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What is $$\mathfrak L (\Gamma(z))$$? And how can it be derived?

$$\Gamma(z)=\int_0^\infty t^{z-1}e^t dt$$

$$\mathfrak L(\Gamma(z)) = \int_0^\infty \int_0^\infty t^{z-1}e^t dt e^{-sz} dz$$

Yes, of course I tried to find on google but I could only find something like laplace transform using the gamma function, and inverse Laplace transform of gamma function, etc..

Also I tried by myself, but no fruit..

This result may be applied to solving gamma functional equation, which is in another question.

Bernard
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KYHSGeekCode
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    It is meaningless. Besides the pole at the origin, $\Gamma(z)$ grows too fast for $\Gamma(z)e^{-sz}$ to be integrable over $\mathbb{R}^+$. – Jack D'Aurizio Oct 01 '18 at 16:49
  • @JackD'Aurizio Yeah, I remember having failed because of that.. – KYHSGeekCode Oct 01 '18 at 16:57
  • @JackD'Aurizio How about unilateral laplace transform at $\mathbb R^-$? – KYHSGeekCode Oct 01 '18 at 16:58
  • Still you have to deal with a lot of poles, so $\int_{\mathbb{R}^-}\Gamma(z)e^{sz},dz$ have to be considered in principal value. In that sense it might be convergent, but what is the actual use of such transform? – Jack D'Aurizio Oct 01 '18 at 17:03
  • @JackD'Aurizio As an attempt to solve the functional equation in this question (https://math.stackexchange.com/q/2937647/553404) – KYHSGeekCode Oct 01 '18 at 22:22
  • I.e. you want to prove Legendre's duplication formula through the Laplace transform. Much better to consider the inverse Laplace transform of the $\Gamma$ function. – Jack D'Aurizio Oct 01 '18 at 22:33
  • Anyway I believe that the Legendre duplication formula is simpler to prove by integrating twice, then exponentiating, the almost-trivial duplication formula for the trigamma function $\psi'(z)=\sum_{n\geq 0}\frac{1}{(n+z)^2}$. – Jack D'Aurizio Oct 01 '18 at 22:36

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You can't find it because it does not exist: $$\lim_{n \to \infty} \frac{n!}{e^{an}}=\infty \quad \forall a \in \mathbb{R}$$

Botond
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