Finding two analytic functions

Question

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 $f(x)=f_+(x)-f_-(x), \frac{1}{i\pi}P.V\int_{-\infty}^{\infty}\frac{f(\xi)}{\xi-x}d\xi=f_+(x)+f_-(x)$

Probably something to adjust. Why can't you put $f_+(z):=1+\frac {1-\cos(z)} z$ and $f_-(z):=1$? — Markiyan Hirnyk, Feb 23 '14 at 17:13
@127.0.9.6: The function $f(z):=0$ is exceptional. For example, its zeros are not isolated. — Markiyan Hirnyk, Feb 23 '14 at 17:28
I edited my post, maybe its clear now what I want to achieve — Alexander, Feb 23 '14 at 17:44
Perhaps you already know this, but I suggest reading up on the Hilbert transform. — froggie, Feb 28 '14 at 21:27

score 0 · Answer 1 · answered Mar 01 '14 at 17:23

0

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 $x\in\mathbb R$.

answered Mar 01 '14 at 17:23

Yiorgos S. Smyrlis

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Finding two analytic functions

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