Digamma Integral Representation

**Anthonny** · May 12th 2010, 07:45 PM

Hi,

I'm trying to figure out how the integral representation of the Digamma function works.

I understand that the Digamma function is the logarithmic derivative of the Gamma function. Therefore

$\Psi(z)=\frac{\Gamma'(z)}{\Gamma(z)}$

Since $\Gamma(z)=\int^\infty_0e^{-t}t^{z-1}dt$

then I would think
$\Gamma'(z)=\int^\infty_0e^{-t}t^{z-1}\ln(t)dt$

Anyway, the integral expression for the Digamma function is given by

$\Psi(z)=\int^\infty_0\frac{e^{-t}}{t}-\frac{e^{-zt}}{1-e^{-t}}dt$

How would I derive this formula?
Should I instead look at another representation of the Gamma function and work with that?

Thank you.

**chisigma** · May 12th 2010, 10:04 PM

May be that the right way is starting from this 'infinite product'...

$\frac{1}{\Gamma(z)} = z \cdot e^{\gamma z} \cdot \prod_{n=1}^{\infty} (1+\frac{z}{n})\cdot e^{-\frac{z}{n}}$ (1)

Now compute the logarithm of (1), change sign, then differentiate...

Kind regards

$\chi$ $\sigma$

**simplependulum** · May 12th 2010, 10:19 PM

We have $\ln(\frac{b}{a}) = \int_0^{\infty} \frac{ e^{-ax} - e^{-bx} }{x}~dx$

Let $a = 1 ~,~ b=t$

Therefore ,

$\Gamma'(z) = \int_0^{\infty} e^{-t} t^{z-1} \int_0^{\infty} \frac{ e^{-x} - e^{-tx} }{x}~dx dt$

By reversing the order , we have

$\int_0^{\infty} \frac{1}{x} \int_0^{\infty} e^{-t} t^{z-1} ( e^{-x} - e^{-tx}) ~dtdx$

Consider $\int_0^{\infty} e^{-t} e^{-tx} t^{z-1}~dt$

$= \int_0^{\infty} e^{-(x+1)t}t^{z-1}~dt$

Sub. $(x+1)t = u$

we find that it is equal to

$\Gamma(z) \frac{1}{(x+1)^z}$

Back to the derivative of Gamma function ,

If we continue we obtain

$\Gamma(z) \int_0^{\infty}\frac{1}{x} \left[ e^{-x} - \frac{1}{(x+1)^z} \right] ~dx$

Sub. $x+1 = e^{u}$ for the second integral ,

we have $\int_0^{\infty}\frac{1}{x(x+1)^z}~dx = \int_0^{\infty} \frac{ e^{-zx} }{1 - e^{-x} }~dx$

Therefore ,

$\frac{ \Gamma'(z) }{\Gamma(z) } = \int_0^{\infty} \left(\frac{e^{-x}}{x} - \frac{ e^{-zx} }{1 - e^{-x} }\right)~dx$

Edit: Here is what i guess , not sure if it is true .

If we differentiate w.r.t. $z$ again , i find something interesting ...

$\frac{ \Gamma(z) \Gamma''(z) - [\Gamma'(z)]^2 }{ \Gamma^2(z)}= \int_0^{\infty} \frac{ x e^{-zx}}{ 1 - e^{-x} } ~dx$

Sub. $z = 1$ , from L.H.S , giving

$\Gamma''(1)- [\Gamma'(1)]^2$

Since $\Gamma'(1) = - \gamma$ , what it gives us is also equal to $\Gamma''(1)- \gamma^2$ , while from R.H.S we have

$\int_0^{\infty} \frac{x}{ e^x - 1}~dx = \Gamma(2) \zeta(2) = \frac{\pi^2}{6}$ so is that ture

$\Gamma''(1) = \gamma^2 + \frac{\pi^2}{6}$ ?

This method might be a way to obtain $\Gamma^{(n)}(1)$ by differentiating the integral $n-1$ times without a great detour ...

**Anthonny** · May 15th 2010, 11:47 PM

Originally Posted by simplependulum

Sub. $x+1 = e^{u}$ for the second integral ,

we have $\int_0^{\infty}\frac{1}{x(x+1)^z}~dx = \int_0^{\infty} \frac{ e^{-zx} }{1 - e^{-x} }~dx$

Therefore ,

$\frac{ \Gamma'(z) }{\Gamma(z) } = \int_0^{\infty} \left(\frac{e^{-x}}{x} - \frac{ e^{-zx} }{1 - e^{-x} }\right)~dx$

Thank you!

I basically understand the whole proof until this part.
I'm not quite sure how this substitution works out into that integral.

As for this at the end of your post:

$\Gamma''(1) = \gamma^2 + \frac{\pi^2}{6}$

It all seems correct. It's pretty interesting seeing the $\zeta(2)$ in there.

Again thank you.

**simplependulum** · May 15th 2010, 11:55 PM

Originally Posted by Anthonny

Thank you!

I basically understand the whole proof until this part.
I'm not quite sure how this substitution works out into that integral.

