solutions04 - Problem Set 4 Solutions Physics 330 M Seifert Due Date November 9 2007 1(MW 3-10 Evaluate I= 0 dx 1 x4 Viewed as a complex function the

solutions04 - Problem Set 4 Solutions Physics 330 M Seifert...

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Problem Set 4: Solutions Physics 330 M. Seifert Due Date: November 9, 2007 1. (MW 3-10) Evaluate I = 0 d x 1 + x 4 . Viewed as a complex function, the integrand has four simple poles, at the roots of the polynomial z 4 + 1 = 0; we denote these values by z n = e (2 n - 1) iπ/ 4 , where n runs from 1 to 4. We choose the contour shown in Figure 1: ˜ I = C d z 1 + z 4 This can be decomposed into three parts: A line running from 0 to R , along which z = x ( x [0 , R ]); A quarter-circle running from R to iR , along which z = Re ( θ [0 , π 2 ]); and A line running from iR to 0, along which z = ix ( x [ R, 0]). Splitting this up into three parts, then, we have ˜ I = R 0 d x 1 + x 4 + π 2 0 iRe d θ 1 + R 4 e 4 + 0 R i d x 1 + ( ix ) 4 = (1 - i ) R 0 d x 1 + x 4 + π 2 0 iRe d θ 1 + R 4 e 4 In the limit where R → ∞ , the second integral will scale as R/R 4 , and so can be neglected: ˜ I = (1 - i ) 0 d x 1 + x 4 = (1 - i ) I Now we simply apply the residue theorem. We have one simple pole enclosed 1
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z 1 z 4 z 2 z 3 Figure 1: Integration contour for Problem 1. in the contour, at z = e iπ/ 4 ; so the residue is Res 1 1 + z 4 z = e iπ/ 4 = z - e iπ/ 4 1 + z 4 z = e iπ/ 4 = 1 ( e iπ/ 4 - e - iπ/ 4 )( e iπ/ 4 - e 3 iπ/ 4 )( e iπ/ 4 - e - 3 iπ/ 4 ) = 1 (2 i ) 3 (sin π 4 )( - e iπ/ 2 sin π 4 )( e - iπ/ 4 sin π 2 ) = - 1 + i 4 2 Thus, ˜ I = (1 - i ) I = 2 πi - (1 + i ) 4 2 I = 2 π 4 2
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! ia ia Figure 2: Integration contour for Problem 2. 2. (MW 3-13) Evaluate I = d 3 x ( a 2 + r 2 ) 3 . Assume that a > 0; the a < 0 case can be obtained by substituting a → - a in the following derivation. In spherical polar coordinates, this is simply I = r 2 d r d θ d φ ( a 2 + r 2 ) 3 = 4 π 0 r 2 d r ( a 2 + r 2 ) 3 = 2 π -∞ r 2 d r ( a 2 + r 2 ) 3 where we’ve used the evenness of the integrand in the last step. We can now use contour integration to evaluate this integral. Viewed as an analytic function, the integrand has poles of order three at z = ± ia . We choose the contour indicated in Figure 2. This consists of two portions, one due to the integration along the real axis and the other due to the upper half-circle: C z 2 d z ( a 2 + z 2 ) 3 = R - R r 2 d r ( a 2 + r 2 ) 3 + π 0 R 2 e 2 iRe d θ ( a 2 + R 2 e 2 ) 3 3
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The integral along the upper half-circle will go as R 3 /R 6 for R 1, and thus in the limit R → ∞ we have C z 2 d z ( a 2 + z 2 ) 3 = -∞ r 2 d r ( a 2 + r 2 ) 3 = I 2 π All that remains is to calculate the residues to obtain the contour integral. We have one pole of order three inside the contour, at z = ia ; the residue there is Res z 2 ( a 2 + z 2 ) 3 z = ia = 1 2 d 2 d z 2 z 2 ( z + ia ) 3 z = ia = 1 2 2 ( z + ia ) 3 - 12 z ( z + ia ) 4 + 12 z 2 ( z + ia ) 5 z = ia = - i 16 a 3 Thus, C z 2 d z ( a 2 + z 2 ) 3 = 2 πi - i 16 a 3 I = π 2 4 a 3 3. (MW 3-15) Evaluate I = 0 x d x 1 + x 5 . As a complex function, the integrand has five simple poles; we will denote them as z n = e (2 n - 1) iπ/ 5 , where n runs from 1 to 5. The symmetry of the denominator of the integrand leads us to choose the contour shown in Figure 3.
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