$ \int_{-\infty}^\infty f(x)e^{-2\pi i\xi x}dx=0$ for all $\xi\in\mathbb R$ with $|\xi|>M $ 
Suppose that $f(z)$ is an entire function on $\mathbb C $, and that there exist constants $M>0$ and $A>0$ such that
  $$ |f(x+iy)|\le\frac{A}{1+x^2}e^{2\pi M|y|}\quad \text{for all}\ x,y\in\mathbb R. $$ Prove that
  $$ \int_{-\infty}^\infty f(x)e^{-2\pi i\xi x}dx=0\quad\text{for all}\ \xi\in\mathbb R\ \text{with}\ |\xi|>M .$$


My attempt:
I have tried to apply the residue theorem. Let $\Gamma$ denote the semi-circle centered at $0$ with radius $R$ lying in the upper half plane. Then
\begin{align}
\int_{\Gamma+[-R,R]}f(z)e^{-2\pi i\xi z}dz&=\int_{-R}^R f(x)e^{-2\pi i\xi x} dx + \int_{\Gamma}f(z)e^{-2\pi i\xi z}dz\\
&=0
\end{align}
since $f(z)e^{-2\pi i\xi z}$ is analytic in $\mathbb C$. We have
$$ \int_{-\infty}^\infty f(x)e^{-2\pi i\xi x}dx=-\lim_{R\to\infty}\int_{\Gamma}f(z)e^{-2\pi i\xi z}dz .$$
However, I failed to show that the right-hand side is zero. 
 A: I suggest using the upper rectangular contour $C(R)$ with vertices at $\pm R$ and $\pm R + Ri$ for $\xi < -M$, and a lower rectangular contour for $\xi > M$. Assume $\xi < -M$. The contour integral $\oint_{C(R)} f(z)e^{-2\pi i \xi z}\, dz$ is zero by Cauchy's theorem, since $f$ is entire. The integrals of $f(z)e^{-2\pi i\xi z}$ along the vertical edges of $C(R)$ are $O(1/R)$ and the integral along the top edge is $O(e^{2\pi(M+\xi)R})$ as $R\to \infty$. Indeed, along the vertical edges, the integral is dominated by $\phi(R) := \frac{A}{1 + R^2} \int_0^R e^{2\pi(M+\xi)y}\, dy$; since $e^{2\pi(M+\xi)y} \le 1$ for $y \ge 0$, $\phi(R) = O(1/R)$. The integral along the upper edge is dominated by $$\int_{-R}^R \frac{A}{1 + x^2} e^{2\pi(M+\xi)R}\, dx = O(e^{2\pi(M + \xi)R})$$ Thus, the integrals along the verticals and top edge are $O(1/R)$ and $O(e^{2\pi(M + \xi)R})$, respectively, as claimed.
Now, in the limit as $R \to \infty$, the result is obtained for $\xi < -M$. A similar argument for the case $\xi > M$ holds using the corresponding lower rectangular contour.
A: Using a rectangular contour gives you better control.  So with a rectangular contour, vertices $\pm R,\pm R+ib$.  We have
\begin{align*}
I&=\int_{-R}^R f(x)\exp(-2\pi i\xi x)\,\mathrm{d}x\\
I_1&=\int_0^b f(R+iy)\exp(-2\pi i\xi (R+iy))i\,\mathrm{d}y\\
I_2&=\int_R^{-R} f(x+ib)\exp(-2\pi i\xi (x+ib))\,\mathrm{d}x\\
I_3&=\int_b^0 f(-R+iy)\exp(-2\pi i\xi (-R+iy))i\,\mathrm{d}y
\end{align*}
and $I=-I_1-I_2-I_3$, so it suffices to show each $I_1,I_2,I_3$ vanish in the limit.
We can easily estimate $I_1, I_3$:
\begin{align*}
\lvert I_1\rvert&\leq\int_0^{\lvert b\rvert} \lvert f(R+i\operatorname{sgn}(b)y)\exp(2\pi\xi (\operatorname{sgn}(b)y-iR))\rvert\,\mathrm{d}y\\
&\leq\int_0^{\lvert b\rvert} \frac{A}{1+R^2}\exp(2\pi M y)\exp(2\pi\xi \operatorname{sgn}(b) y)\,\mathrm{d}y\to 0\\
\lvert I_3\rvert&\leq\int_0^{\lvert b\rvert} \lvert f(-R+i\operatorname{sgn}(b)y)\exp(2\pi\xi (\operatorname{sgn}(b)y+iR))\rvert\,\mathrm{d}y\\
&\leq\int_0^{\lvert b\rvert} \frac{A}{1+R^2}\exp(2\pi M y)\exp(2\pi\xi \operatorname{sgn}(b)y)\,\mathrm{d}y\to 0
\end{align*}
as $R\to\infty$, for every $b$.
Finally, for $I_2$, we have
\begin{align*}
\lvert I_2\rvert&\leq\int_{-R}^R \lvert f(x+ib)\exp(-2\pi i\xi (x+ib))\rvert\,\mathrm{d}x\\
&\leq\int_{-R}^R \frac{A}{1+x^2}\exp(2\pi (\operatorname{sgn}(b)M+\xi)b)\,\mathrm{d}x\\
&\leq\pi A\exp(2\pi (\operatorname{sgn}(b)M+\xi)b)
\end{align*}
Thus $\lvert I\rvert\leq \pi A\exp(2\pi (\operatorname{sgn}(b)M+\xi)b)$, and so taking limit $b\to\pm\infty$ (opposite sign to $\xi$), we have $I=0$ for $\lvert\xi\rvert>M$.
