The usual derivation involves integration by parts technique, while I want a derivation using differentiation of definite integral. Is there a way? $$\Gamma(x) = \int_0^\infty t^{x-1}e^{-t}dt\,= (x-1)!$$ Wikipedia has the usual derivation using integration by parts, but can't this be done using differentiation of definite integral?

  • $\begingroup$ @Rohan come on man. I have seen this already. This is not what I am asking. What I want is a derivation directly starting from the expression $$\int_0^\infty t^{x-1}e^{-t}dt\,$$ and not starting from arbitrary function $$\int_0^{\infty}e^{-ax}dt$$ $\endgroup$ – Vivek Dec 23 '17 at 12:25
  • $\begingroup$ You could try to differentiate the parameter $x$. However the tools involved with integral with parameters are not so easy to use, let alone those involved with improper integrals, since you have to check so many things, which is tedious in more general cases. P.S. we usually do not write $\Gamma(x) = (x-1)!$, since this only holds for integers which are commonly denoted by $n$. $\endgroup$ – xbh Dec 23 '17 at 12:25
  • $\begingroup$ I tried but I am getting an additional term of log(x-1) $\endgroup$ – Vivek Dec 23 '17 at 12:28
  • $\begingroup$ I am convinced that you cannot prove that property by differentiating with respect to $x$. That kind of reasoning would work if Gamma satisfied a differential equation, but as you can see here there are no algebraic differential equations satisfied by Gamma. $\endgroup$ – Giuseppe Negro Dec 23 '17 at 12:42
  • $\begingroup$ That's unfortunate, but the definition of $\Gamma(x)$ involves $t^x$ whose derivative with respect to $x$ is not so easy to integrate or differentiate, so this computational method may not work for large numbers. $\endgroup$ – xbh Dec 23 '17 at 13:03

First note that $$\int_0^\infty e^{-tr}dt\,= \frac 1r$$

Now differentiate $n-1$ times with respect to $r$ (within the integral sign on the left) to obtain:$$(-1)^{n-1}\int_0^\infty t^{n-1}e^{-rt}dt\,= (-1)^{n-1} \frac {(n-1)!}{r^n}$$ and evaluate at $r=1$.

  • $\begingroup$ I have seen this already. This is not what I am asking. What I want is a derivation directly starting from the expression $$\int_0^\infty t^{n-1}e^{-rt}dt\,$$ and not starting from arbitrary function $$\int_0^{\infty}e^{-rt}dt$$ Is there a way? $\endgroup$ – Vivek Dec 23 '17 at 12:31
  • $\begingroup$ @Vivek: come on, this answer is exactly what you are looking for. Does starting with "first note that ..." annoy you? Then just move it to the end of the post, so the computation starts with the differentiating of $\int_0^\infty t^{n-1}e^{-rt}\, dt$. $\endgroup$ – Giuseppe Negro Dec 23 '17 at 12:38
  • $\begingroup$ @GiuseppeNegro If I start from $$ \int_0^\infty t^{n-1}e^{-rt}\, dt $$ how many times should I have to differentiate it with respect to 'r'? Can you show me the step by step process? I am not a PhD like you. I am still in school. $\endgroup$ – Vivek Dec 23 '17 at 12:55
  • $\begingroup$ @GiuseppeNegro Don't use the fact that $$ \int_0^\infty e^{-rt}dt\,= \frac 1r $$ in doing so you are using integration by parts in a trivial manner. $\endgroup$ – Vivek Dec 23 '17 at 13:04
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    $\begingroup$ @Vivek The whole point is to start with a simple expression on the right-hand side which you can differentiate easily lots of times. If you start with an integral involving $t^{n-1}$ the elementary way to reduce the power of $t$ is to use integration by parts. Differentiation increases the power. If you want to use differentiation, therefore, you have to start with a low power of $t$ like $t^0=1$. $\endgroup$ – Mark Bennet Dec 23 '17 at 14:14

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