Consider a rectangle $R_\epsilon=\{x+iy\in\mathbb{C}\mid 1-\epsilon\leq x\leq 1+\epsilon,0\leq y\leq H\}$ where $\epsilon>0$ is sufficiently small, $H>0$ and $\epsilon$ depends on $H$.
If $f$ is an analytic function on and inside the rectangle $R_\epsilon$ with no zeros or poles on the boundary of $R_\epsilon$ and $f(x+iH)$ is non zero for $1-\epsilon\leq x\leq1+\epsilon$ then prove that there exists an $M>0$ such that$$ \int_{1-\epsilon}^{1+\epsilon}\Im \left\{ \log f(x+iH)\right\}dx\leq 2\epsilon M\tag{1}$$ where $\Im$ denotes the imaginary part.
My try: Since $$ \int_{1-\epsilon}^{1+\epsilon}\Im \left\{ \log f(x+iH)\right\}dx\leq \left| \int_{1-\epsilon}^{1+\epsilon}\Im \left\{ \log f(x+iH)\right\}dx\right|$$ So by triangle inequality $$ \int_{1-\epsilon}^{1+\epsilon}\Im \left\{ \log f(x+iH)\right\}dx\leq \int_{1-\epsilon}^{1+\epsilon}|\Im \left\{ \log f(x+iH)\right\}|dx $$ Take $$M=\max_{1-\epsilon\leq \sigma \leq 1+\epsilon} \left|\Im \left\{ \log f(x+iH)\right\}\right| $$ But note that $\epsilon$ is a function of $H$, so $\epsilon=\epsilon(H)$. So can we write equation (1) when $\epsilon$ depends on $H$? I mean can we write the following $$ \int_{1-\epsilon}^{1+\epsilon}\Im \left\{ \log f(x+iH)\right\}d\sigma\leq M \int_{1-\epsilon}^{1+\epsilon} dx $$
Thank you.