\u03c0 until it is inside ln z and z 3 4 I 3 \u02c6 ln xe i 2 \u03c0 xe i 2 \u03c0 3 4 1 x dx i I \u02c6

# Π until it is inside ln z and z 3 4 i 3 ˆ ln xe i 2

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π until it is inside ln z and z 3 / 4 I 3 = 0 ˆ ln ( xe i 2 π ) ( xe i 2 π ) 3 / 4 (1 + x ) dx = - i I + ˆ 0 2 πi x 3 / 4 (1 + x ) dx (7) where I used e - i 3 π/ 2 = i , and where the minus sign in the 2nd line arises from flipping the integration limits from the first line. Note that a new integral has popped up on the right- hand-side above, along with the integral that we want to evaluate. So it looks like we are stuck with evaluating the extra integral I extra = ˆ 0 1 x 3 / 4 (1 + x ) dx
7 before we can recover the one we set out to evaluate. But, before we get too upset with having an additional task to perform, let’s substitute Eqs. (6) and (7) into Eqs. (4) and (5), to obtain I C = (1 - i ) × I + 2 πI extra = 2 π 2 (1 + i ) (8) Now here comes a very clever idea, due to Amar Masalmeh: we can simply take the imaginary part of the two sides of the above equation, which eliminates the term with the extra integral, and allows us to immediately read off the value of the integral that we want! We thereby obtain the desired result I = - 2 π 2 At this point, one could also deduce the extra integral, but taking the real part of Eq. (8), and using the result we just obtained for I , to find I extra = 2 π It turns out that we already evaluated an integral of the form of I extra , in class, which provides a cross-check of the above ˆ 0 x p - 1 1 + x dx = π sin πp 0 < p < 1 Setting p = 1 / 4 , we recover our previous result for I extra !

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• Winter '20
• Physics, Unit Circle, Methods of contour integration, Complex Plane, Pole, Branch point

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