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Another exercise from Apostol's book, this time we're supposed to prove

$$\mathrm{Li}(x)=\frac{x}{\log x}+\int_2^x \frac{dt}{\log^2t}-\frac{2}{\log 2}.$$

which is easy to do via integration by parts. But then he goes on to write,

[...] and that, more generally, $$\mathrm{Li}(x)=\frac{x}{\log x} \left(1+ \sum_{k=1}^{n-1} \frac{k!}{\log^k x} \right)+n! \int_2^x \frac{dt}{\log^{n+1}t}+C_n,$$ where $C_n$ is independent of $x$.

My question may come across as stupid, but what's going on here? Is this a comment, or should I be able to prove this as well? If so, I'd very much like some small hint on how to approach this! Thanks!

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Have you tried repeatedly applying integration by parts? –  J. M. Dec 3 '11 at 8:25
    
@J.M.: Ouch, embarrassing! Thanks for the hint! –  Carolus Dec 3 '11 at 8:36
    
Related: math.stackexchange.com/questions/7793/… –  Eric Naslund Apr 9 '13 at 23:14
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1 Answer 1

up vote 7 down vote accepted

To settle this:

$$\int_2^x \frac1{\log^k t} \mathrm dt=\frac{x}{\log^k x}-\frac2{\log^k 2}+k \int_2^x\frac1{\log^{k+1}t} \mathrm dt$$

is the identity you should be applying. Repeatedly.

For instance...

$$\small\begin{align*}\int_2^x \frac1{\log\,t}\mathrm dt&=\frac{x}{\log\,x}-\frac2{\log\,2}+\int_2^x \frac1{\log^2 t} \mathrm dt\\&=\frac{x}{\log\,x}-\frac2{\log\,2}+\left(\frac{x}{\log^2 x}-\frac2{\log^2 2}+2\int_2^x \frac1{\log^3(t)} \mathrm dt\right)\\&=\frac{x}{\log\,x}-\frac2{\log\,2}+\left(\frac{x}{\log^2 x}-\frac2{\log^2 2}+2\left(\frac{x}{\log^3 x}-\frac2{\log^3 2}+3\int_2^x \frac1{\log^4 t}\mathrm dt\right)\right)\\&=\frac{x}{\log\,x}-\frac2{\log\,2}+\left(\frac{x}{\log^2 x}-\frac2{\log ^2(2)}+2\left(\frac{x}{\log^3 x}-\frac2{\log^3 2}+3\left(\frac{x}{\log^4 x}-\frac2{\log^4 2}+\right.\right.\right.\\ &\qquad\qquad\left.\left.\left.4\int_2^x \frac1{\log^5 t} \mathrm dt\right)\right)\right)\end{align*}$$

I hope the pattern is somewhat transparent at this point...

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I'm having trouble figuring out where $\sum_{k=1}^{n-1}k!/\log^kx$ fits in. Specifically the term $k!$. Using the identity you supplied I end up with $\sum_{k=1}^{n-1}1/\log^kx$. –  Carolus Dec 3 '11 at 10:08
    
I should try harder before asking questions... Thank you for your patience! –  Carolus Dec 3 '11 at 10:43
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