This question was part of my assignment in complex analysis .
Find the largest open set on which $\displaystyle \int_{0}^{1} \frac{1}{1+tz} dt $ is analytic .
I wrote $F(t)= \displaystyle \int_{0}^{1} \frac{1}{1+tz} dt $ and then computing $\dfrac{F(t+h)-F(t)}{h}$ . Then in $F(t+h)$ I will get $\mathrm{d}(t+h)$ which I put equal to $\mathrm{d}t+\mathrm{d}t$ . So , I am getting $3$ integrals.
But there is a confusion: the limit of $F(t)$ is $0$ to $1$ over $\mathrm{d}t$ but due to $\mathrm{d}(t+h)$ inside the integral I am getting limit of $\mathrm{d}h$ also equal to $0$ to $1$ and then I will put the limit $h \rightarrow0$.
After that only calculations are left. So, is my approach correct? If not, kindly tell me what is the mistake and what would be the right approach.
Thanks!!
You can use the Leibniz rule for parametric integrals: If $D\subseteq\mathbb C$ is open, $f:[a,b]\times D\to\mathbb C$ is continuous, and $f_t(z):=f(t,z)$ is analytic on $D$ for all $t\in[a,b]$, then
$$F(z):=\int_a^b f(t,z)\mathrm dt$$
is analytic on $D$. In your specific example, $f(t,z)=\frac{1}{1+tz}$, which is analytic on $\mathbb C\backslash(-\infty,-1]$ for all $t\in[0,1]$, since $f_t$ is analytic everywhere except at $z=-\frac{1}{t}$. So the integral in question is analytic on the domain I mentioned, and it is not defined outside that domain, so that domain is also the largest one on which it is analytic.