I want to show that the following functions (in the picture) are absolutely and locally uniformly convergent if real part of complex number $s$ is bigger than 1.
Absolute part for zeta function is easy but I am not sure about the absolute convergence of phi function so i tried to prove it as in picture please check it for me. Is it the right name of the theorem that i used in the end. I am not sure about this theorem.
And also tell me how we will check the local uniform convergence for both of them.
Thanks
The name of the theorem is wrong, the Cauchy integral theorem says that the integral of holomorphic functions over certain (null-homologous) closed paths vanishes.
The theorem you use is, as far as I remember, commonly called "integral comparison theorem". However, since your sum extends only over the primes, and not over all natural numbers, you could not (simply) conclude divergence of the sum when the integral diverges. Example:
$$\sum_{p\in \mathbb{P}} \frac{1}{p\log p} < \infty = \int_2^\infty \frac{dt}{t\log t}.$$
But since you conclude the convergence of the sum from the convergence of the integral, everything is okay.
The locally uniform convergence is an easy consequence of the absolute convergence. For every fixed $x_0 > 1$, both sums converge uniformly in the closed half plane $\{ z : \operatorname{Re} z \geqslant x_0\}$.
That follows since the absolute value of each term is monotonically decreasing in the real part of $z$, so
$$\left\lvert\zeta(s) - \sum_{n=1}^N \frac{1}{n^s}\right\rvert =\left\lvert\sum_{n > N}\frac{1}{n^s}\right\rvert \leqslant \sum_{n > N} \frac{1}{n^{\operatorname{Re}z}} \leqslant \sum_{n > N}\frac{1}{n^{x_0}},$$
and the latter sum can be made arbitrarily small by choosing $N$ appropriately. The proof for the locally uniform convergence of $\phi$ is analogous.