781 Method 1 Consider the related integral contintegraldisplay C log z 1 z 3 dz

781 method 1 consider the related integral

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7.8.1 Method 1 Consider the related integral contintegraldisplay C log z 1 + z 3 dz, (7.44) over the contour shown in Fig. 7.6. Here we have chosen the branch line of the logarithm to lie along the + z axis; the discontinuity across it is disc log z = log ρ log ρe 2 = 2 iπ. (7.45) The integral over the large circle is zero, as is the integral over the little circle: lim ρ →∞ , 0 integraldisplay 2 π 0 log ρe 1 + ρ 2 e 3 ρe i dθ = 0 . (7.46) Therefore, I = 1 2 πi contintegraldisplay C log z 1 + z 3 dz = summationdisplay poles inside C (residues) . (7.47)
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72 Version of October 26, 2011 CHAPTER7. THE CALCULUS OF RESIDUES To find the sum of the residues, we note that the poles occur at the three cube roots of 1, namely, e iπ/ 3 , e , and e 5 iπ/ 3 , so summationdisplay (residues) = log e iπ/ 3 parenleftbigg 1 e iπ/ 3 e 1 e iπ/ 3 e i 5 π/ 3 parenrightbigg + log e parenleftbigg 1 e e iπ/ 3 1 e e i 5 π/ 3 parenrightbigg + log e 5 iπ/ 3 parenleftbigg 1 e 5 iπ/ 3 e iπ/ 3 1 e 5 iπ/ 3 e parenrightbigg = i π 3 1 parenleftBig 1+ 3 i 2 + 1 parenrightBig 1 parenleftBig 1+ 3 i 2 1 - 3 i 2 parenrightBig + 1 parenleftBig 1 1+ 3 i 2 parenrightBig 1 parenleftBig 1 1 - 3 i 2 parenrightBig + i 5 π 3 1 parenleftBig 1 - 3 i 2 1+ 3 i 2 parenrightBig 1 parenleftBig 1 - 3 i 2 + 1 parenrightBig = 3 1 3 i 2 3 + 3 i + 4 9 + 3 + i 5 π 3 1 3 i 2 3 3 i = π 12 bracketleftbigg 2 3 3 (3 3 i ) + 4 i 10 3 3 (3 + 3 i ) bracketrightbigg = 2 π 3 3 , (7.48) or I = 2 π 3 3 . (7.49) 7.8.2 Method 2 An alternative method which is simpler algebraically is the following. Consider contintegraldisplay C dz z 3 + 1 , (7.50) where the contour C is shown in Fig. 7.7. The integral over the arc of the circle at “infinity,” C 2 , evidently vanishes as the radius of that circle goes to infinity. The integral over C 1 is the integral I . The integral over C 3 is integraldisplay C 3 dz z 3 + 1 = integraldisplay 0 d ( xe 2 iπ/ 3 ) ( xe 2 iπ/ 3 ) 3 + 1 = e 2 iπ/ 3 I, (7.51) since ( e 2 iπ/ 3 ) 3 = 1. Thus contintegraldisplay C dz z 3 + 1 = I parenleftBig 1 e 2 πi/ 3 parenrightBig = Ie iπ/ 3 2 i sin π 3 . (7.52)
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7.9. EXAMPLE 6 73 Version of October 26, 2011 2 π/ 3 C 1 C 2 C 3 Figure 7.7: Contour used in the evaluation of Eq. (7.50). The only pole of 1 / ( z 3 + 1) contained within C is at z = e iπ/ 3 , the residue of which is 1 e iπ/ 3 e 1 e iπ/ 3 e i 5 π/ 3 = e - 2 πi/ 3 e - iπ/ 3 e iπ/ 3 e - 3 iπ/ 3 e - 2 iπ/ 3 e 2 iπ/ 3 , (7.53) so I = 2 πie - 6 πi/ 3 (2 i ) 3 ( sin π 3 ) 2 sin 2 π 3 , (7.54) or since sin π 3 = sin 2 π 3 = 3 2 , I = π 4 parenleftbigg 2 3 parenrightbigg 3 = 2 π 3 3 , (7.55) the same result (7.49) as found by method 1. 7.9 Example 6 Consider I = integraldisplay 0 x μ - 1 1 + x dx, 0 < μ < 1 . (7.56) We may use the contour integral contintegraldisplay C ( z ) μ - 1 1 + z dz = integraldisplay 0 e - ( μ - 1) x μ - 1 dx 1 + x integraldisplay 0 e ( μ - 1) x μ - 1 dx 1 + x , (7.57) where C is the same contour shown in Fig. 7.6, and because μ is between zero
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