By Alexander O. Gogolin, Elena G. Tsitsishvili, Andreas Komnik

The speculation of complicated features is a strikingly appealing and strong region of arithmetic. a few rather interesting examples are doubtless advanced integrals that are without difficulty computed after reshaping them into integrals alongside contours, in addition to it appears tricky differential and imperative equations, which are elegantly solved utilizing comparable tools. to exploit them is sometimes routine yet in lots of circumstances it borders on an artwork. The target of the booklet is to introduce the reader to this gorgeous quarter of arithmetic and to coach her or him tips to use those tips on how to remedy a number of difficulties starting from computation of integrals to fixing tricky critical equations. this can be performed with a aid of various examples and issues of designated options.

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**Additional info for Lectures on Complex Integration (Undergraduate Lecture Notes in Physics)**

2. 7 Generalized Hypergeometric sequence I = q i=1 p j=1 ninety nine (a j ) p Fq [(a); (b); z]. (bi ) For |z| > 1 we stream the combination line to −∞ as a substitute. Then we need to look after the poles at s = −ak − n, so ∞ p p i=1 q j=1 I = k=1 n=0 (ai − ak − n) (ak + n) (b j − ak − n) (−1)n (−z)−ak −n , n! the place the primed product doesn't include the time period i = okay. Now we wish to exploit the id (2. 6) within the shape (c − n) = π(−1)n /[sin(πc) (1 − c + n)], so that it will rewrite the above expression as ∞ p (−z)−ak I = p n=0 i=1 k=1 q π(−1)n 1 sin[π(ai − ak )] (1 − ai + ak + n) (−1)n sin[π(b j − ak )] (1 − b j + ak − n) π × j=1 (ak + n) (1/z)n . n! extra we use the truth that p = q + 1 and take away n from the Gamma capabilities through advantage of (c + n) = (c)(c)n to reach at ∞ p (−z)−ak I = n=0 i=1 k=1 ⎡ p −ak = (−z) k=1 ∞ × p ⎣ ⎡ p n=0 i=1 p i=1 (ak ) p Fq q j=1 q j=1 ⎡ (1 − b j + ak )n (b j − ak ) q (ai − ak )⎦ (ak ) ⎣ (ak )n ⎣ k=1 (ai − ak ) (1 − ai + ak )n ⎤ i=1 p = × p ⎤⎡ 1 ⎦⎣ (1 − ai + ak )n (ai − ak ) (b j − ak ) j=1 q ⎤ 1 ⎦ (b j − ak ) ⎤ (1 − b j + ak )n ⎦ j=1 (ak )(ak )n (1/z)n n! 1 (1/z)n n! (−z)ak ak , 1 − b1 + ak , . . . , 1 − bq + ak ; 1 + a1 + ak , . . . , 1 − a p + ak ; 1/z . As required the F-function has 1 + q = p parameters of the a and b-type and p − 1 = q parameters of the c-type (because 1 = 1 − ak + ak parameter is absent). To summarize we receive the next relation for the analytic continuation 100 2 Hypergeometric sequence with functions p p Fq [(ai ); (b j ); z] p = k=1 i=1 × p Fq (ai − ak ) (ai ) q (b j ) (−z)−ak (b j − ak ) j=1 (2. seventy three) ak , 1 − b1 + ak , . . . , 1 − bq + ak ; 1 + a1 + ak , . . . , 1 − a p + ak ; 1/z , the place bear in mind that p = q + 1. particularly, for q = 1 and p = 2 we instantly recuperate the end result for the normal hypergeometric sequence (2. 55). 2. eight Examples from Mathematical Physics 2. eight. 1 Momentum Distribution functionality of Interacting platforms The calculation of particle distribution features with admire to their quantum numbers is among the most crucial initiatives in many-body difficulties. whereas natural noninteracting bosonic and fermionic structures are topic to Bose and Fermi distribution capabilities, the placement adjustments dramatically once the debris begin to engage. The momentum distribution functionality n(k) is generally calculated as a Fourier remodel of the equivalent time Green’s functionality [10]. in a single size the matter of interacting fermions is less than yes stipulations analytically solvable and the corresponding Green’s functionality is given via G(x) = 1 1 , 2π (−i x + δ)α (2. seventy four) the place δ is a favorable infinitesimal and α is a dimensionless interplay parameter. it really is α = 1 in a non-interacting approach after which n(k) = ∞ −∞ d x eikx G(x) = ∞ −∞ d x eikx 2π −i x + δ might be evaluated through final within the upper/lower half-plane for positive/negative okay. the end result, identified from the speculation of non-interacting Fermi fuel, is the Heaviside step-function n(k) = (−k) outlined in (1. 26). 18 For conventional α one is faced with a multivalued integrand.