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March 2nd, 2019, 04:00 PM   #1
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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?
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March 2nd, 2019, 04:05 PM   #2
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It's just the product rule.

Last edited by skipjack; March 2nd, 2019 at 06:59 PM.
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March 2nd, 2019, 05:03 PM   #3
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Quote:
Originally Posted by SDK View Post
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?
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March 2nd, 2019, 08:16 PM   #4
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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 $F$.
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March 3rd, 2019, 04:09 AM   #5
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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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