Suppose that $f$ is a holomorphic function over an open neighbourhood of $\{z\in\mathbb{C}:\operatorname{Im}(z)\in[0,1]\}$, $f|_\mathbb{R}\in L^1(\mathbb{R})$ and $$ \lim_{R\to\pm\infty}\sup_{t\in[0,1]}|f(R+{\rm i}t)|=0. $$ Can we conclude that $x\mapsto f(x+(1-\varepsilon){\rm i})\in L^1(\mathbb{R})$ for $\varepsilon\in(0,1]$?
By Cauchy's integral theorem, we know that $\displaystyle\int^{+\infty}_{-\infty}f(x+(1-\varepsilon){\rm i}){\rm d}x$ exists as an improper integral, but this is not enough to deduce that $x\mapsto f(x+(1-\varepsilon){\rm i})\in L^1(\mathbb{R})$. Also, we cannot conclude that $x\mapsto f(x+{\rm i})\in L^1(\mathbb{R})$, as shown by the example $$ f(z)=\dfrac{\sin(z^2)}{z\cosh(2z)} $$ (please correct me there is something wrong with this example). So what about the function $f(x+(1-\varepsilon){\rm i})$? Any help appreciated.
Using Stirling approximation one can easily show that for $c>0$ we have $g(w)= \Gamma(cw)/\Gamma(1-cw) \sim |t|^{2c \sigma-1}, w=\sigma+it, |t| \to \infty$ uniformly in any strip $\alpha \le \sigma \le \beta$ and $g$ is analytic in the strip as long as $\Gamma(cw)$ is so as long as $\Re (\alpha) >0$
So if we let $w=-i(z+iK)$ for some $K>0$ we get that $h(z)=g(w)=g(-i(z+iK))$ is analytic in a strip containing $\{z\in\mathbb{C}:\operatorname{Im}(z)\in[0,1]\}$ and on any horizontal line $\Im z =t$ we have $|h(z)| \sim |z|^{2c(t+K)-1}$
Now considering another parameter $a>0$ and the function $f(z)=h(z)(z+iK)^{-a}$ which is analytic in our strip using the principal branch of the logarithm since $z+iK$ stays away from the real axis, we get that $|f(z)| \sim |z|^{2c(t+K)-1-a}$ as $|z| \to \infty$
So if $b>0$ we can get $2cK-1-a=-1-b$ so $2cK-a=-b$ and then $2c+2cK-1-a=2c-1-b$ which we can make negative but close to $0$ if we choose $2c$ close but slightly smaller than $1+b$ while $K=(a-b)/(2c)>0$ as long as $a>b$ so $|h| \sim |z|^{-1-b}, |z| \to \infty$ on the real axis, $|h| \to 0, |z| \to \infty$ in our original strip $\{z\in\mathbb{C}:\operatorname{Im}(z)\in[0,1]\}$ uniformly in $\Im z$, but $h$ won't be integrable on horizontal lines in the strip from some $t<1$ to $1$
This shows that we need some growth condition on $f$ for the wanted result to hold (exponential type essentially, so order at most $1$ and finite type if order $1$, since the counterexample above is of order $1$ but maximal infinity type as $\Gamma$ is such)