What is the integral representation of the Digamma function?

[DreamWeaver]

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:

Spoiler

Evaluate the double integral

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

in two different ways, and equate the results.

 

[alyafey22]

Spoiler

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 $t\to 0$

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

 

[DreamWeaver]

Spoiler

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)

 

[alyafey22]

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'

 

[DreamWeaver]

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)