Consider the following integral,
∫01x−1xs−1ln(x)dx=ψ1(s)
Then applying the Taylor series of ln(x)(hereψ1(s) denotes the first derivative of diagamma function),
∫01x−1xs−1n=1∑∞(−1)n−1n1(x−1)ndx
Then putting the sigma notation outside and simplifying as such,
n=1∑∞(−1)n−1n1∫01xs−1(x−1)n−1dx
Which is same as,
n=1∑∞(−1)2n−2n1∫01xs−1(1−x)n−1dx
Realising the integral is B(s,n),we have the following series,
n=1∑∞n1B(s,n)=ψ1(s)
Then entering the general case,
n=1∑∞n1Ba(s,n)=ψa+1(s)
(HereBa(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)
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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.
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 are you 15?
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Of course
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