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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?


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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

1 Answer 1

up vote 5 down vote accepted

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 $s=1$.

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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

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