Why is $f(s)=\int^{b}_{a}\frac{1}{t^s} dt$ holomorphic?

Question

In complex analysis, let $a, b>0$ in $\mathbb R$, $f(s)=\int^{b}_{a}1/t^s dt$, then $f$ is holomorphic for $Re(s)>0$.

If $s\neq 1$, then $f(s)=\frac{a^{1-s}}{(1-s)}-\frac{b^{1-s}}{(1-s)}$, but if $s=1$, then $f(s)=\ln\big(\frac{b}{a}\big)$, they seems quite different in the form, how to prove that it is holomorphic?

Answer 1

For problems like this, you can use Morera's theorem (and Fubini).

For your particular case, check that the discontinuity of $\frac{a^{1-s}}{(1-s)}-\frac{b^{1-s}}{(1-s)}$ at $s=1$ is removable and $\lim_{s\to 1}\cdots=\ln(b/a)$.

Answer 2

Assume $0<a<b$ and write $$f(s):=\int_a^b {1\over {\mathstrut t}^s}\>dt=\int_a^b e^{-s\,\log t}\>dt=\int_{\log a}^{\log b} e^{(1-s)u}\>du\ .$$ Now its obvious that $f$ is an entire function: We can differentiate under the integral sign.