2}\frac{1}{iz} \,dz \\
&= \oint_C \frac{-i}{ z+\frac34 z\left(z+\frac{1}{z}\right)^2}\,dz \\
&= -i \oint_C \frac{dz}{ z+\frac34 z\left(z^2+2+\frac{1}{z^2}\right)} \\
&= -i \oint_C \frac{dz}{ z+\frac34 \left(z^3+2z+\frac{1}{z}\right)} \\
&= -i \oint_C \frac{dz}{ \frac34 z^3+\frac52 z+\frac{3}{4z}} \\
&= -i \oint_C \frac{4}{ 3z^3+10z+\frac{3}{z}}\,dz \\
&= -4i \oint_C \frac{dz}{ 3z^3+10z+\frac{3}{z}} \\
&= -4i \oint_C \frac{z}{ 3z^4+10z^2+3 } \,dz \\
&= -4i \oint_C \frac{z}{ 3\left(z+\sqrt{3}i\right)\left(z-\sqrt{3}i\right)\left(z+\frac{i}{\sqrt 3}\right)\left(z-\frac{i}{\sqrt 3}\right)}\,dz \\
&= -\frac{4i}{3} \oint_C \frac{z}{\left(z+\sqrt{3}i\right)\left(z-\sqrt{3}i\right)\left(z+\frac{i}{\sqrt 3}\right)\left(z-\frac{i}{\sqrt 3}\right)}\,dz.
\end{align}</math>

The singularities to be considered are at <math>\tfrac{\pm i}{\sqrt{3}}.</math> Let {{math|''C''<sub>1</sub>}} be a small circle about <math>\tfrac{i}{\sqrt{3}},</math> and {{math|''C''<sub>2</sub>}} be a small circle about <math>\tfrac{-i}{\sqrt{3}}.</math> Then we arrive at the following:
<math display=block>\begin{align}
& -\frac{4i}{3} \left [\oint_{C_1} \frac{\frac{z}{\left(z+\sqrt{3}i\right)\left(z-\sqrt{3}i\right)\left(z+\frac{i}{\sqrt 3} \right)}}{z-\frac{i}{\sqrt 3}}\,dz +\oint_{C_2} \frac{\frac{z}{\left(z+\sqrt{3}i\right)\left(z-\sqrt{3}i\right)\left(z-\frac{i}{\sqrt 3}\right)}}{z+\frac{i}{\sqrt 3}} \, dz \right ] \\
= {} & -\frac{4i}{3} \left[ 2\pi i \left[\frac{z}{\left(z+\sqrt{3}i\right)\left(z-\sqrt{3}i\right)\left(z+\frac{i}{\sqrt 3}\right)}\right]_{z=\frac{i}{\sqrt 3}} + 2\pi i \left[\frac{z}{\left(z+\sqrt{3}i\right)\left(z-\sqrt{3}i\right)\left(z-\frac{i}{\sqrt 3}\right)} \right]_{z=-\frac{i}{\sqrt 3}}\right] \\
= {} & \frac{8\pi}{3} \left[\frac{\frac{i}{\sqrt 3}}{\left(\frac{i}{\sqrt 3}+\sqrt{3}i\right)\left(\frac{i}{\sqrt 3}-\sqrt{3}i\right)\left(\frac{i}{\sqrt 3}+\frac{i}{\sqrt 3}\right)} + \frac{-\frac{i}{\sqrt 3}}{\left(-\frac{i}{\sqrt 3}+\sqrt{3}i\right)\left(-\frac{i}{\sqrt 3}-\sqrt{3}i\right)\left(-\frac{i}{\sqrt 3}-\frac{i}{\sqrt 3}\right)} \right] \\
= {} & \frac{8\pi}{3} \left[\frac{\frac{i}{\sqrt 3}}{\left(\frac{4}{\sqrt 3}i\right)\left(-\frac{2}{i\sqrt{3}}\right)\left(\frac{2}{\sqrt{3}i}\right)}+\frac{-\frac{i}{\sqrt 3}}{\left(\frac{2}{\sqrt 3}i\right)\left(-\frac{4}{\sqrt 3}i\right)\left(-\frac{2}{\sqrt 3}i\right)}\right] \\
= {} & \frac{8\pi}{3}\left[\frac{\frac{i}{\sqrt 3}}{i\left(\frac{4}{\sqrt 3}\right)\left(\frac{2}{\sqrt 3}\right)\left(\frac{2}{\sqrt 3}\right)}+\frac{-\frac{i}{\sqrt 3}}{-i\left(\frac{2}{\sqrt 3}\right)\left(\frac{4}{\sqrt 3}\right)\left(\frac{2}{\sqrt 3}\right)}\right] \\
= {} & \frac{8\pi}{3}\left[\frac{\frac{1}{\sqrt 3}}{\left(\frac{4}{\sqrt 3}\right)\left(\frac{2}{\sqrt 3}\right)\left(\frac{2}{\sqrt 3}\right)}+\frac{\frac{1}{\sqrt 3}}{\left(\frac{2}{\sqrt 3}\right)\left(\frac{4}{\sqrt 3}\right)\left(\frac{2}{\sqrt 3}\right)}\right] \\
= {} & \frac{8\pi}{3}\left[\frac{\frac{1}{\sqrt 3}}{\frac{16}{3\sqrt{3}}}+\frac{\frac{1}{\sqrt 3}}{\frac{16}{3\sqrt{3}}} \right] \\
= {} & \frac{8\pi}{3}\left[\frac{3}{16} + \frac{3}{16} \right] \\
= {} & \pi.
\end{align}</math>

