Simple expression for an integral function with complex parameter

Question

Consider the following integral: $f(z) = \int_0^{\infty} \frac{e^{-t} - e^{-zt}}{t} dt$

(a) Find a simple expression for the analytic function with which $f(z)$ agrees for $|z-1|<1$

I have other questions about this problem. Is this function $f(z)$ analytic? And what is the idea to find this simple expression? Expansion then find the derivatives?

Answer 1

As you have a Frullani integral, namely $$\int_0^\infty \frac{g(at) - g(bt)}{t} \, dt = [g(0^+) - g(\infty)] \ln \left (\frac{b}{a} \right ),$$ with $g(t) = e^{-t}, a = 1$, and $b = z$, one can immediately write down $$f(z) = \ln (z).$$

Answer 2

For $|z-1 | < 1$ and even for $Re(z) > 0$ everything converges and we can say

$$f(z) = \int_0^{\infty} \frac{e^{-t} - e^{-zt}}{t} dt=\int_0^{\infty}\int_1^z e^{-st}ds dt=\int_1^z \int_0^{\infty} e^{-st} dtds= \int_1^z \frac1{s}ds = \log(z)$$

Another way to say it: $f(1) = 0, f'(z) = \int_0^\infty e^{-zt}dt = \frac{1}{z}$

Analyticity is obtained by expanding $\frac{1}{a+z} = \frac{1}{1+z/a}$ as a geometric series, and integrating.