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I know this may seem like a really low level/silly question, I apologize in advance.

I do not know how to differentiate the gamma,beta,digamma function, for ex: $$ \frac{ d^2}{d^2 x} \frac{\Gamma(x+1) }{\Gamma(x+3)} $$ how would we do this? Thank you for the help.

I am not sure what to do. I know $$ \frac{d}{d x}\frac{ \Gamma'(x+1)\Gamma(x+3)-\Gamma(x+1)\Gamma'(x+3)}{\big(\Gamma(x+3)^2\big)}=... $$ I can do the second derivative too, I just am not familiar with how to do this and I notice it a lot. THanks

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  • $\begingroup$ I do not know how to differentiate $\Gamma(x)$. That is my point @achillehui $\endgroup$ – user143444 May 13 '14 at 2:53
  • $\begingroup$ @WillJagy I am looking for a proof to the question I posted. I do not follow . Thanks a lot $\endgroup$ – user143444 May 13 '14 at 3:01
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You don't need anything that complicated. Just show that by integration by parts

$$\frac{\Gamma(x+1)}{\Gamma(x+3)} = \frac{1}{(x+1)(x+2)},$$

and then differentiate.

Yes, this works for any real $x>0$.

Detail: From the definition of the Gamma function,

$$\forall x \geq 0: \Gamma(x+1)=\int_0^\infty t^x e^{-t} dt = [-t^x e^{-t}]_0^\infty + x \int_0^\infty t^{x-1} e^{-t} dt = x \Gamma(x).$$

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  • $\begingroup$ How do you integrate that by parts ? Thanks $\endgroup$ – user143444 May 13 '14 at 3:16
  • $\begingroup$ I've edited my response to answer your question. $\endgroup$ – JPi May 13 '14 at 3:19
  • $\begingroup$ Okay @Jpi But how does this apply to $\Gamma(x+3)$, it seems $\Gamma(x+1)$ is simplest. Thanks +1 I will check it as answer if you can elaborate this for me thank you $\endgroup$ – user143444 May 13 '14 at 3:27
  • $\begingroup$ It works for any $x$. You can simply replace $x+1$ with $x+3$ and get $\Gamma(x+3)=(x+2)\Gamma(x+2)$ and hence $\Gamma(x+3)=(x+2)\Gamma(x+2)=(x+2)(x+1)\Gamma(x+1)$. $\endgroup$ – JPi May 13 '14 at 3:29
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By, one of the many, definition the digamma function is: \begin{align} \psi(x) = \partial_{x} \ \ln\Gamma(x) = \frac{\Gamma^{'}(x)}{\Gamma(x)}. \end{align} Now consider \begin{align} f(x,y,t) &= \ \frac{\Gamma(ax+by)}{\Gamma(cx+dt)} \end{align} for which the following derivatives are seen: \begin{align} \partial_{x} \ f(x,y,t) &= \frac{1}{\Gamma^{2}(cx+dt)} \left[ a \Gamma^{'}(ax+by) \Gamma(cx+dt) - c \Gamma(ax+by) \Gamma^{'}(cx+dt) \right] \\ &= \frac{\Gamma(ax+by)}{\Gamma(cx+dt)} \left[ a \psi(ax+by) - c \psi(cx+dt) \right] \\ &= f(x,y,t) \left[ a \psi(ax+by) - c \psi(cx+dt) \right]. \end{align} Differentiation with respect to $y$ is \begin{align} \partial_{y} \ f(x,y,t) = b \ f(x,y,t) \ \psi(ax+by) \end{align} and differentiation with respect to $t$ is \begin{align} \partial_{t} \ f(x,y,t) &= - d \ f(x,y,t) \ \psi(cx+dt). \end{align}

In the case proposed the following is seen. \begin{align} \partial_{x}^{2} \ \frac{\Gamma(x+a)}{\Gamma(x+a+2)} &= \partial_{x}^{2} \ \frac{\Gamma(x+a)}{(x+a+1)(x+a)\Gamma(x+a)} \\ &= \partial_{x}^{2} \ \frac{1}{(x+a+1)(x+a)} = \partial_{x}^{2} \left[ \frac{1}{x+a} - \frac{1}{x+a+1} \right] \\ &= \frac{1}{(x+a)^{2}} - \frac{1}{(x+a+1)^{2}} = \frac{2x + 2a+1}{(x+a)^{2}(x+a+1)^{2}} \\ &= (2x+2a+1) \left( \frac{\Gamma(x+a)}{\Gamma(x+a+2)} \right)^{2}. \end{align}

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  • $\begingroup$ This is awesome thank you so much $\endgroup$ – user143444 May 13 '14 at 5:47

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