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)

#Calculus

Note by Aruna Yumlembam
11 months, 4 weeks ago

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Comments

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 - 11 months, 4 weeks 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.

A Former Brilliant Member - 11 months, 4 weeks ago

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Thanks,plus I love them because Ramanujan did.

Aruna Yumlembam - 11 months, 4 weeks ago

@Aruna Yumlembam are you 15?

Mahdi Raza - 11 months, 1 week ago

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Of course

Aruna Yumlembam - 11 months, 1 week ago

Did you understand the above discussion?

Aruna Yumlembam - 11 months, 1 week ago

Thank you very much for explaining in detail how these examples are solved, for me this is important. Recently, when there is not enough time to solve homework on my own, I turn to the service https://assignmentbro.com/uk/homework-help for help, which helps me with homework. I know it's not worth doing this all the time, but sometimes when there is no time at all, and a very important assessment is at stake, I can use their services.

Kieran Williams - 6 months ago
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