Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Consider the function $$F(s)=\int_{1}^{\infty}\frac{\text{Li}(x)}{x^{s+1}}dx$$ where $\text{Li}(x)=\int_2^x \frac{1}{\log t}dt$ is the logarithmic integral. What is the series expansion around $s=1$?

It has a logarithmic singularity at $s=1$, and I am fairly certain (although I cannot prove it) that it should expand as something of the form $$\log (1-s)+\sum_{n=0}^\infty a_n (s-1)^n.$$ (An expansion of the above form is what I am looking for) I also have a guess that the constant term is $\pm \gamma$ where $\gamma$ is Euler's constant. Does anyone know a concrete way to work out such an expansion?

Thanks!

share|improve this question
    
Are we allowed to make the integral defining $F$ start at $x=2$ (as in the Li function) or are you only interested in the case when it starts at $x=1$? –  Did Nov 12 '11 at 16:05
    
@DidierPiau: I don't think it should make much of a different, but I am interested in both. –  Eric Naslund Nov 12 '11 at 16:44
add comment

1 Answer 1

Note that the integral $F(s)$ diverges at infinity for $s\leqslant1$ and redefine $F(s)$ for every $s\gt1$ as $$ F(s)=\int\limits_2^{+\infty}\frac{\text{Li}(x)}{x^{s+1}}\mathrm dx. $$ An integration by parts yields $$ F(s)=\frac1s\int\limits_2^{+\infty}\frac{\mathrm dx}{x^s\log x}, $$ and the change of variable $x^{s-1}=\mathrm e^t$ yields $sF(s)=G((s-1)\log2)$ with $$ G(z)=\int\limits_z^{+\infty}\mathrm e^{-t}\frac{\mathrm dt}t. $$ For every $z\leqslant1$, $$ G(z)=-\log(z)-\int\limits_z^{1}(1-\mathrm e^{-t})\frac{\mathrm dt}t+\int\limits_1^{+\infty}\mathrm e^{-t}\frac{\mathrm dt}t. $$ Thus $F(s)=-\log(s-1)+H(s-1)$ where $H(z)$ can be expanded as a series in $z^n$ and $z^n\log(z)$ for nonnegative $n$, and $$ H(0)=-\int\limits_0^{1}(1-\mathrm e^{-t})\frac{\mathrm dt}t+\int\limits_1^{+\infty}\mathrm e^{-t}\frac{\mathrm dt}t=-\mathrm{Ein}(1)+\mathrm{E}_1(1)=-\gamma, $$ where $\mathrm{Ein}$ and $\mathrm{E}_1$ are related to the exponential integral function. More generally, $$ G(z)=-\log(z)-\mathrm{Ein}(z)+\mathrm{E}_1(1)=-\log(z)-\gamma+\mathrm{Ein}(1)-\mathrm{Ein}(z), $$ with $$ \mathrm{Ein}(z) = \sum\limits_{k=1}^\infty \frac{(-1)^{k+1}}{k\,k!}z^k. $$ Finally, using the expansion $\dfrac1s=\sum\limits_{n=0}^{+\infty}(-1)^n(s-1)^n$, one gets $$ F(s)=-\log(s-1)+\sum\limits_{n=1}^{+\infty}(-1)^{n+1}(s-1)^n\log(s-1)-\gamma+\sum\limits_{n=1}^{+\infty}c_n(s-1)^n, $$ for some coefficients $(c_n)_{n\geqslant1}$. Due to the log terms, this is a slightly more complicated expansion than the one suggested in the question, in particular $s\mapsto F(s)+\log(s-1)$ is not analytic around $s=1$.

share|improve this answer
    
I think this is what was looking for. I am pretty sure that starting from $1$ might clean it up more. –  Eric Naslund Nov 12 '11 at 18:08
    
Thanks. Corrected a typo. –  Did Nov 12 '11 at 20:38
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.