Digamma Integral Representation

Apr 2010
50
2
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

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

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

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

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

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

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

Thank you.
 

chisigma

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

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

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

Kind regards

\(\displaystyle \chi\) \(\displaystyle \sigma\)
 
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simplependulum

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

Let \(\displaystyle a = 1 ~,~ b=t \)

Therefore ,

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

By reversing the order , we have

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

Consider \(\displaystyle \int_0^{\infty} e^{-t} e^{-tx} t^{z-1}~dt\)

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

Sub. \(\displaystyle (x+1)t = u \)

we find that it is equal to

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

Back to the derivative of Gamma function ,

If we continue we obtain

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

Sub. \(\displaystyle x+1 = e^{u} \) for the second integral ,

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

Therefore ,

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

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

If we differentiate w.r.t. \(\displaystyle z \) again , i find something interesting ...


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

Sub. \(\displaystyle z = 1 \) , from L.H.S , giving

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

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

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

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

This method might be a way to obtain \(\displaystyle \Gamma^{(n)}(1) \) by differentiating the integral \(\displaystyle n-1\) times without a great detour ...
 
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Apr 2010
50
2
Sub. \(\displaystyle x+1 = e^{u} \) for the second integral ,

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

Therefore ,

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

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

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

Again thank you.
 

simplependulum

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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:

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

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

Again thank you.
Let me see ...

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

Sub. \(\displaystyle x + 1 = e^u \)

\(\displaystyle dx = e^u du \)

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

so it is equal to

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

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

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

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

Then change the "dummy" variable

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

(Happy)

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

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

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

because the integral \(\displaystyle \int_0^{\infty} \frac{e^{-x}}{x} ~dx \) is divergent .
 
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Apr 2010
50
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\(\displaystyle = \int_0^{\infty} \frac{ e^{-zx} }{ 1- e^{-x} }~dx \)

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

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

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

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

Where \(\displaystyle H_{n}\) are the Harmonic numbers and \(\displaystyle \gamma\) 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

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

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

Substituting, I get

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

which simplifies to

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

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

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

\(\displaystyle H_n=\int^1_0 \frac{1-x^n}{1-x}dx\), 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 \(\displaystyle \gamma\) arises.

Thank you.(Happy)
 
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simplependulum

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You have done a great job !

Your substitution is perfect (Clapping) !

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


Also \(\displaystyle H_n = \int_0^1 \frac{1 - x^n}{1-x}~dx \)

We need to make a further substitution ,

\(\displaystyle x = e^{-t} \)

\(\displaystyle dx = -e^{-t}~dt\)

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

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

Therefore ,

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

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

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

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

\(\displaystyle n > 1\)
 
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Apr 2010
50
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\(\displaystyle = \Psi(n+1) \) !!

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

\(\displaystyle n > 1\)

Thanks!

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

Thanks on the help on the Digamma function.