Editing Contour integration (section)

====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>