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I want to find two analytic functions (the first one is analytic in the upper half plane the second one in the lower half plane) $f_+(z)$ and $f_-(z)$ which satisfy $f_+(x)-f_-(x)=\frac{1-\cos x}{x}$ and $f_{+-}(x)=\lim_{\epsilon\rightarrow 0}f_{+-}(x+-i\epsilon)$ with $x\in\mathbb R$

Whats the best way to start with these kind of problems? Integrate the RHS over an appropirate contour and see if this gives me something?

EDIT: I want to solve this problem in order to have some idea how to solve the integral equation $f(x)+\frac{\alpha}{i\pi}P.V\int_{-\infty}^{\infty}\frac{f(\xi)}{\xi-x}d\xi=\frac{1-\cos x}{x}$ where $\alpha$ is a constant different that +-1 and $f(x)=f_+(x)-f_-(x), \frac{1}{i\pi}P.V\int_{-\infty}^{\infty}\frac{f(\xi)}{\xi-x}d\xi=f_+(x)+f_-(x)$

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    $\begingroup$ Probably something to adjust. Why can't you put $f_+(z):=1+\frac {1-\cos(z)} z$ and $f_-(z):=1$? $\endgroup$
    – Markiyan Hirnyk
    Feb 23, 2014 at 17:13
  • $\begingroup$ @127.0.9.6: The function $f(z):=0$ is exceptional. For example, its zeros are not isolated. $\endgroup$
    – Markiyan Hirnyk
    Feb 23, 2014 at 17:28
  • $\begingroup$ I edited my post, maybe its clear now what I want to achieve $\endgroup$
    – Alexander
    Feb 23, 2014 at 17:44
  • $\begingroup$ Perhaps you already know this, but I suggest reading up on the Hilbert transform. $\endgroup$
    – froggie
    Feb 28, 2014 at 21:27

1 Answer 1

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Unless there is something I miss, the functions $$ f_+(z)=\frac{1-\cos z}{2z}\quad \text{and}\quad f_-(z)=-\frac{1-\cos z}{2z}, $$ are both entire analytic, and $$ f_+(x)-f_-(x)=\frac{1-\cos x}{x}, $$ for every $x\in\mathbb R$.

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