- Let \(f ( z ) = e ^ { a z } e ^ { - e ^ { z } }\) where \(a > 0\) . Observe that in the strip \(\{ x + i y : ~ | y | < \pi / 2 \}\) the function \(f ( x + i y )\) is exponentially decreasing as \(| x |\) tends to infinity. Prove that
\[ \hat {f} (\xi) = \Gamma (a - 2 \pi i \xi), \quad \text { for all } \xi \in \mathbb {R}. \]
Solution
With the Fourier-transform convention used in this text, \[ \widehat f(\xi)=\int_{-\infty}^{\infty}f(x)e^{-2\pi i x\xi}\,dx. \] For \(f(x)=e^{ax}e^{-e^x}\) and \(a>0\), \[ \widehat f(\xi) =\int_{-\infty}^{\infty}e^{(a-2\pi i\xi)x}e^{-e^x}\,dx. \] Set \(t=e^x\). Then \(dx=dt/t\) and \(x\) runs from \(-\infty\) to \(\infty\) exactly as \(t\) runs from \(0\) to \(\infty\). Therefore \[ \widehat f(\xi) =\int_0^\infty t^{a-2\pi i\xi-1}e^{-t}\,dt =\Gamma(a-2\pi i\xi), \] since \(\Re(a-2\pi i\xi)=a>0\).
The asserted decay of \(f\) on every closed substrip \(|y|\le \pi/2-\delta\) follows directly from \[ |f(x+iy)|=e^{ax}e^{-e^x\cos y}. \] As \(x\to-\infty\) this is \(O(e^{ax})\). On \(|y|\le\pi/2-\delta\) we have \(\cos y\ge\sin\delta>0\), so as \(x\to+\infty\) the factor \(e^{-e^x\cos y}\) gives superexponential decay. Thus the stated strip decay holds.