Show that $\lim\limits_{N\rightarrow \infty} \int_0^N f(x)dx$ exists and relate the limit $\lim\limits_{N\rightarrow \infty}\int_0^N f(iy)dy.$ Let $f$ be a continuous function on the quadrant $\{z : \Re z \geq 0, \Im z \geq 0\}$ satisfying

*

*$f$ is analytic on $\{z : \Re z>0, \Im z>0\}$;

*$|f(x+iy)|\leq 10e^{−y}$;

*for each $y>0$ we have $f(x+iy)→0$ as $x\to \infty$.

(a) Show that $\lim\limits_{N\rightarrow \infty} \int_0^N f(x)dx$ exists and relate the limit $\lim\limits_{N\rightarrow \infty}\int_0^N f(iy)dy.$
(b) Prove or disprove: under the given hypotheses $f(x)$ must in fact be absolutely integrable on $[0, ∞).$
I am not so sure how to begin this problem.  I can't think of a way to even bound the first integral appropriately.  All I see is that it is less than $10.$  Some help would be appreciated.  Thanks.
 A: Consider the triangles $\Delta_N$ with vertices $0,N,iN$. By Cauchy's integral theorem,
$$\int_{\partial \Delta_N} f(z)\,dz = 0.$$
The estimate $\lvert f(z)\rvert \leqslant 10 e^{-\operatorname{Im} z}$ shows that
$$\int_0^\infty \lvert f(iy)\rvert\,dy < \infty.$$
It thus suffices to see that the integral over the side from $N$ to $iN$ tends to $0$ as $N\to\infty$. For that, fix $\varepsilon > 0$, and $y_0 > 0$ such that
$$\int_{y_0}^\infty 10 e^{-y}\,dy < \frac{\varepsilon}{3}.$$
Let $g_N(y) := f(N-y+iy)$. By the dominated convergence theorem, and the assumption on $f$,
$$\int_0^{y_0} g_N(y)\,dy \to 0.$$
Thus, for all large enough $N$, we have
$$\left\lvert \int_N^{iN} f(z)\,dz\right\rvert < \varepsilon.$$
Put the pieces together to obtain a). For b), I suggest looking for a counterexample.
The given bound suggests that $f$ should contain the factor $e^{iz}$. Now we need to arrange that $f(x+iy) \to 0$ as $x \to \infty$ for every $y$. Dividing by $z$ would do that, but destroy continuity at $0$. But if we shift a little, all is well, thus we can take for example
$$f(z) = \frac{e^{iz}}{1+z}$$
as a function satisfying the conditions that isn't absolutely integrable on $[0,\infty)$.
