Famous mathematician and scientist Carl Gauss developed the Pi function

$$\Pi(x)=\int_{t=0}^\infty t^x e^{-t}\,dt$$

which has the property that $$\Pi(n)=\prod_{k=1}^{n}\left(k\right)=1\cdot2\cdot3\cdots n=n!$$ for all $n\in\mathbb{N}$. In other words, the Pi function generalizes the factorial operator to the real numbers (and complex numbers also).

Wikipedia says the factorial function grows faster than all exponential functions (and therefore faster than all polynomial functions) but not as fast as hyper-exponential functions, otherwise known as tetration or towering. Well, I would like to know exactly how fast it grows.

What is the rate of growth of the Pi function?

$$\Pi^{\prime}(x)=\frac{d}{dx}\int_{t=0}^\infty t^x e^{-t}\,dt$$

Is the FTC applicable here?


The $\Pi$ function is one of my favourites. Using differentiation under the integral sign,

$$\Pi'(x)=\frac{\text d}{\text dx}\int_0^\infty t^x e^{-t}\,\text dt = \int_0^\infty \frac{\partial }{\partial x}t^x e^{-t}\,\text dt = \int_0^\infty t^{x}e^{-t}\log t\,\text dt \tag{1} = \Pi(x) \Psi (x+1)$$

where $\Psi(x)$ is the digamma function.

Repeated differentiation yields

$$\Pi^{(n)}(x)=\int_0^\infty t^x e^{-t}\log^n t \,\text dt \tag{2}$$

An interesting special case of $(1)$ can be found by letting $x = 0$:

$$\Pi'(0) = \int_0^\infty e^{-t}\log t\, \text dt = -\gamma = -0.577216\dots$$

where $\displaystyle \gamma = \lim_{n \to\infty} 1+\frac{1}{2} + \cdots + \frac{1}{n} - \log n\;\;$ is Euler's Constant.


Recall that, if you define $$ \Gamma (z)=\int_0^\infty t^{z-1}e^{-t}\, dt, $$ there results (see here) $$ \Pi(z)=\Gamma (z+1)=z \Gamma (z), $$ and since $$ \frac{d}{dz}\Gamma (z)= \int_0^\infty t^{z-1}e^{-t} \log t \, dt, $$ you deduce that $$ \frac{d}{dz}\Pi (z) = \Gamma (z) + z \int_0^\infty t^{z-1}e^{-t} \log t \, dt. $$

  • $\begingroup$ Please explain line 3. $\endgroup$ – chharvey Nov 10 '12 at 17:44
  • $\begingroup$ Just differentiate under the integral sign. $\endgroup$ – Siminore Nov 10 '12 at 18:31
  • $\begingroup$ Got it. If you can directly differentiate $\Gamma$ then is it possible to directly differentiate $\Pi$ without putting it in terms of $\Gamma$? $\endgroup$ – chharvey Nov 10 '12 at 21:47
  • $\begingroup$ Of course. I proposed this approach since the Gamma function is usually well-known. $\endgroup$ – Siminore Nov 11 '12 at 9:10

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