This is a question based on the method here: http://en.wikipedia.org/wiki/Methods_of_contour_integration#Example_.28V.29_.E2.80.93_the_square_of_the_logarithm
The author chose a contour which requires a bit of a "magical" replacement of the original function $\displaystyle f(x)=\frac{\log(x)}{(1+x^2)^2}$ with analysis of $\displaystyle f(z)=\left(\frac{\log(z)}{(1+z^2)}\right)^2$, the need for which becomes apparent later
I chose to instead pick a contour that only involves the upper half of the figure drawn, discarding the lower line N and cutting the curve through with the section of the real line $\mathbb R^+$
Numerically I obtain the same result:
\begin{align}
\int_\gamma f(z)dz &= \int_{\small M}+\int_{\small R^+} \\
&=\int_0^\infty\frac{\log(-x+i\epsilon)}{(1+(-x+i\epsilon)^2)^2}dx+\int_0^\infty f(x)dx \\
2\pi i\cdot\operatorname{Res}[f(z),i]&=\int_0^\infty\frac{\log(x)+i\pi}{\left(1+x^2\right)^2}dx+\int_0^\infty f(x)dx&\epsilon\to0 \\
2\pi i\cdot (\frac18(\pi+2i))&=\int_0^\infty\frac{i\pi}{\left(1+x^2\right)^2}dx+2\int_0^\infty f(x) dx\\
-\frac\pi2+\frac{i\pi^2}4&=\frac{i\pi^2}4+2\int_0^\infty f(x) dx\\
\int_0^\infty f(x)dx&=-\frac\pi4
\end{align}
Is it just coincidence that I have the same result? Was there any flaw in my method? If not, why did the author pick a contour that requires a nonintuitive analysis which "as it turns out … is a multiple of the initial integral that we wish to calculate?"
Best Answer
No fundamental flaw in your method, although you still need to be careful about the origin. You should keep in mind, though, that your method applies only to otherwise even functions. That is, integrals of the form
$$\int_0^{\infty} dx\, g(x) \log{x}$$
where $g$ is an even function of $x$. I applaud you for finding a very nifty shortcut that works well in these special cases.
I disagree with you, though, about the "magic" involved, or that it is nonintuitive - it sure is intuitive to me. Multiplying by $\log{x}$ is a standard method of dealing with integrals of the form
$$\int_0^{\infty} dx \, f(x)$$
where $f$ is not even with respect to $x$. The fact is that we deal with the integral by exploiting the multivaluedness of the log.
You should also see whether this contour works for integrals of the form
$$\int_0^{\infty} dx \, g(x) \log^2{x}$$