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paddo
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How would you integrate sin(z)/(z-1)^2 using Cauchy's Integral Formula? 1 is in C.
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To integrate sin(z)/(z-1)^2 using Cauchy's Integral Formula, it's essential to recognize that the contour must enclose both z = 1 and z = i, although the focus is primarily on z = 1. The formula indicates that the term 1/(z-1)^2 can be confusing, but it is crucial for determining the coefficients in the expansion. The integration requires identifying the coefficient a_{-1} from the series expansion of sin(z) around z = 1. Understanding these coefficients is key to successfully applying the integral formula.
First, I tried to show that ##f_n## converges uniformly on ##[0,2\pi]##, which is true since ##f_n \rightarrow 0## for ##n \rightarrow \infty## and ##\sigma_n=\mathrm{sup}\left| \frac{\sin\left(\frac{n^2}{n+\frac 15}x\right)}{n^{x^2-3x+3}} \right| \leq \frac{1}{|n^{x^2-3x+3}|} \leq \frac{1}{n^{\frac 34}}\rightarrow 0##. I can't use neither Leibnitz's test nor Abel's test. For Dirichlet's test I would need to show, that ##\sin\left(\frac{n^2}{n+\frac 15}x \right)## has partialy bounded sums...
It's easy to see that any polynomial is function of that class. Also it seems that composition of exponent and polynomial is good as well. G_f(a, b) should be equal G_f(b, a) as the f(a+b) is symmetrical.
Does anyone have information about this or at least related to it?