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March 2nd, 2019, 04:00 PM  #1 
Newbie Joined: Oct 2016 From: none Posts: 3 Thanks: 0  How was the second step in this integral solved?
$$(sF(s))' = \int_2^\infty x^{s}dx= \frac{2^{1s}}{s1}, F(s)= \frac{\log(s1)}{s}+ \frac{\int_2^s \frac{2^{1u}1}{u}du+ 2 F(2)}{s}$$ I'm confused by the second step. Anyone? 
March 2nd, 2019, 04:05 PM  #2 
Senior Member Joined: Sep 2016 From: USA Posts: 645 Thanks: 406 Math Focus: Dynamical systems, analytic function theory, numerics 
It's just the product rule.
Last edited by skipjack; March 2nd, 2019 at 06:59 PM. 
March 2nd, 2019, 05:03 PM  #3 
Newbie Joined: Oct 2016 From: none Posts: 3 Thanks: 0  
March 2nd, 2019, 08:16 PM  #4 
Senior Member Joined: Sep 2016 From: USA Posts: 645 Thanks: 406 Math Focus: Dynamical systems, analytic function theory, numerics  
March 3rd, 2019, 04:09 AM  #5 
Newbie Joined: Oct 2016 From: none Posts: 3 Thanks: 0 
I get $$F(s)=\frac{d}{ds} sF(s)  sF'(s)$$ $$=\frac{2^{1s}}{s1}sF'(s)$$ The previous line that I didn't show has F: $$F(s) = \int_1^\infty Li(x)x^{s1}dx, F(s) = \int_2^\infty \frac{x^{s}}{s \ln x}dx$$ which gives: $$F(s) =\frac{2^{1s}}{s1} \int_2^\infty \frac{sx^{s1}}{ \ln x}dx$$ what am I doing wrong? 

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