Lectures on Complex Integration by A. O. Gogolin (auth.), Elena G. Tsitsishvili, Andreas Komnik

By A. O. Gogolin (auth.), Elena G. Tsitsishvili, Andreas Komnik (eds.)

The concept of advanced services is a strikingly attractive and strong quarter of arithmetic. a few rather attention-grabbing examples are probably complex integrals that are easily computed after reshaping them into integrals alongside contours, in addition to it seems that tough differential and indispensable equations, which are elegantly solved utilizing related tools. to take advantage of them is usually regimen yet in lots of instances it borders on an artwork. The objective of the publication is to introduce the reader to this pretty zone of arithmetic and to coach her or him find out how to use those the way to clear up quite a few difficulties starting from computation of integrals to fixing tough vital equations. this is often performed with a support of various examples and issues of exact solutions.

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The latter choice is more convenient here. To compute F(s) for s > 0 observe that (see Fig. 25) d x eisx f (x) = 0, C+ +C +C1 +C2 +P and then d x eisx f (x) = F(s). P Next we estimate the integral along the circle path C around the upper branching point z = i, which we parametrise by z = r eiφ , 2π dx e isx dz √ ∼ z f (x) ∼ C r eiφ dφ √ ∼ r. √ r 0 C Therefore it vanishes for r → 0. For s > 0 d x eisx f (x) = 0, C+ as C+ can be pushed upwards. On C1 we put x = i(1 + η) and then obtain 0 d x eisx f (x) = C1 ∞ idηe−s−ηs = −e−s √ i η(η + 2) ∞ √ 0 dη e−ηs = + η(η + 2) In this way we achieve a very convenient representation C2 d x eisx f (x).

17 Schematic representation of the contour used in the method (ii) of Sect. 4 √ for the function f (z) = z 2 − 1 e±iπ/2 1 √ 1− x2 = ∓i √ 1 1 − x2 , 28 1 Basics and therefore ± C± = 2i I . Now equating the result of the calculation (i) with the result of the calculation (ii), 2πi = 2i I , we recover the elementary result I = π. Let us now consider another elementary integral, π/2 1 1 − x 2d x = I = −1 π/2 dθ cos2 θ = −π/2 −π/2 1 1 π + cos(2θ) dθ = , 2 2 2 to illustrate a separate point. We integrate the function f (z) = (z 2 − 1)1/2 over the same contour C R .

Now f (z) = (z + 1)−a (z − 1)a−1 =O z − x0 1 z2 so that a−1 (a; x0 ) = 0. Again, there are no residues, and the answer is: Ia (x0 ) = π cot(πa)(1 + x0 )−a (1 − x0 )a−1 . In particular, for a = 1/2, we have 1 I1/2 (x0 ) = P √ −1 1 1− x2 dx =0 x − x0 for all x0 ∈ (−1, 1). We conclude this section by discussing the case when there is a residue, R(x) = 1/(x − z 0 ), where z 0 is an arbitrary point in the complex plane except on the branch cut: 1 Ia (x0 ) = P −1 (1 + x)−a (1 − x)a d x. (x − x0 )(x − z 0 ) For large z, f (z) = O(1/z 2 ), hence a−1 (a; x0 ) = 0.

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