# How do I evaluate $\lim_{\Delta z \to 0} \frac{1}{\Delta z}\int_{z_1}^{z_1 + \Delta z}f(t)dt$

How do I evaluate $$\lim_{\Delta z \to 0} \frac{1}{\Delta z}\int_{z_1}^{z_1 + \Delta z}f(t)dt$$ in order to finish this proof?

Given $$f(z)$$ is a continuous function on domain $$D$$, and given that for any contour $$C$$ connecting $$z_1,z_2\in D$$

$$\int_C f(z)dz = \int_{z_1}^{z_2}f(t)dt = F(z_2) - F(z_2)$$

Then $$\forall z\in D$$, $$F'(z) = f(z)$$

My work

1.) Take any two points $$z_1, z_2 \in D$$. Let $$z_2 = z_1 +\Delta z$$.

2.) Fix any countour $$C$$ connecting $$z_1$$ and $$z_2$$

Then by hypothesis

$$\int^{z_1 + \Delta z}_{z_1} f(t)dt = F(z_1 + \Delta z) - F(z_1)$$

and

$$\frac{1}{\Delta z}\int^{z_1 + \Delta z}_{z_1} f(t)dt = \frac{1}{\Delta z}[F(z_1 + \Delta z) - F(z_1)]$$

which gives the difference quotient on the RHS. Therefore

$$\lim_{\Delta z \to 0} \frac{1}{\Delta z}\int^{z_1 + \Delta z}_{z_1} f(t)dt = F'(z_1)$$

end of my work

It looks like I'm really close. If this were a real valued function, I could use the IVT for integrals to finish this proof. But since we haven't proved the complex analog of the IVT in my class, I'm unsure as how to make this final leap.

edit: my professor says I can "temporarily fix $$t$$" in the last line and therefore move $$f(t)$$ outside of the integrand... Which I see yields the desired result. But I'm pretty skeptical because he couldn't justify why fixing what is already a dummy variable is allowed.

edit 2: The source of this question comes from page 142 of Complex Variables and Applications 8th edition by Brown and Churchill. 3 statements are made and claimed to be equivalent. They show a implies b, b implies c, and c implies a. The proof in this question is me trying to prove b implies a, which is not done in the textbook.

Beware that your function $$F$$ is not uniquely defined if $$f$$ is not holomorphic! If $$f$$ was holomorphic, you could simply use the fact that $$F'=f$$ to conclude.
To get the continuous case, you could also start by using the continuity of $$f$$ at $$z_1$$. Indeed, $$\forall \varepsilon>0,\exists \eta>0, \forall z \in \mathbb C, \quad |z-z_1|<\eta \Rightarrow |f(z)-f(z_1)|<\varepsilon$$ Let $$\varepsilon>0$$, let such a $$eta>0$$ ; for all $$|\Delta z |< \eta$$, $$\left| \frac{1}{\Delta z} \int_{z_1}^{z_1+\Delta z} f(z)\mathrm d z -f(z_1)\right| = \left| \frac{1}{\Delta z} \int_{z_1}^{z_1+\Delta z} (f(z)-f(z_1))\mathrm d z\right| \leq \left|\frac{1}{\Delta z}\int_{z_1}^{z_1+\Delta z_1} \eta\right| = \eta$$ Thus $$\lim_{\Delta z\rightarrow 0} \left|\frac{1}{\Delta z} \int_{z_1}^{z_1+\Delta z} f(z)\mathrm d z -f(z_1)\right| =0$$ and thus $$\lim_{\Delta z\rightarrow 0} \frac{1}{\Delta z} \int_{z_1}^{z_1+\Delta z} f(z)\mathrm d z =f(z_1)$$
• I cannot see how the first equality holds. You brought a constant $f(z_1)$ which was previously subtracted from the integrand on the LHS to be included in the integrand on the RHS. From that step on, I understand completely! Can you just clarify why the equality holds. Thanks! – rocksNwaves Oct 31 '18 at 18:03