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5. Show that if f(2) is analytic for |2| < 1+ €, then for any z = reio with r < 1, eit z dt = 0 1– žeit 1 (a) 27 eit -dt = f(z) 1 (b) 27 eit

5. Show that if f(2) is analytic for |2| < 1+ €, then for any z = reio with r < 1, eit z dt = 0 1– žeit 1 (a) 27 eit -dt = f(z) 1 (b) 27 eit

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Trigonometric Integral via Cauchy's Formula • Basic Idea: Switch trigonometric function into rational function, and then use Cauchy's Formula. • Procedure: Given an integral of the type 1 a cos 0 + b sin 0 +c° -d0, a, b, с € C 1. Use the change of variables sin 0 = (: -), dz where z = et0 cos 0 (z+ OP iz and plug in the integral, making it into a curve integral 1 ;dz, y(t) = e"t, t e [0,1] pz² + qz +l' 2. Perform the factorisation pz² + qz +l = p(z – 21)(z – 22) (you can do this thanks to the fundamental theorem of algebra). 3. Identify the position of 21, 2 with respect to the unit circle (the closed contour along which your curve integral is evaluated). 4. Now there are three cases: 1 pz2 + qz +1 (a). if [21|,|22| > 1 then the function f(2) is analytic and the integral vanishes. 1 (note that g(2) (b). if w.l.o.g. [21| > 1 > |22], then take g(z) as a function of z, and 21 is a constant parameter). g is analytic in the unit circle, and apply Cauchy's formula for g(z) at z = 2, i.e. p(z – 21) g(2) z – 22 (c). if [21| < 1,|22| < 1, then the cases is slightly more complicated, and we g(22) = don't discuss this in detail here.
Transcribed Image Text:Trigonometric Integral via Cauchy's Formula • Basic Idea: Switch trigonometric function into rational function, and then use Cauchy's Formula. • Procedure: Given an integral of the type 1 a cos 0 + b sin 0 +c° -d0, a, b, с € C 1. Use the change of variables sin 0 = (: -), dz where z = et0 cos 0 (z+ OP iz and plug in the integral, making it into a curve integral 1 ;dz, y(t) = e"t, t e [0,1] pz² + qz +l' 2. Perform the factorisation pz² + qz +l = p(z – 21)(z – 22) (you can do this thanks to the fundamental theorem of algebra). 3. Identify the position of 21, 2 with respect to the unit circle (the closed contour along which your curve integral is evaluated). 4. Now there are three cases: 1 pz2 + qz +1 (a). if [21|,|22| > 1 then the function f(2) is analytic and the integral vanishes. 1 (note that g(2) (b). if w.l.o.g. [21| > 1 > |22], then take g(z) as a function of z, and 21 is a constant parameter). g is analytic in the unit circle, and apply Cauchy's formula for g(z) at z = 2, i.e. p(z – 21) g(2) z – 22 (c). if [21| < 1,|22| < 1, then the cases is slightly more complicated, and we g(22) = don't discuss this in detail here.
5. Show that if f(2) is analytic for |2| < 1+ €, then for any z = reio with r < 1, (a) eit z dt = 0 1– žeit 27 eit -dt = f(2) (b) f(e“)- eit
Transcribed Image Text:5. Show that if f(2) is analytic for |2| < 1+ €, then for any z = reio with r < 1, (a) eit z dt = 0 1– žeit 27 eit -dt = f(2) (b) f(e“)- eit
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