Let $I \subseteq \mathbb{R}$ be an interval and $g: I \to \mathbb{C}$ continuous. Define $f: \mathbb{C} \backslash \overline{Im(f)} \to \mathbb{C}$ by

$f(z) := \int_I \frac{1}{g(x) - z} dx$

(with $\overline{Im(f)}$ being the closure of $Im(f)$.)

I now want to show that $f$ is analytic, and rewrite it into a power series that's (locally) defined for each $z_0 \in \mathbb{C} \backslash \overline{Im(f)}$.

Now I must admit that I don't really know how to get started. I haven't dealt much with analytic functions before. I know that a complex function is per definition analytic iff it can be written as a power series (therefore, by completing the second part of the task, the first one would follow, although I don't really know how I could write the function a), and iff it is differentiable once (hence differentiable infinitely often).

Therefore, it would also be sufficient for the first part to show that $f$ is differentiable, I think? How do I show that though? I would need to differentiate by $z$, whereas $f$ is defined as an integral with respect to $x$. I'm rather confused by this function.

  • $\begingroup$ Just a soft comment: $Im( f)$ is not great notation because in complex analysis, the first thing that comes to mind is the imaginary part of $f.$ $\endgroup$ – zhw. Oct 21 '15 at 23:35
  • $\begingroup$ OP: Did you actually check the approach in the answer you accepted? If you did so seriously, you probably met some problems... $\endgroup$ – Did Oct 28 '15 at 23:28
  • $\begingroup$ @Did Well I must admit that I just read the first part briefly, as the Cauchy-Riemann equations weren't introduced to me at that point, and seeing as the second part of the post answered my question in a shorter way (it showed that $f$ is analytic, since it has a power series representation). The Cauchy-Riemann equations were introduced to me this week and I planned to also work through the first part of the answer, but seeing as the second part was sufficient to answer the question, I didn't want to keep the poster waiting with accepting the answer. Thanks for pointing that out, though. $\endgroup$ – moran Oct 29 '15 at 18:35
  • $\begingroup$ @moran, I fix the problem and now it is perfectly right. $\endgroup$ – hermes Nov 5 '15 at 1:33

Let $g(t)=g_1(t)+ig_2(t)$ $$ f(z) = \int_I \frac{1}{g(t) - z} dt=\int_I \frac{1}{g(t) - (x+iy)} dt=u+vi $$ where $$ u=\int_I \frac{g_1(t) - x}{(g_1(t) - x)^2+(g_2(t) - y)^2} dt,\quad v=-\int_I \frac{g_2(t) - y}{(g_1(t) - x)^2+(g_2(t) - y)^2} dt $$ By Cauchy-Riemann equation $$ u_x=v_y=\int_I\frac{(g_1(t) - x)^2-(g_2(t) - y)^2}{((g_1(t) - x)^2+(g_2(t) - y)^2)^2}dt $$ $$ u_y=-v_x=\int_I\frac{2(g_1(t) - x)(g_2(t) - y)}{((g_1(t) - x)^2+(g_2(t) - y)^2)^2}dt $$ So $f$ is analytic. And $$ f(z) = \int_I \frac{1}{g(t)(1 - z/g(t))} dt=\int_I\sum_{n=0}^{\infty} \frac{z^n}{g(t)^{n+1}} dt=\sum_{n=0}^{\infty}\int_I \frac{dt}{g(t)^{n+1}} z^n $$

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  • $\begingroup$ First, this assumes that $g$ is real-valued to identify the real part and the imaginary part of $f$, although the question states that $g$ is complex-valued. Second, none of the four derivatives $u_x$, $v_y$, $v_x$, $u_y$ is what is written (and what would be $t$ in these formulas anyway?) $\endgroup$ – Did Oct 28 '15 at 23:27
  • $\begingroup$ ((Gotta love the sound of silent edits in the evening.)) $\endgroup$ – Did Oct 29 '15 at 10:03

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