The Series Representation of Polygamma Function

Consider the following integral,

01xs1ln(x)x1dx=ψ1(s)\int_0^1\frac{x^{s-1}\ln (x)}{x-1}dx=\psi^1(s) Then applying the Taylor series of ln(x)\ln (x)(hereψ1(s)\psi^1(s) denotes the first derivative of diagamma function),

01xs1x1n=1(1)n11n(x1)ndx\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,

n=1(1)n11n01xs1(x1)n1dx\sum_{n=1}^{\infty}(-1)^{n-1}\frac{1}{n}\int_0^1x^{s-1}(x-1)^{n-1}dx Which is same as,

n=1(1)2n21n01xs1(1x)n1dx\sum_{n=1}^{\infty}(-1)^{2n-2}\frac{1}{n}\int_0^1x^{s-1}(1-x)^{n-1}dx

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

n=11nB(s,n)=ψ1(s)\boxed{\sum_{n=1}^{\infty}\frac{1}{n}\Beta(s,n)=\psi^1(s)} Then entering the general case,

n=11nBa(s,n)=ψa+1(s)\boxed{\sum_{n=1}^{\infty}\frac{1}{n}\Beta^a(s,n)=\psi^{a+1}(s)} (HereBa(s,n)\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.

Aruna Yumlembam - 3 months ago

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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.

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Thanks,plus I love them because Ramanujan did.

Aruna Yumlembam - 3 months ago

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@Aruna Yumlembam are you 15?

Mahdi Raza - 2 months, 3 weeks ago

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Of course

Aruna Yumlembam - 2 months, 3 weeks ago

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Did you understand the above discussion?

Aruna Yumlembam - 2 months, 3 weeks ago

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