Editing Contour integration (section)

===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>