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The gamma function is expressed as $n! =\Gamma(n+1)= \int_{0}^{\infty}x^ne^{-x}dx$

which reminds me of the Taylor series for $e^x = \sum_{n=0}^{\infty}\frac{x^n}{n!}$

Unmathematically, this makes the gamma function make sense to me: you're effectively doing a 'rearrangment' to get $n!$ However, if I am right about my intuition, I'm feeling particularly foolish about it, as I can't for the life of me properly explain what is precisely going on to get you from one to the other.

So, to you more experienced mathematicians out there, am I correct about my intuition, and if so, could I possibly trouble you to tell me exactly how you get from the Taylor series to the gamma function?

fruitless fruit juice
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    By Cauchy's formula, $$ \frac{1}{{n!}} = \frac{1}{{2\pi i}}\oint_{(0 + )} {e^t t^{ - n - 1} dt} . $$ You can expand the contour of integration to an infinite loop around the negative axis (Hankel-type contour). After collapsing the contour on the negative axis, employing the reflection formula for the gamma function, you will arrive at Euler's integral for $n!$. – Gary Mar 18 '21 at 14:08
  • The Gamma integral is easily computed by integration by parts. –  Mar 18 '21 at 14:30
  • Related question: https://math.stackexchange.com/q/3917532 – pregunton Mar 18 '21 at 14:51
  • @Gary would you mind me asking you to provide a slightly more in-depth explanation of your answer? – fruitless fruit juice Mar 18 '21 at 15:02
  • Alright, so after a little digging, I'm lead to believe that you can get from one to the other using Laplace transforms. I've put writing out the proper answer on my todo list, and will happily post it unless someone does something similar before me. – fruitless fruit juice Mar 18 '21 at 20:12
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    @ArmelFrançois Here is a detailed derivation: https://www.damtp.cam.ac.uk/user/md327/fcm_3.pdf – Gary Mar 19 '21 at 08:56

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It is true that on one hand,

$$\int_0^\infty\frac{x^n}{n!}e^{-x}dx=1$$ and on the other

$$\sum_{n=0}^\infty\frac{x^n}{n!}e^{-x}=1.$$

But the integral is on $x$, while the sum is on $n$, and I just see a coincidence here.

You can modify the argument of the exponential to ensure convergence and establish for $s>1$

$$\sum_{n=0}^\infty\int_0^\infty\frac{x^n}{n!}e^{-sx}dx=\sum_{n=0}^\infty s^{1-n}=\frac1{s-1}=\int_0^\infty e^{(1-s)x}dx=\int_0^\infty\sum_{n=0}^\infty\frac{x^n}{n!}e^{-sx}dx.$$

  • (+1) It is just coincidence. – Mark Viola Mar 18 '21 at 14:31
  • Perhaps you're right about it being coincidence. The potential (pseudomathematical) explanation I had in mind was the along the lines of the fact that for every $x$ you have a corresponding $n!$ sum, and that you only need to go from $0$ to $\infty$ because of the way $x$ and $-x$ are used that leads the formula to encompass all $x$ values. Effectively, you're sort of doing $x^n(\frac{n!}{x^n})$ to give $n!$ – fruitless fruit juice Mar 18 '21 at 14:46
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    It is hard to say if this is just coincidence unless we know the general solution to the equation system $\sum a_n x^n=\frac1{f(x)}$, $\int x^n f(x) dx=\frac1{a_n}$. – user Mar 18 '21 at 14:48
  • @user. Right. With $f(x)$ a unit square, $\int_0^\infty x^nf(x) dx=\int_0^1 x^n dx=\frac1{n+1}$. But $\sum_{n=0}^\infty(n+1)x^n=\frac1{(x-1)^2}$ so it does not work with any function. –  Mar 18 '21 at 15:35
  • @user: by the way, the inverse coefficients of an entire series are not known to be given by the integral of $x^n$ over the function it defines, $$\dfrac1{a_n}=\displaystyle\int_0^\infty\dfrac{x^n}{\displaystyle\sum_{k=0}^\infty a_kx_k}dx\ ?$$ –  Mar 18 '21 at 15:45
  • It is an interesting question. Certainly not in a general case. Also the function should tend to infinity rather strong to ensure convergence. – user Mar 18 '21 at 15:54
  • @user: not in a general case = coincidence. –  Mar 18 '21 at 16:15