===Example 3a – trigonometric integrals, the general procedure===
The above method may be applied to all integrals of the type
<math display=block> \int_0^{2\pi} \frac{P\big(\sin(t),\sin(2t),\ldots,\cos(t),\cos(2t),\ldots\big)}{Q\big(\sin(t),\sin(2t),\ldots,\cos(t),\cos(2t),\ldots\big)}\, dt</math>

where {{mvar|P}} and {{mvar|Q}} are polynomials, i.e. a rational function in trigonometric terms is being integrated. Note that the bounds of integration may as well be {{pi}} and −{{pi}}, as in the previous example, or any other pair of endpoints 2{{pi}} apart.

The trick is to use the substitution {{math|1=''z'' = ''e<sup>it</sup>''}} where {{math|1=''dz'' = ''ie<sup>it</sup> dt''}} and hence
<math display=block> \frac{1}{iz} \,dz = dt.</math>

This substitution maps the interval {{closed-closed|0, 2π}} to the unit circle. Furthermore,
<math display=block> \sin(k t) = \frac{e^{i k t} - e^{- i k t}}{2 i} = \frac{z^k - z^{-k}}{2i}</math>
and
<math display=block> \cos(k t) = \frac{e^{i k t} + e^{- i k t}}{2} = \frac{z^k + z^{-k}}{2}</math>
so that a rational function {{math|''f''(''z'')}} in {{mvar|z}} results from the substitution, and the integral becomes
<math display=block> \oint_{|z|=1} f(z) \frac{1}{iz}\, dz </math>
which is in turn computed by summing the residues of {{math|''f''(''z''){{sfrac|1|''iz''}}}} inside the unit circle.

[[Image:TrigonometricToComplex.png|right]]
The image at right illustrates this for
<math display=block> I = \int_0^\frac{\pi}{2} \frac{1}{1 + (\sin t)^2}\, dt,</math>
which we now compute. The first step is to recognize that
<math display=block> I = \frac14 \int_0^{2\pi} \frac{1}{1 + (\sin t)^2} \,dt.</math>

The substitution yields
<math display=block> \frac{1}{4} \oint_{|z|=1} \frac{4 i z}{z^4 - 6z^2 + 1}\, dz = \oint_{|z|=1} \frac{i z}{z^4 - 6z^2 + 1}\, dz.</math>

The poles of this function are at {{math|1 ± {{sqrt|2}}}} and {{math|−1 ± {{sqrt|2}}}}. Of these, {{math|1 + {{sqrt|2}}}} and {{math|−1 − {{sqrt|2}}}} are outside the unit circle (shown in red, not to scale), whereas {{math|1 − {{sqrt|2}}}} and {{math|−1 + {{sqrt|2}}}} are inside the unit circle (shown in blue). The corresponding residues are both equal to {{math|−{{sfrac|''i''{{sqrt|2}}|16}}}}, so that the value of the integral is
<math display=block> I = 2 \pi i \; 2 \left( - \frac{\sqrt{2}}{16} i \right) = \pi \frac{\sqrt{2}}{4}.</math>

===Example 4 – branch cuts===
Consider the real integral
<math display=block>\int_0^\infty \frac{\sqrt x}{x^2+6x+8}\,dx.</math>

We can begin by formulating the complex integral
<math display=block>\int_C \frac{\sqrt z}{z^2+6z+8}\,dz=I.</math>

