Complex function as integral

Question

I have $$f _k(z)=\int_0^1e^{-t^2}t^{\frac z k}\frac {dt} t$$ I must show that $f_k$ is holomorphic on the set $\Pi= \{z:\mathrm {Re} (z)\gt0\}$. There is a theorem stating that if $e^{-t^2}t^{\frac z k}\frac {1} t $ was jointly continuous in $\Pi ×[0,1]$, and holomorphic in $z $ for every $t\in [0,1]$, then $f_k$ would be holomorphic in $\Pi$.

However here we have a function which is not continuous in $t=0,z=\frac 1 2$ (for example), while on the other side, for each fixed $t $, it is entire in $z$. So I would say that there is an other way to show that $f_k $ is holomorphic in $\Pi $. Thank you in advance

Answer 1

Let $h_n \to 0$ a sequence if complex numbers and $w \in \Bbb{C}$

$$\frac{f_k(w+h_n)-f_k(w)}{h_n}=\int_0^1 \frac{1}{h_n}e^{-t^2}[t^{\frac{w+h_n}{k}}-t^{\frac{w}{k}}]\frac{dt}{t}=:\int_0^1g_n(t)dt$$

Note that $g_n(t) \to \frac{e^{-t^2}}{t}t^{\frac{w}{k}}\frac{\ln{t}}{k}$

Now note that $|t^{w/k}|=|e^{\frac{w\ln{t}}{k}}|=e^{\Re{w}\frac{\ln{t}}{k}}=t^{\Re{w}/k}$

Thus $\frac{|t^{w/k}|}{t}=t^{\frac{\Re{w}}{k}-1}$ and since $\Re{w}>0$ we have that $\frac{\Re{w}}{k}-1>-1$

Now $\frac{|t^{\frac{h_n}{k}}-1|}{|h_n|}=\frac{|e^{\frac{h_n}{k}\ln{t}}-1|}{|h_n|} \leq \frac{1}{|h_n|} |\frac{h_n}{k}||\ln{t}|e^{|\frac{h_n}{k}||\ln{t}|}=\frac{1}{k}|\ln{t}|t^{\frac{M}{k}}$

Here we used that $|h_n| \leq M$ since it converges and also the inequality $|e^z-1| \leq|z|e^{|z|}$

Combining all these estimations we have that $|g_n(t)| \leq \frac{1}{k}e^{-t^2}t^{\frac{M+\Re{w}}{k}-1}|\ln{t}| \in L^1(0,1)$

So by dominated convergence we have the conclusion.