Derivative of the Gamma Function

[Unit]

A very vague question:

What is the derivative of the gamma function?
Here's what I've got, using differentiation under the integral. Can anybody tell me if I'm on the right track? What does my answer mean?

$$\Gamma(z) = \int_0^{\infty} t^{z - 1} \: e^{-t} \; dt$$

The integrand can be expressed as a function $f(z, t) = t^{z - 1} \: e^{-t}$:

$$\frac{d}{dz} \int_{y_0}^{y_1} f(z, t) \; dt = \int_{y_0}^{y_1} \frac{\partial}{\partial z} f(z, t) \; dt$$

Because $e^{-t}$ is a constant of sorts (correct me if I'm wrong), with respect to the partial derivative to z, all this turns into

$$\frac{d}{dz} \Gamma(z) = \int_0^{\infty} t^{z-1} \: e^{-t} \: \ln(t) \; dt$$

But this is useless! Does anybody have any thoughts?

(I'm trying to find all the extrema of the gamma function, ... they look like the follow an exponential curve and I want to see if there is an expression for it)

 

[g_edgar]

$$\frac{d}{dz} \Gamma(z) = \int_0^{\infty} t^{z-1} \: e^{-t} \: \ln(t) \; dt$$

But this is useless!

nevertheless correct. $$\Gamma'$$ cannot be written in simpler ways. You often see $$\Psi(x) = \Gamma'(x)/\Gamma(x)$$ called the digamma function.

(I'm trying to find all the extrema of the gamma function, ... they look like the follow an exponential curve and I want to see if there is an expression for it)

These extrema are for negative x, right? But the integral form converges only for positive x, so it is of no use in this.

 

[n1person]

Yeah, although it would be cool if THIS non-analytical function could have an analytical derivative... alas... :'(

 

[Unit]

wait, n1person, are you hinting that there exist OTHER non-analytical functions that could have an analytical derivative?

p.s. thank you both for your replies

 

[Orion1]

Derivative of the Gamma Function...

What is the derivative of the gamma function?

Derivative to the gamma function according to Mathematica 6:
$$\frac{d}{dz} \Gamma(z) = \Gamma(z) \cdot \psi^{(m)}(z)} = \left( \int_0^\infty t^{z-1} e^{-t} \, dt \right) \left( \int_0^{\infty}\left(\frac{e^{-t}}{t} - \frac{e^{-zt}}{1 - e^{-t}}\right) \, dt \right) \; \; \; m = 0$$

$$\boxed{\frac{d}{dz} \Gamma(z) = \left( \int_0^\infty t^{z-1} e^{-t} \, dt \right) \left( \int_0^{\infty}\left(\frac{e^{-t}}{t} - \frac{e^{-zt}}{1 - e^{-t}}\right) \, dt \right) \; \; \; m = 0}$$

$$\Gamma(z)$$ - Gamma function
$$\psi^{(m)}(z)$$ - Polygamma function
The m^th derivative of the digamma function.
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Reference:
http://en.wikipedia.org/wiki/Gamma_function"
http://en.wikipedia.org/wiki/Digamma_function"
http://en.wikipedia.org/wiki/Polygamma_function"

 

[n1person]

Well any function that is used to define an anti-derivative/integral of an analytical function would have an analytical derivative, right?
$$\operatorname{erf}(x) = \frac{2}{\sqrt{\pi}}\int_0^x e^{-t^2} dt.$$

so the derivative of error function;

$$\frac{\rm d}{{\rm d}x}\,\mathrm{erf}(x)=\frac{2}{\sqrt{\pi}}\,e^{-x^2}.$$