Editing Contour integration (section)

===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>