Editing Contour integration (section)

===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.