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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? $$(sF(s))' = \int_2^\infty x^{-s}dx= \frac{2^{1-s}}{s-1}, F(s)= \frac{\log(s-1)}{s}+ \frac{\int_2^s \frac{2^{1-u}-1}{u}du+ 2 F(2)}{s}$$ 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
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 Originally Posted by SDK Its just the product rule.
How was $$F(s)= \frac{\log(s-1)}{s}+ \frac{\int_2^s \frac{2^{1-u}-1}{u}du+ 2 F(2)}{s}$$ arrived at through the product rule?

March 2nd, 2019, 08:16 PM   #4
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 Originally Posted by sticker592 How was $$F(s)= \frac{\log(s-1)}{s}+ \frac{\int_2^s \frac{2^{1-u}-1}{u}du+ 2 F(2)}{s}$$ arrived at through the product rule?
Express $\frac{d}{ds} sF(s)$ using the product rule. What do you get? Now solve for $F$.

 March 3rd, 2019, 04:09 AM #5 Newbie   Joined: Oct 2016 From: none Posts: 3 Thanks: 0 I get $$F(s)=\frac{d}{ds} sF(s) - sF'(s)$$ $$=\frac{2^{1-s}}{s-1}-sF'(s)$$ The previous line that I didn't show has F: $$F(s) = \int_1^\infty Li(x)x^{-s-1}dx, F(s) = \int_2^\infty \frac{x^{-s}}{-s \ln x}dx$$ which gives: $$F(s) =\frac{2^{1-s}}{s-1}- \int_2^\infty \frac{sx^{-s-1}}{ \ln x}dx$$ what am I doing wrong?

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