Show that $\int\limits_{-\infty}^\infty f(t)dt=0$ where $f\in H^\infty(\mathbb{H})$

Question

The problem is stated as follows: Let $\mathbb{H}$ denote the open upper half plane. Let $f \in H^{\infty}(\mathbb{H})$ Suppose $f$ can be extended to be continuous on $\overline{\mathbb{H}}$ with $$\int_{-\infty}^\infty |f(t)|dt<\infty$$

Show that $$\int_{-\infty}^\infty f(t)dt=0$$

and, the following hint is provided: Remark: Let $C_R$ denote the upper semicircle with radius $R$.

It is not so obvious why $\int_{C_R}f(z)dz\longrightarrow 0$ as $R\longrightarrow \infty$. A technique to overcome this is to first consider the functions $f_\epsilon(z)=f(z)e^{i\epsilon z}$ where $\epsilon>0$

It is straightforward to show that $\int_{C_R}f_\epsilon(z)dz\longrightarrow 0$ as $R\longrightarrow \infty$. The result for $f_\epsilon$ follows immediately. One then lets $\epsilon \longrightarrow 0$ and shows the convergence of the integrals."

The problem that I am having is that,Assuming $$\underset{z\in \mathbb{H}}{\max}|f(z)|=M,$$ we get

$$\begin{align} \left|\int_{C_R}f_\epsilon(z)dz\right|&=\left| \int_{C_R}f(z)e^{i\epsilon z}dz\right| \\ &= \left|\int_{C_R}f(z)e^{i\epsilon(x+iy)}dz\right| \\ &\le \int_{C_R}\left| f(z)e^{-\epsilon y+i\epsilon x}\right|dz \\ &=\int_{C_R}\left| f(z)e^{-\epsilon y}\right|dz \\ &\le \int_{C_R} \frac{M}{e^{\epsilon y}}dz \end{align}$$

My Question:

From here, I just don't see how to make the R.H.S. go to zero. I tried parameterizing it, but then the best I can do for an upper bound is $$ \int_0^\pi \frac{MR}{e^{\epsilon R \sin{t}}}dt$$

And, while this looks a little more promising, I still cant get this integral to go to zero either because, when $t=0$, $\sin{t}=0$. I asked my professor for clarification, and he replied that ``In any case, a straightforward application of ML to the integrand from 0 to π won't work. We had the same integral (after putting in absolute value etc) for one of the contour integral examples we did in class. We got a usable bound by making two very elementary observations on the sin function."

I tried to find this in my notes, but the only integral that looks similar to this one that I can find was quite a bit different, and, there, we were able to apply ML.

Answer 1

Basically from what you did at the end it suffices to prove that , $$\lim_{R\to\infty}\int_0^\pi \frac{R}{e^{\epsilon R \sin{t}}}dt = \lim_{R\to\infty}\int_0^\pi Re^{-\epsilon R\sin t}dt = 0$$

This is true!!! see below First split the integral involves at $t=\frac{\pi}{2}$ then take the change of variable $t'=\pi -t$

$$ \int_0^\pi e^{-\epsilon R\sin t}dt =2\int_0^{\frac{\pi}{2}} e^{-\epsilon R\sin t}dt$$

Afterward shows by studying the function $[0,\frac{\pi}{2}]\ni t \mapsto\frac{\sin t}{t}$ that

$$ \color{blue}{\frac{2}{\pi}t\le \sin\theta \le t,~~~~ ~~ \forall t \in [0,\frac{\pi}{2}] } $$ therefore we get that $$\lim_{R\to\infty}\int_0^{\pi} Re^{-\epsilon R\sin t}dt =\lim_{R\to\infty}2R\int_0^{\frac{\pi}{2}} e^{-\epsilon R\sin t}dt\\\leq\lim_{R\to\infty}2R\int_0^{\frac{\pi}{2}} e^{-\frac{2\epsilon R}{\pi}t}dt =\lim_{R\to\infty}\frac{\pi}{\epsilon}(1-e^{-\epsilon R}) =0$$

see also here: https://math.stackexchange.com/q/2498579