May be that the right way is starting from this 'infinite product'...
(1)
Now compute the logarithm of (1), change sign, then differentiate...
Kind regards
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
Since
then I would think
Anyway, the integral expression for the Digamma function is given by
How would I derive this formula?
Should I instead look at another representation of the Gamma function and work with that?
Thank you.
We have
Let
Therefore ,
By reversing the order , we have
Consider
Sub.
we find that it is equal to
Back to the derivative of Gamma function ,
If we continue we obtain
Sub. for the second integral ,
we have
Therefore ,
Edit: Here is what i guess , not sure if it is true .
If we differentiate w.r.t. again , i find something interesting ...
Sub. , from L.H.S , giving
Since , what it gives us is also equal to , while from R.H.S we have
so is that ture
?
This method might be a way to obtain by differentiating the integral times without a great detour ...
Thanks! I think I just overlooked the substitution process. I completely understand it now though.
Anyway, now I understand the integral representation for
However, I have one more question (and this question is open to anyone). I want to see how
Where are the Harmonic numbers and 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
Let , therefore and
Substituting, I get
which simplifies to
I know that
However, I'm not sure what to do from here. I also know the integral representation for the Harmonic numbers is given by:
, 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 arises.
Thank you.