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I have a theory that uses the gamma function:

$$\Gamma(n)=\int_0^\infty x^{n-1}e^{-x} \space dx$$

Then I was inclined to think that perhaps the derivative is:

$$x^{n-1}e^{-x}$$

But I'm not sure we can just drop the integral along with the bounds to get the derivative. Then I thought about taking the limit:

$$\lim_{x\to\infty}x^{n-1}e^{-x}$$

But now we can't specify at what $x$ value we want to get the rate of change of. At this point I feel like I can't get any further on my own and would appreciate some insight.

EDIT: Looking for derivative in terms of $n$ actually.

Badr B
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  • You should take the derivative with respect to $n$ and not $x$, however you won't be able to solve it. – kingW3 Jan 20 '18 at 00:50
  • are you sure you don't mean the derivative in $n$? – lulu Jan 20 '18 at 00:50
  • @Davy M Thank you very much. Checking it out right now. – Badr B Jan 20 '18 at 00:51
  • @kingW3 Oh yeah you're right haha. Didn't think of that. – Badr B Jan 20 '18 at 00:51
  • @lulu Yep I do haha. Fixing it now. – Badr B Jan 20 '18 at 00:52
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    In statistical physics, Stirling's approximation is often used $x! \approx \sqrt{2\pi x} \left( \frac{x}{e}\right)^x$ to approximate the factorial as being continuous. –  Jan 20 '18 at 00:53
  • Look again in your calculus textbook about the fundamental theorem of calculus. You have applied it incorrectly. It will not help with this derivative. – GEdgar Jan 20 '18 at 00:59
  • @DavyM Just looked through the duplicate post and was surprised to find the Harmonic numbers as well as Euler's constant involved. I was interested in the derivative of $x!$ so I could try deriving a formula that calculated the partial sum of the Harmonic series up until the $nth$ term. – Badr B Jan 20 '18 at 00:59
  • @Alex That's a very nice approximation I've never seen before. Thanks for mentioning it! – Badr B Jan 20 '18 at 01:01
  • @GEdgar Sorry I haven't taken calculus yet (as many can probably tell haha). If I'm not mistaken, this would work if it was an indefinite integral, correct? – Badr B Jan 20 '18 at 01:02
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    My recommendation: wait until you have taken calculus before attempting to compute derivatives. – GEdgar Jan 20 '18 at 01:04
  • @GEdgar Sadly that'll be in a few years from now, but I'm still fascinated with calculus and its applications. I try doing a lot of researching and studying on my own time and I think I've gotten fairly decent at differentiation and integration, it was just this particular concept I was unsure of. – Badr B Jan 20 '18 at 01:08
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    If you have $\displaystyle f(n) = \int_\cdots^n g(x),dx,$ then you can "drop the integral" as follows $ f'(n) = g(n).$ But you don't have anything like that here. If you were to "drop the integral," you would get something depending not only on $n$ but also on something called $x.$ What would this thing called $x$ be? By contrast, $\displaystyle \int_0^\infty x^{n-1} e^{-x},dx$ does not depend on anything called $x. \qquad$ – Michael Hardy Jan 20 '18 at 02:31
  • I would offer a similar objection to those offered in the linked duplicate. $x!$ is a function on the integers, and thus talking about its derivative doesn't make sense. $\Gamma(x)$ is a different matter. The fact that it coincides with $(x-1)!$ on the integers doesn't mean $x!$ has a derivative. So while I don't have a problem with any of the derivations here I would suggest your title should be corrected. – Michael Grant Jan 22 '18 at 03:23

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Here is how to calculate it: you have to move the derivative into the integral: \begin{align} \frac{d}{dn}\Gamma(n) &=\frac{d}{dn}\int_0^\infty x^{n-1}e^{-x}\,dx\\ &=\int_0^\infty \frac{d}{dn}x^{n-1}e^{-x}\,dx\\ &=\int_0^\infty e^{-x}\frac{d}{dn}e^{(n-1)\ln(x)}\,dx\\ &=\int_0^\infty e^{-x}\cdot e^{(n-1)\ln(x)}\ln(x)\,dx\\ &=\int_0^\infty x^{n-1}e^{-x}\ln(x)\,dx\\ \end{align} and so we have $$\Gamma'(n)=\int_0^\infty x^{n-1}e^{-x}\ln(x)\,dx$$

Michael Hardy
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Franklin Pezzuti Dyer
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