Let $$f(z)=\int_0 ^1 t^z dt.$$ Prove that $f$ is holomorphic on $\{\Re(z)>-1\}$.
My attempt: First notice that $$|t^z|=|e^{z\log(t)}|=e^{\Re(z\log(t))}=e^{\log(t)\Re(z)}=t^{\Re(z)},$$ and thus $$\int_0^1 |t^z|dt=\int_0^1 t^{\Re(z)}dt<\infty$$ for $\Re(z)>-1$. We know $f$ is well defined. Now, if we knew $f$ was continuous, we could apply Morera's theorem and Fubini's theorem and obtain that $f$ is holomorphic - I have already worked that out. However, I'm having trouble showing $f$ is continuous. I started with $$\left|f(z_2)-f(z_1)\right|=\left|\int_0^1 (t^{z_2}-t^{z_1})dt\right|\leq\int_0^1 \left|t^{z_2}-t^{z_1}\right|dt=\int_0^1 \left|t^{z_1}\right|\cdot\left|t^{z_2-z_1}-1\right|dt$$ and got stuck. What is a good way to continue from here? Thank you.
If anyone is interested in the solution, I think I found it $$\left|f(z_2)-f(z_1)\right|=\left|\int_0^1 (t^{z_2}-t^{z_1})dt\right|\leq\int_0^1 \left|t^{z_2}-t^{z_1}\right|dt=\int_0^1 \left|t^{z_1}\right|\left|t^{z_2-z_1}-1\right|dt$$ Expand $t^{z_2-z_1}-1$: $$\left|t^{z_2-z_1}-1\right|=\left|e^{\left(z_2-z_1\right)\log{t}}\right|=\left|\sum_{n=0}^{\infty}\frac{(z_2-z_1)^n\log^nt}{n!}-1\right|\leq\sum_{n=1}^{\infty}\left|\frac{(z_2-z_1)^n\log^n{t}}{n!}\right|\leq|\log{t}||z_2-z_1|\sum_{n=1}^{\infty}\frac{\left|(z_2-z_1)\log{t}\right|^{n-1}}{n!}\leq|\log{t}||z_2-z_1|\left(e^{|z_2-z_1|\log{t}}-1\right)=-\log{t}\cdot|z_2-z_1|\left(t^{|z_2-z_1|}-1\right)$$ and therefore $$\int_0^1 \left|t^{z_1}\right|\left|t^{z_2-z_1}-1\right|dt\leq|z_2-z_1|\int_0^1 -\log{t}\cdot\left|t^{z_1}\right|\left(t^{|z_2-z_1|}-1\right)dt\longrightarrow0$$ for $|z_2-z_1|\to0$.