We've got a random sample of iid $X_1,\dots,X_n$. We're testing the mean of $X \sim \mathcal{N}(\mu,\sigma^2)$, where $\sigma^2$ is known. The size of the test $\alpha=0.05$.
$H_0: \mu=0$
$H_1: \mu=v$
By the Neyman-Pearson lemma the Most Powerful test is $\phi(X) = \mathbf{1}_A$ where the set $A =\{ x: \prod_{i=1}^n \frac{f(\mu_1,\sigma^2)}{f(\mu_0,\sigma^2)} > k \} $
simplyfying we can reduce the test to: \begin{equation} \frac{1}{n}\sum_{i=1}^n X_i > \frac{2\sigma^2(\log k+\mu_1^2 n)}{n \mu_1} \end{equation}
where $\mu_1 = v$. Calculating the critical value $k$ we evaluate: $$ \begin{align} \alpha=\mathbb{E}[\phi(X) |H_0] &= \frac{1}{\sqrt{2 \pi \sigma^2}} \int_{-\infty}^{\infty}\phi(x) e^{-\frac{x^2}{2\sigma^2}} dx\\ &=\frac{1}{\sqrt{2 \pi \sigma^2}} \int_{-\infty}^{\infty}\mathbf{1}_A e^{-\frac{x^2}{2\sigma^2}} dx\\ &= \frac{1}{\sqrt{2 \pi \sigma^2}} \int_{k}^{\infty}e^{-\frac{(x-\mu_1)^2}{2\sigma^2}} dx\\ &= \frac{1}{\sqrt{2 \pi }} \int_{\frac{k-\mu_1}{\sigma}}^{\frac{\infty-\mu_1}{\sigma}}e^{-\frac{x^2}{2}} dx\\ &=1-\Phi\Bigg(\frac{k-\mu_1}{\sigma} \Bigg) \end{align} $$
therefore we can derive $k=\mu_1 + \sigma \Phi^{-1}(1-\alpha)$
Is my approach correct, or did I mess up the calculation of the critical value?