# Digamma Integral Representation

#### Anthonny

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

MHF Hall of Honor
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

$$\displaystyle \chi$$ $$\displaystyle \sigma$$

• Anthonny

#### simplependulum

MHF Hall of Honor
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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• chiph588@ and Anthonny

#### Anthonny

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

MHF Hall of Honor
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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• Anthonny

#### Anthonny

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

(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

#### simplependulum

MHF Hall of Honor
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 $$\displaystyle \Psi(n) = H_{n-1} - \gamma$$

$$\displaystyle n > 1$$

• Anthonny

#### Anthonny

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

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.