MHB What is the integral representation of the Digamma function?

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The integral representation of the Digamma function is derived from the Gamma function, expressed as $$\psi_0(x) = \frac{d}{dx}\log \Gamma(x) = \frac{\Gamma'(x)}{\Gamma(x)}$$. Dirichlet's integral representation for the Digamma function is given by $$\psi_0(x) = \int_0^{\infty} \frac{1}{z}\left( e^{-z} - \frac{1}{(1+z)^x} \right)\, dz$$. The discussion includes a light-hearted exchange about the speed of proving the representation, with participants joking about the time it takes to respond. Overall, the focus remains on the mathematical derivation and properties of the Digamma function. The conversation highlights both the complexity and the humor in engaging with advanced mathematical concepts.
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For the Gamma function:

$$\Gamma(x) = \int_0^{\infty}t^{x-1}e^{-t}\, dt$$And the Digamma function:

$$\psi_0(x) = \frac{d}{dx}\log \Gamma(x) = \frac{\Gamma'(x)}{\Gamma(x)}$$Prove Dirichlet's integral representation for the Digamma function:$$\psi_0(x) = \int_0^{\infty} \frac{1}{z}\left( e^{-z} - \frac{1}{(1+z)^x} \right)\, dz$$Hint:

Evaluate the double integral

$$\int_{0}^{\infty}\int_{1}^{q}e^{-tz}\, dt\, dz$$

in two different ways, and equate the results.
 
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Consider

$$f(t) = \int^\infty_0 z^{t-1}\left(e^{-z}-\frac{1}{(1+z)^x} \right)dz = \Gamma(t)-\frac{ \Gamma(t) \Gamma(x-t)}{ \Gamma(x)}$$

$$f(t)= \frac{\Gamma(1+t)}{\Gamma(x)}\frac{ \Gamma(x) -\Gamma(x-t)}{t}$$

Now take the limit as $t\to 0$

$$ \frac{1}{\Gamma(x)}\lim_{t \to 0}\frac{ \Gamma(x) -\Gamma(x-t)}{t}= \frac{\Gamma'(x)}{\Gamma(x)}=\psi(x)$$
 
ZaidAlyafey said:
Consider

$$f(t) = \int^\infty_0 z^{t-1}\left(e^{-z}-\frac{1}{(1+z)^x} \right)dz = \Gamma(t)-\frac{ \Gamma(t) \Gamma(x-t)}{ \Gamma(x)}$$

$$f(t)= \frac{\Gamma(1+t)}{\Gamma(x)}\frac{ \Gamma(x) -\Gamma(x-t)}{t}$$

Now take the limit as $t\to 0$

$$ \frac{1}{\Gamma(x)}\lim_{t \to 0}\frac{ \Gamma(x) -\Gamma(x-t)}{t}= \frac{\Gamma'(x)}{\Gamma(x)}=\psi(x)$$
Crikey! That was a very quick proof... Very impressive! (Rock)(Rock)(Rock)
 
Not that quick. Especially if I had used my phone. Don't post lots of interesting questions for otherwise I'll spend the whole day typing... Just kiddin'
 
ZaidAlyafey said:
Not that quick. Especially if I had used my phone. Don't post lots of interesting questions for otherwise I'll spend the whole day typing... Just kiddin'

He he! Just for that, I'm going to post more, not less... (Hug)
 
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