Problem: Let $f(z) = e^{az}e^{-e^z}$ where $a>0$. Observe that in the strip $\{x+iy:|y|<\pi\}$ the function $f(x+iy)$ is exponentially decreasing as $|x|$ tends to infinity. Prove that $$\hat{f}(\xi)= \Gamma(a+i\xi), \quad\text{for all }\xi\in\mathbb{R}.$$
My attempt:
$$\hat{f}(\xi) = \int_{-\infty}^\infty f(x)e^{-2\pi ix\xi}\,\mathrm{d}x = \int_{-\infty}^\infty e^{ax}e^{-e^x}e^{-2\pi ix\xi}\,\mathrm{d}x.$$ Let $u = e^x$, so $x = \log u$, and $$\hat{f}(\xi) = \int_0^\infty u^{a- 1} e^{-u}u^{-2\pi i\xi}\,\mathrm{d}u = \Gamma(a-2\pi i\xi ).$$
I am confused with the discrepancy and do not know whether there was a mistake in my calculation or the two expressions are actually equivalent.