[[Image:Keyhole contour.svg|right|180px]]
We can use the Cauchy integral formula or residue theorem again to obtain the relevant residues. However, the important thing to note is that {{math|1=''z''<sup>1/2</sup> = ''e''<sup>(Log ''z'')/2</sup>}}, so {{math|''z''<sup>1/2</sup>}} has a [[branch cut]]. This affects our choice of the contour {{mvar|C}}. Normally the logarithm branch cut is defined as the negative real axis, however, this makes the calculation of the integral slightly more complicated, 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 {{mvar|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 {{math|1=''z'' = −2}} and {{math|1=''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 {{math|1=''z'' = 0}} was avoided by detouring around the origin.
{{clear}}

Let {{mvar|γ}} be the small circle of radius {{mvar|ε}}, {{math|Γ}} the larger, with radius {{mvar|R}}, then
<math display=block>\int_C = \int_\varepsilon^R + \int_\Gamma + \int_R^\varepsilon + \int_\gamma.</math>

It can be shown that the integrals over {{math|Γ}} and {{mvar|γ}} both tend to zero as {{math|''ε'' → 0}} and {{math|''R'' → ∞}}, by an estimation argument above, that leaves two terms. Now since {{math|1=''z''<sup>1/2</sup> = ''e''<sup>(Log ''z'')/2</sup>}}, on the contour outside the branch cut, we have gained 2{{pi}} in argument along {{mvar|γ}}. (By [[Euler's identity]], {{math|''e''<sup>''i''π</sup>}} represents the unit vector, which therefore has {{pi}} as its log. This {{pi}} is what is meant by the argument of {{mvar|z}}. The coefficient of {{sfrac|1|2}} forces us to use 2{{pi}}.) So
<math display=block>\begin{align}
\int_R^\varepsilon \frac{\sqrt z}{z^2+6z+8}\,dz&=\int_R^\varepsilon \frac{e^{\frac12 \operatorname{Log} z}}{z^2+6z+8}\,dz \\[6pt]
&=\int_R^\varepsilon \frac{e^{\frac12(\log|z|+i \arg{z})}}{z^2+6z+8}\,dz \\[6pt]
& = \int_R^\varepsilon \frac{ e^{\frac12\log|z|}e^{\frac12(2\pi i)}}{z^2+6z+8}\,dz\\[6pt]
&=\int_R^\varepsilon \frac{ e^{\frac12\log|z|}e^{\pi i}}{z^2+6z+8}\,dz \\[6pt]
& = \int_R^\varepsilon \frac{-\sqrt z}{z^2+6z+8}\,dz\\[6pt]
&=\int_\varepsilon^R \frac{\sqrt z}{z^2+6z+8}\,dz.
\end{align}</math>

Therefore:
<math display=block>\int_C \frac{\sqrt z}{z^2+6z+8}\,dz=2\int_0^\infty \frac{\sqrt x}{x^2+6x+8}\,dx.</math>

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
<math display=block>\pi i \left(\frac{i}{\sqrt 2}-i\right)=\int_0^\infty \frac{\sqrt x}{x^2+6x+8}\,dx = \pi\left(1-\frac{1}{\sqrt 2}\right).\quad\square</math>

===Example 5 – the square of the logarithm===
[[Image:KeyholeContourLeftTikz.tif|thumbnail|upright=2|right]]

This section treats a type of integral of which
<math display=block>\int_0^\infty \frac{\log x}{\left(1+x^2\right)^2} \, dx</math>
is an example.

To calculate this integral, one uses the function
<math display=block>f(z) = \left (\frac{\log z}{1+z^2} \right )^2</math>
and the branch of the logarithm corresponding to {{math|−π < arg ''z'' ≤ π}}.

We will calculate the integral of {{math|''f''(''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 and by the Cauchy residue theorem we have

<math>\begin{align}
\left( \int_R + \int_M + \int_N + \int_r \right) f(z) \, dz
= &\  2 \pi i \big( \operatorname{Res}_{z=i} f(z) + \operatorname{Res}_{z=-i} f(z) \big) \\
= &\  2 \pi i \left( - \frac{\pi}{4} + \frac{1}{16} i \pi^2 - \frac{\pi}{4} - \frac{1}{16} i \pi^2 \right) \\
= &\  - i \pi^2. \end{align}</math>

Let {{mvar|R}} be the radius of the large circle, and {{mvar|r}} the radius of the small one. We will denote the upper line by {{mvar|M}}, and the lower line by {{mvar|N}}. As before we take the limit when {{math|''R'' → ∞}} and {{math|''r'' → 0}}. The contributions from the two circles vanish. For example, one has the following upper bound with the [[ML lemma|{{mvar|ML}} lemma]]:
<math display=block>\left| \int_R f(z) \, dz \right| \le 2 \pi R \frac{(\log R)^2 + \pi^2}{\left(R^2-1\right)^2} \to 0.</math>

