The power series for $e^x$ is as follows.

$$e^{x} =\sum ^{\infty }_{n=0}\frac{x^{n}}{n!}$$

If we define $n! = \Gamma(n+1)$, then we have

$$n!=\int ^{\infty }_{0} x^{n} e^{-x} dx.$$

An extremely surface level observation is that if we compare the summand/integrand of each equation to their corresponding left-hand-sides, we end up with the following correspondences.

$$e^x \longleftrightarrow \frac{x^n}{n!} \hspace{2 in} n! \longleftrightarrow \frac{x^n}{e^x}$$

Very naïvely, one could make the observation that the correspondence on the left looks like an algebraic manipulation of the correspondence on the right (multiply both sides by $n!$ and divide by $e^x$).

Is this merely coincidence or is there some deeper connection here?


1 Answer 1


The exponential (actually $e^{-x}$ for convergence reasons) and the $\Gamma$ are Mellin pairs so one transforms into the other with the Mellin and the Mellin inverse transforms.

The second equation above for arbitrary $s-1,\hspace{.1 in} \Re(s) > 0$ rather than $n$ is the definition of the Mellin transform, so it shows that "$(s-1)!" = \Gamma(s)=\int_{0}^{\infty}x^{s-1}e^{-x}dx, \hspace{.1 in} \Re(s) > 0$ is the direct Mellin transform of $e^{-x}$ and one of the reasons for the shift in definition ($n! = \Gamma(n+1)$).

For the inverse it exactly follows by showing the coefficients of the resultant Taylor series are the coefficient of the exponential ones which is analytically reflected in $e^{-x}=\frac{1}{2\pi i}\int_{c-\infty}^{c+\infty}x^{-s}\Gamma(s)ds, \hspace{.1 in} x>0, \hspace{.1 in}$$c>0$ arbitrary.


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