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Straight to question, how do you analytically integrate factorial function like the following: $\int_0^\infty 1/(n-1)!\, \mathrm dn\ $. This is equivalent to $\int_0^\infty 1/Γ(n)\, \mathrm dn\ $ but how do you integrate something like this. Using grapher shows a nice smooth graph that converges quickly.

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  • $\begingroup$ Use the reflection formula for the Gamma function and then its integral definition $\endgroup$ – Yuriy S Jul 6 '16 at 5:59
  • $\begingroup$ I still don't get it, can you please show me. 🐸 $\endgroup$ – akira kato Jul 6 '16 at 6:14
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    $\begingroup$ The Fransén–Robinson constant. $\endgroup$ – nospoon Jul 6 '16 at 6:54
  • $\begingroup$ Never heard of it, thanks for introducing it to me. $\endgroup$ – akira kato Jul 6 '16 at 12:22
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A hint.

Your integral is:

$$I=\int_0^\infty \frac{dx}{\Gamma(x)}$$

By reflection formula for the Gamma function:

$$\frac{1}{\Gamma(x)}=\frac{1}{\pi} \Gamma(1-x) \sin \pi x$$

$$I=\frac{1}{\pi} \int_0^\infty \Gamma(1-x) \sin \pi x~ dx$$

Now by the integral definition of the Gamma function we get:

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

So now the integral becomes:

$$I=\frac{1}{\pi} \int_0^\infty \int_0^\infty \sin (\pi x)~ \frac{e^{-t}}{t^x}~dt ~ dx$$

$$I=\frac{1}{\pi} \int_0^\infty \int_0^\infty \sin (\pi x)~ e^{- \ln(t) x} e^{-t}~dt ~ dx$$

See @nospoon's comment. No closed form is available, apparently, but maybe this hint might still help.

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  • $\begingroup$ I get it now, thanks. $\endgroup$ – akira kato Jul 6 '16 at 12:22

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