In order to compute the contributions of {{mvar|M}} and {{mvar|N}} we set {{math|1=''z'' = −''x'' + ''iε''}} on {{mvar|M}} and {{math|1=''z'' = −''x'' − ''iε''}} on {{mvar|N}}, with {{math|0 < ''x'' < ∞}}:

<math>\begin{align} -i \pi^2 &= \left( \int_R + \int_M + \int_N + \int_r \right) f(z) \, dz \\[6pt]
&= \left( \int_M + \int_N \right) f(z)\, dz && \int_R, \int_r \mbox{ vanish} \\[6pt]
&=-\int_\infty^0 \left (\frac{\log(-x + i\varepsilon)}{1+(-x + i\varepsilon)^2} \right )^2\, dx - \int_0^\infty \left (\frac{\log(-x - i\varepsilon)}{1+(-x - i\varepsilon)^2}\right)^2 \, dx \\[6pt]
&= \int_0^\infty \left (\frac{\log(-x + i\varepsilon)}{1+(-x + i\varepsilon)^2} \right )^2 \, dx - \int_0^\infty \left (\frac{\log(-x - i\varepsilon)}{1+(-x - i\varepsilon)^2} \right )^2 \, dx \\[6pt]
&= \int_0^\infty \left (\frac{\log x + i\pi}{1+x^2} \right )^2 \, dx - \int_0^\infty \left (\frac{\log x - i\pi}{1+x^2} \right )^2 \, dx && \varepsilon \to 0 \\
&= \int_0^\infty \frac{(\log x + i\pi)^2 - (\log x - i\pi)^2}{\left(1+x^2\right)^2} \, dx \\[6pt]
&= \int_0^\infty \frac{4 \pi i \log x}{\left(1+x^2\right)^2} \, dx \\[6pt]
&= 4 \pi i \int_0^\infty \frac{\log x}{\left(1+x^2\right)^2} \, dx \end{align}</math>

which gives
<math display=block>\int_0^\infty \frac{\log x}{\left(1+x^2\right)^2} \, dx = - \frac{\pi}{4}.</math>

===Example 6 – logarithms and the residue at infinity===
[[Image:ContourLogs.png|right]]

We seek to evaluate
<math display=block>I = \int_0^3 \frac{x^\frac34 (3-x)^\frac14}{5-x}\,dx.</math>

This requires a close study of
<math display=block>f(z) = z^\frac34 (3-z)^\frac14.</math>

We will construct {{math|''f''(''z'')}} so that it has a branch cut on {{closed-closed|0, 3}}, shown in red in the diagram. To do this, we choose two branches of the logarithm, setting
<math display=block>z^\frac34 = \exp \left (\frac34 \log z \right ) \quad \mbox{where } -\pi \le \arg z < \pi </math>
and
<math display=block>(3-z)^\frac14 = \exp \left (\frac14 \log(3-z) \right ) \quad \mbox{where } 0 \le \arg(3-z) < 2\pi. </math>

The cut of {{math|''z''<sup>{{3/4}}</sup>}} is therefore {{open-closed|−∞, 0}} and the cut of {{math|(3 − ''z'')<sup>1/4</sup>}} is {{open-closed|−∞, 3}}. It is easy to see that the cut of the product of the two, i.e. {{math|''f''(''z'')}}, is {{math|[0, 3]}}, because {{math|''f''(''z'')}} is actually continuous across {{open-open|−∞, 0}}. This is because when {{math|1=''z'' = −''r'' < 0}} and we approach the cut from above, {{math|''f''(''z'')}} has the value
<math display=block> r^\frac34 e^{\frac34 \pi i} (3+r)^\frac14 e^{\frac24 \pi i} = r^\frac34 (3+r)^\frac14 e^{\frac54 \pi i}.</math>

When we approach from below, {{math|''f''(''z'')}} has the value
<math display=block> r^\frac34 e^{-\frac34 \pi i} (3+r)^\frac14 e^{\frac04 \pi i} = r^\frac34 (3+r)^\frac14 e^{-\frac34 \pi i}.</math>

But
<math display=block>e^{-\frac34 \pi i} = e^{\frac54 \pi i},</math>

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 {{math|''z''<sup>{{3/4}}</sup>}} and {{math|(3 − ''z'')<sup>1/4</sup>}}.

