# Is there an integral representation for $\frac{1}{n!}$?

I know that $n!$ has various integral representations, for instance the $\Gamma$ function. I was wondering if $\frac{1}{n!}$ has an integral representation.

• An integral representation in which the integrated is NOT in a factorial form!! – Ben Jun 24 '15 at 18:03
• Maybe through the Fourier transform? – ogogmad Jun 24 '15 at 18:08
• I don't see how. – Ben Jun 24 '15 at 18:09
• How about where the limits are a factorial. I.e. $\int_0^{1/n!} dx$ :) – user223391 Jun 24 '15 at 19:19

$$\dfrac{1}{2\pi} \int_0^{2\pi} e^{e^{i\theta}} e^{-in\theta}\; d\theta$$

• How can I show that? – Irddo Jun 24 '15 at 18:08
• Express it as a contour integral around the unit circle, and use the Residue Theorem. – Robert Israel Jun 24 '15 at 18:09
• Ohh yes, of course! Great idea! – Irddo Jun 24 '15 at 18:09
• what exactly gives you trouble? – user190080 Jun 24 '15 at 18:35

Following @RobertIsrael, we have

$$\frac{1}{2\pi}\int_0^{2\pi}e^{e^{i\phi}}e^{-in\phi}d\phi=\oint_{|z|=1}e^{z}z^{-n}\frac{dz}{iz}\tag 1$$

We note that the integrand on the right-hand side of $(1)$ has a pole of order $n+1$ at $z=0$. The residue is given by

$$\text{Res}\left(-i\frac{e^z}{z^{n+1}},z=0\right)=\frac{1}{n!}\lim_{z\to 0}\frac{d^n}{dz^n}\left(z^{n+1}\frac{-ie^z}{z^{n+1}}\right)=\frac{-i}{n!}$$

Putting it all together reveals that

$$\frac{1}{2\pi}\int_0^{2\pi}e^{e^{i\phi}}e^{-in\phi}d\phi=2\pi i \frac{-i}{2\pi n!}=\frac{1}{n!}$$

as expected!

• @Ben You're welcome. My pleasure. And thanks to Robert Israel for the inspiration. – Mark Viola Jun 24 '15 at 18:55

Take Cauchy's differentiation formula: $$f^{(n)}(a) = \frac{n!}{2\pi i} \oint_\gamma \frac{f(z)}{(z-a)^{n+1}}\, dz$$ and plug a holomorphic $f$ such that $f^{(n)}(a)=1$.

For example, $f(z)=\exp(z)$, $a=0$, and $\gamma$ the unit circle: $$\frac{1}{n!} = \frac{1}{2\pi i} \oint_\gamma \frac{e^z}{z^{n+1}}\, dz$$ Does that count?

• The last formula is the one given by @RobertIsrael. – lhf Jun 24 '15 at 19:32

You can write the inverse Laplace transform of $1/s^{n+1}$, evaluated at $t=1$, as $1/n!$. The integral is $$\int_{c-i\infty}^{c+i\infty}\frac{1}{2\pi is^{n+1}}e^s\,ds=\frac{1}{n!},$$ for suitable real $c$.