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We know $f(x)$ is in Schwartz space if for given $m,k $ non-negative integers the supremum of $ \left|x^{m } f^{(k)}(x) \right| $ over the real numbers is finite. Let $a $ be positive real and $f(t) = (|\Gamma (a+ i t )| )^2 $ is in Schwarz space.

Xetrov
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    Hint: Use the integral representation of the $\Gamma$ function. – Jose27 Mar 22 '17 at 12:31
  • then how to differentiate general times –  Mar 22 '17 at 12:34
  • i am unable to proceed can u hint little more please... –  Mar 22 '17 at 12:39
  • The idea is to differentiate under the integral sign. You will be getting some logarithmic factors each time you differentiate. – Jose27 Mar 22 '17 at 12:48
  • wil i have double integral –  Mar 22 '17 at 12:50
  • because mod power two is there so we have gamma into gamma bar –  Mar 22 '17 at 12:51
  • The Fourier transform sends $S(\mathbb{R}) \to S(\mathbb{R})$ and $\Gamma(s) = \frac{\Gamma(s+1)}{s}$ and $\Gamma(s) = \int_{-\infty}^\infty e^{-e^{-x}} e^{-sx}dx$ for $Re(s) > 0$. Overall for $-\sigma \not \in \mathbb{N}, g(t) = \Gamma(\sigma+it)$ is Schwartz and hence $|g(t)|^2$ too – reuns Mar 22 '17 at 13:48

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