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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? Tags integral, solved, step Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post idontknow Calculus 5 December 12th, 2015 12:57 PM king.oslo Calculus 9 December 15th, 2013 04:26 PM ananda Calculus 3 August 8th, 2013 02:31 AM layd33foxx Calculus 1 December 7th, 2011 04:59 PM auntjamima Calculus 4 October 23rd, 2010 01:47 AM

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