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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.

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    $\begingroup$ It is meaningless. Besides the pole at the origin, $\Gamma(z)$ grows too fast for $\Gamma(z)e^{-sz}$ to be integrable over $\mathbb{R}^+$. $\endgroup$ Oct 1, 2018 at 16:49
  • $\begingroup$ @JackD'Aurizio Yeah, I remember having failed because of that.. $\endgroup$ Oct 1, 2018 at 16:57
  • $\begingroup$ @JackD'Aurizio How about unilateral laplace transform at $\mathbb R^-$? $\endgroup$ Oct 1, 2018 at 16:58
  • $\begingroup$ 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? $\endgroup$ Oct 1, 2018 at 17:03
  • $\begingroup$ @JackD'Aurizio As an attempt to solve the functional equation in this question (math.stackexchange.com/q/2937647/553404) $\endgroup$ Oct 1, 2018 at 22:22

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

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  • $\begingroup$ Yeah, I remember having failed because of that.. $\endgroup$ Oct 1, 2018 at 16:56
  • $\begingroup$ How about unilateral laplace transform at $\mathbb R^-$? $\endgroup$ Oct 1, 2018 at 16:59

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