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Introduction to the Residue Theorem: (With Solution to the Problem 5B of Exercise 2)

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Introduction to the Residue Theorem: (With Solution to the Problem 5B of Exercise 2) Introduction to the Residue Theorem: (with solution to the problem 5b of exercise 2) Some good references out of the countless mathematics books can be: • George Arfken: Mathematical methods for physicists. • C. R. Wylie and L. C. Barrett: Advanced engineering mathematics. We skip the entire introduction about the complex analysis and begin at the residue theorem: Let D ⊂ C be a connected area in the complex plane, and z1; :::; zk 2 D a nite number of points. Furthermore let (means function maps the area minus some points, were the function is not f : D − fz1; :::; zkg ! C f D analytical at them, to the complex plane) be an analytic function and α :[a; b] ! D − z1; :::; zk a closed curved (piecewise smooth). Then residue theory states that: k X f(ζ)dζ = 2πi · Res(f; z ): ˛ j α i=1 It should be noted that this integral should be performed counter-clockwise in the complex plane. Here the Laurent expansion becomes handy, which is the generalization of the Taylor expansion to functions that are not analytical(the expansion is valid for regions around the pole, but the function value at the pole remains undened): 1 X n f(z) = an(z − z0) : n=−∞ If this series truncate from the lower index on n = −m, then it is said that f has a pole of the mth order at z0. If it does not truncate, it is said that it has an essential singularity. The expansion coecient a−1 is called the residue of f(z) at the point z0: Res(f(z); z0) = a−1: If this coecient is zero, then the result of the integral is zero. The case were a function is analytical, or has no poles, is one of these cases. Some rules for nding the residue: • If f has a pole of the rst order: Res(f(z); z0) = lim ((z − z0)f(z)) z!z0 h m−1 i If has a pole of the order : Res 1 @ m • f(z) m (f(z); z0) = (m−1)! lim @zm−1 ((z − z0) f(z)) z!z0 When has a root of the rst order at , and is analytical: Res g(z) g(z0) • f(z) z0 g(z) ; z0 = 0 f(z) f (z0) • When f has an essential singularity, the residue cannot be found like this, and has to be found directly from the Laurent expansion. For example, 1=z 1 1 1 1 , has e = 1 + z + 2!z2 + 3!z3 + 4!z4 + ::: an essential singularity at z = 0, but its Laurent expansion can be easily found from the Taylor expansion of ez and then substituting z with 1=z. b) To solve this integral, we can write it as a complex integral 1 sin(ax) 1 1 eiax dx = Im P dx ˆ0 x 2 ˆ−∞ x This complex integral, could be part of a contour integral of the function eiaz over the contour z shown in the picture below 1 This function is analytical in the area inside the contour (gray area), so the integral over the closed contour is zero. Splitting this integral over its dierent parts, will show that what we need is the rst two parts of it that are going over the real axis: " # eiaz − eiax R eiax eiaz eiaz dz = lim dx + dx + dz + dz = 0 ˛ z R!1 ˆ x ˆ x ˆ z ˆ z !0 −R C1 C2 To nd the rst two parts of the integral we need to evaluate the last two part, namely the integrals over the half circles. As can be noticed, we avoided the rst order pole at zero, by bypassing it with a clockwise contour with an innitesimally small radius . It can easily be shown that such an integral results in −iπ: iθ eiaz 0 eiae 0 dz = lim eiθidθ = idθ = −iπ !0 iθ ˆC1 z ˆπ e ˆπ Where we used z = eiθ. If this integral was counter-clockwise, the result would have been iπ, which matches with the fact that if this small contour was passing below this pole (hence containing it in the closed loop) then the closed loop integral would have been bigger by the amount 2πi. More generally, for any semicircle with innitely small radius, the contribution to the closed loop integral would be ±πia−1, where + is for a counter-clockwise semicircle and − for a clockwise one. For the last part of the integral, over the semicircle with innitely big radius, we could have easily said this integral was zero, if the function inside the integral was approaching zero slightly faster than 1=z (because the integral would have multiplied it by something in the order of πR and the limit of R ! 1 would have resulted in zero). However, here we have exactly 1=z speed of approaching zero. But it can be shown that even for this case, if a is real and positive, the integral will again go to zero (for more details on the proof look for Jordan's lemma). All of these will give us: 1 eiax P dx = iπ for a > 0 ˆ−∞ x Which in the end gives us: 1 sin(ax) π dx = for a > 0 ˆ0 x 2 You can easily show that the for this integral gives −π , either by doing the whole complex a < 0 2 integration over a contour that goes in the lower half part of the complex plane, or, just using the fact that sin(ax) is an odd function of a. 2.
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