# The Series Representation of Polygamma Function

Consider the following integral,

$\int_0^1\frac{x^{s-1}\ln (x)}{x-1}dx=\psi^1(s)$ Then applying the Taylor series of $\ln (x)$(here$\psi^1(s)$ denotes the first derivative of diagamma function),

$\int_0^1\frac{x^{s-1}}{x-1}\sum_{n=1}^{\infty}(-1)^{n-1}\frac{1}{n}(x-1)^ndx$ Then putting the sigma notation outside and simplifying as such,

$\sum_{n=1}^{\infty}(-1)^{n-1}\frac{1}{n}\int_0^1x^{s-1}(x-1)^{n-1}dx$ Which is same as,

$\sum_{n=1}^{\infty}(-1)^{2n-2}\frac{1}{n}\int_0^1x^{s-1}(1-x)^{n-1}dx$

Realising the integral is $\Beta(s,n)$,we have the following series,

$\boxed{\sum_{n=1}^{\infty}\frac{1}{n}\Beta(s,n)=\psi^1(s)}$ Then entering the general case,

$\boxed{\sum_{n=1}^{\infty}\frac{1}{n}\Beta^a(s,n)=\psi^{a+1}(s)}$ (Here$\Beta^a(s,n)$ denotes the ath derivative of beta function respect to one variable.)(Using the above result to solve my problem in calculus section name Fairly Impossible#2) Note by Aruna Yumlembam
3 months ago

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Here's a thing about me I really ,really ,really love sums that includes repeating gamma's and beta's like the one above so if you have any such series then you are always welcome to show me.

- 3 months ago

You really like Beta and Gamma Functions, right? Talk to @Gandoff Tan - he posts a lot of proofs - he may not be very active though.

Thanks,plus I love them because Ramanujan did.

- 3 months ago

@Aruna Yumlembam are you 15?

- 2 months, 3 weeks ago

Of course

- 2 months, 3 weeks ago

Did you understand the above discussion?

- 2 months, 3 weeks ago