Abstract
where \(x\) and \(y\) are real numbers and \(i^2=-1\). Here \(x={\mathrm{Re}}z\) and \(y={\mathrm{Im}}z\) are the real and imaginary parts of the complex number \(z\). For complex conjugated variable the notation
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Notes
- 1.
Although this condition can be relaxed.
- 2.
Clearly the function \(f(z)/(z-a)^{n+1}\) is analytic between \(C'\) and \(C\) for negative \(n\), so the \(C'\) contour for these terms is deformed into \(C\). The assumption that the function \(f(z)\) is analytic between \(C\) and \(C'\) means for the Laurent series that the terms with non-negative powers of \(h\) should have a radius of convergence (around \(a\)) that encircles \(C\). On the other hand the terms with the negative powers of \(h\) converge outside a smaller radius that lies inside \(C'\). One way to look at the Laurent expansion is then that it provides the means of splitting the function into two: the one analytic outside a smaller circle and the other analytic inside a bigger circle.
- 3.
We would like to remark, that cutting along the negative real axis is not the only option. The function can be uniquely defined in any open simply connected set obtained from the whole complex plane \({\mathbb {C}}\) by cutting from the branching point to infinity.
- 4.
This integral can be reduced to an integral of the type studied in Sect. 1.3.2 by substitution
$$\begin{aligned} y=\frac{1-x}{1+x}\;\;\Rightarrow I(a)=2\int \limits _0^\infty \frac{y^a dy}{(y+1)^2}, \end{aligned}$$the double pole here explains why the result is proportional to \(a\).
- 5.
It is instructive to investigate the \(a\rightarrow 0\) limit of this formula. The \(O(1/a)\) term must vanish telling us that
$$\begin{aligned} \sum \mathrm{Res}R(z)|_{{\mathrm{finite}}\;\;z}=a_{-1}(0), \end{aligned}$$that is the sum of all finite residues is equal to the residue at infinity for an arbitrary rational function. For non-rational functions this is, generally, not true. Indeed the constant, \(O(a^0)\) term produces an interesting expression:
$$\begin{aligned} \int \limits _{-1}^1R(x)dx=\left[ \sum \mathrm{Res}|_{{\mathrm{finite}}\;\;z}-\mathrm{Res}|_{z=\infty } \right] \ln \left( \frac{z-1}{z+1}\right) R(z). \end{aligned}$$ - 6.
Alternatively, instead of shifting the contours we can think about closing the straight part with semicircles to the left/right. The convergence in both cases is given as \((-z)^\xi \rightarrow 0\) as Re \(\xi \rightarrow -\infty \) in the former case and for Re \(\xi \rightarrow \infty \) in the latter.
- 7.
We would like to point out that the derivative with respect to \(x\) (\(f_+^{'}(x)\)) is not equivalent to multiplying the Fourier transform by \(ik\) because of its ‘one-sidedness’, since the boundary contributions when integrating by part are non-trivial.
- 8.
In many practical applications both Fourier and Laplace transforms can become very complicated. Usually it is useful to consult the tables of integral transforms, see for example [6].
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© 2014 Springer International Publishing Switzerland
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Gogolin, A.O., Tsitsishvili, E.G., Komnik, A. (2014). Basics. In: Tsitsishvili, E., Komnik, A. (eds) Lectures on Complex Integration. Undergraduate Lecture Notes in Physics. Springer, Cham. https://doi.org/10.1007/978-3-319-00212-5_1
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