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

Let {{math|1=''z'' = ''r''}} (in the limit, i.e. as the two green circles shrink to radius zero), where {{math|0 ≤ ''r'' ≤ 3}}. Along the upper segment, we find that {{math|''f''(''z'')}} has the value
<math display=block>r^\frac34 e^{\frac04 \pi i} (3-r)^\frac14 e^{\frac24 \pi i} = i r^\frac34 (3-r)^\frac14</math>
and along the lower segment,
<math display=block>r^\frac34 e^{\frac04 \pi i} (3-r)^\frac14 e^{\frac04 \pi i} = r^\frac34 (3-r)^\frac14.</math>

It follows that the integral of {{math|{{sfrac|''f''(''z'')|5 − ''z''}}}} along the upper segment is {{math|−''iI''}} in the limit, and along the lower segment, {{mvar|I}}.

If we can show that the integrals along the two green circles vanish in the limit, then we also have the value of {{math|I}}, by the [[Cauchy residue theorem]]. Let the radius of the green circles be {{mvar|ρ}}, where {{math|''ρ'' < 0.001}} and {{math|''ρ'' → 0}}, and apply the [[Estimation lemma|{{mvar|ML}} inequality]]. For the circle {{math|''C''<sub>L</sub>}} on the left, we find
<math display=block>\left| \int_{C_\mathrm{L}} \frac{f(z)}{5-z} dz \right| \le 2 \pi \rho \frac{\rho^\frac34 3.001^\frac14}{4.999} \in \mathcal{O} \left( \rho^\frac74 \right) \to 0.</math>

Similarly, for the circle {{math|''C''<sub>R</sub>}} on the right, we have
<math display=block>\left| \int_{C_\mathrm{R}} \frac{f(z)}{5-z} dz \right| \le 2 \pi \rho \frac{3.001^\frac34 \rho^\frac14}{1.999} \in \mathcal{O} \left( \rho^\frac54 \right) \to 0.</math>

Now using the [[Cauchy residue theorem]], we have
<math display=block>(-i + 1) I = -2\pi i \left( \operatorname{Res}_{z=5} \frac{f(z)}{5-z} + \operatorname{Res}_{z=\infty} \frac{f(z)}{5-z} \right).</math>
where the minus sign is due to the clockwise direction around the residues. Using the branch of the logarithm from before, clearly
<math display=block>\operatorname{Res}_{z=5} \frac{f(z)}{5-z} = - 5^\frac34 e^{\frac14 \log(-2)}.</math>

The pole is shown in blue in the diagram. The value simplifies to
<math display=block>-5^\frac34 e^{\frac14(\log 2 + \pi i)} = -e^{\frac14 \pi i} 5^\frac34 2^\frac14.</math>

We use the following formula for the residue at infinity:
<math display=block>\operatorname{Res}_{z=\infty} h(z) = \operatorname{Res}_{z=0} \left(- \frac{1}{z^2} h\left(\frac{1}{z}\right)\right).</math>

Substituting, we find
<math display=block>\frac{1}{5-\frac{1}{z}} = -z \left(1 + 5z + 5^2 z^2 + 5^3 z^3 + \cdots\right)</math>
and
<math display=block>\left(\frac{1}{z^3}\left (3-\frac{1}{z} \right )\right)^\frac14 = \frac{1}{z} (3z-1)^\frac14 = \frac{1}{z}e^{\frac14 \pi i} (1-3z)^\frac14, </math>
where we have used the fact that {{math|1=−1 = ''e''<sup>π''i''</sup>}} for the second branch of the logarithm. Next we apply the binomial expansion, obtaining
<math display=block>\frac{1}{z} e^{\frac14 \pi i} \left( 1 - {1/4 \choose 1} 3z + {1/4 \choose 2} 3^2 z^2 - {1/4 \choose 3} 3^3 z^3 + \cdots \right). </math>

The conclusion is that
<math display=block>\operatorname{Res}_{z=\infty} \frac{f(z)}{5-z} = e^{\frac14 \pi i} \left (5 - \frac34 \right ) = e^{\frac14 \pi i}\frac{17}{4}.</math>

Finally, it follows that the value of {{mvar|I}} is
<math display=block> I = 2 \pi i \frac{e^{\frac14 \pi i}}{-1+i} \left(\frac{17}{4} - 5^\frac34 2^\frac14 \right) = 2 \pi 2^{-\frac12} \left(\frac{17}{4} - 5^\frac34 2^\frac14 \right)</math>
which yields
<math display=block>I = \frac{\pi}{2\sqrt 2} \left(17 - 5^\frac34 2^\frac94 \right) = \frac{\pi}{2\sqrt 2} \left(17 - 40^\frac34 \right).</math>