As for this at the end of your post:

$\Gamma''(1) = \gamma^2 + \frac{\pi^2}{6}$

It all seems correct. It's pretty interesting seeing the $\zeta(2)$ in there.

Again thank you.

Let me see ...

$\int_0^{\infty}\frac{1}{x(x+1)^z}~dx = \int_0^{\infty} \frac{ e^{-zx} }{1 - e^{-x} }~dx$

Sub. $x + 1 = e^u$

$dx = e^u du$

the bound changes to $(0,\infty)$

so it is equal to

$\int_0^{\infty} \frac{ 1}{ (e^u - 1) (e^u)^z } (e^u du)$

$= \int_0^{\infty} \frac{ e^{-zu}}{ e^u - 1 } (e^u du)$

$= \int_0^{\infty} \frac{ e^{-zu}}{ e^u ( 1- e^{-u}) } (e^u du)$

$= \int_0^{\infty} \frac{ e^{-zu} }{ 1- e^{-u} }~du$

Then change the "dummy" variable

$= \int_0^{\infty} \frac{ e^{-zx} }{ 1- e^{-x} }~dx$

To have a more formal proof , i think we need to show that

$\int_0^{\infty} [\frac{e^{-x}}{x} - \frac{1}{x(x+1)^z}]~dx$

$= \int_0^{\infty} [\frac{e^{-x}}{x} - \frac{ e^{-zx}}{ 1 - e^{-x}}]~dx$

because the integral $\int_0^{\infty} \frac{e^{-x}}{x} ~dx$ is divergent .

**Anthonny** · May 16th 2010, 10:04 AM

Originally Posted by simplependulum

$= \int_0^{\infty} \frac{ e^{-zx} }{ 1- e^{-x} }~dx$

Thanks! I think I just overlooked the substitution process. I completely understand it now though.

Anyway, now I understand the integral representation for $\Psi(z)$

However, I have one more question (and this question is open to anyone). I want to see how

$\Psi(z)=H_{n-1}-\gamma$

Where $H_{n}$ are the Harmonic numbers and $\gamma$ is the Euler-Mascheroni constant.

I want to derive this relationship using the integral representation of the Digamma function. I tried doing substitution, but I'm stuck. Here is my work so far:

For

$\Psi(x)=\int_0^{\infty} [\frac{e^{-t}}{t} - \frac{ e^{-xt}}{ 1 - e^{-t}}]~dt$

Let $u=e^{-t}$ , therefore $-\ln(u)=t$ and $-\frac{1}{u}du=dt$

Substituting, I get

$\int^0_1 [\frac{u}{-\ln(u)}-\frac{e^{-x\ln(u)}}{1-u}][-\frac{1}{u}]du$

which simplifies to

$-\int^1_0 [\frac{1}{\ln(u)}+\frac{u^{-x}u^{-1}}{1-u}]du$

I know that $\gamma=\int^1_0 [\frac{1}{\ln(u)}+\frac{1}{1-u}du$

However, I'm not sure what to do from here. I also know the integral representation for the Harmonic numbers is given by:

$H_n=\int^1_0 \frac{1-x^n}{1-x}dx$ , but I don't see that occurring in the integral.

Of course, my substitution might not be the right choice. The main thing to do is to get from the Digamma integral representation into a form in which the Harmonic integral representation and an integral representing $\gamma$ arises.

Thank you.

**simplependulum** · May 16th 2010, 07:43 PM

Your substitution is perfect

!

so $-\gamma = \Psi(1) = \int_0^{\infty} [\frac{e^{-t}}{t} - \frac{ e^{-t}}{ 1 - e^{-t}}]~ dt$

Also $H_n = \int_0^1 \frac{1 - x^n}{1-x}~dx$

We need to make a further substitution ,

$x = e^{-t}$

$dx = -e^{-t}~dt$

$H_n = \int_0^{\infty} \frac{1 - e^{-nt}}{1 - e^{-t}} e^{-t}~dt$

$= \int_0^{\infty} \frac{e^{-t} - e^{-(n+1)t}}{1 - e^{-t}}~dt$

Therefore ,

$H_n - \gamma = \int_0^{\infty} \frac{e^{-t} - e^{-(n+1)t}}{1 - e^{-t}}~dt + \int_0^{\infty} [\frac{e^{-t}}{t} - \frac{ e^{-t}}{ 1 - e^{-t}}]~ dt$

$= \int_0^{\infty} [\frac{ e^{-t}}{t} - \frac{e^{-(n+1)t}}{1 - e^{-t}}]~dx$

$= \Psi(n+1)$ !!

Or $\Psi(n) = H_{n-1} - \gamma$

$n > 1$

**Anthonny** · May 16th 2010, 07:52 PM

Originally Posted by simplependulum

$= \Psi(n+1)$ !!

Or $\Psi(n) = H_{n-1} - \gamma$

$n > 1$

Thanks!

I see how to apply the substitution even further and understand your explanation.

Thanks on the help on the Digamma function.

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