Editing Contour integration (section)

===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>