# Series around $s=1$ for $F(s)=\int_{1}^{\infty}\text{Li}(x)\,x^{-s-1}\,dx$

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!

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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

Note that the integral $F(s)$ diverges at infinity for $s\leqslant1$ and redefine $F(s)$ for every $s\gt1$ as $$F(s)=\int_2^{+\infty}\frac{\text{Li}(x)}{x^{s+1}}\mathrm dx.$$ An integration by parts yields $$sF(s)=\int_2^{+\infty}\frac{\mathrm dx}{x^s\log x},$$ and the change of variable $x^{s-1}=\mathrm e^t$ yields $$sF(s)=\mathrm{E}_1(u\log2),\qquad u=s-1,$$ where the exponential integral function $\mathrm{E}_1$ is defined, for every complex $z$ not a nonpositive real number, by $$\mathrm{E}_1(z)=\int_z^{+\infty}\mathrm e^{-t}\frac{\mathrm dt}t.$$ One knows that, for every such $z$, $$\mathrm{E}_1(z) = -\gamma-\log z-\sum\limits_{k=1}^\infty \frac{(-z)^{k}}{k\,k!}.$$ On the other hand, $$\frac1s=\frac1{1+u}=\sum_{n\geqslant0}(-1)^nu^n,$$ hence $$F(s)=\frac1s\mathrm{E}_1(u\log2)=\sum_{n\geqslant0}(-1)^nu^n\cdot\left(-\gamma-\log\log2-\log u-\sum\limits_{k=1}^\infty \frac{(-1)^{k}(\log2)^k}{k\,k!}u^k\right).$$ One sees that $F(1+u)$ coincides with a series in $u^n$ and $u^n\log u$ for nonnegative $n$, and that $G(u)=F(1+u)+\log u$ is such that $$G(0)=-\gamma-\log\log2.$$ Finally, $$F(s)=-\gamma-\log\log2-\log(s-1)-\sum\limits_{n=1}^{+\infty}(-1)^{n}(s-1)^n\log(s-1)+\sum\limits_{n=1}^{+\infty}c_n(s-1)^n,$$ for some coefficients $(c_n)_{n\geqslant1}$. Due to the logarithmic terms, this is a slightly more complicated expansion than the one suggested in the question, in particular $s\mapsto G(s-1)=F(s)+\log(s-1)$ is not analytic around $s=1$.
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
I believe that the $\log 2$ disappeared somewhere. In particular, the final constant should be $-\gamma-\log\log 2$. Also the equation directly after "More Generally" appears to have some typos in it as the constants are different for different parts of that line. – Eric Naslund Dec 21 '15 at 14:58
By looking at the sum $$\sum_{k=2}^N\frac{1}{k\log k}=\log \log N+K+O\left(\frac{1}{N}\right)$$ in this question, I found another way to prove that as $s\rightarrow 0$ $$\int_2^\infty \frac{x^{-s-1}}{\log x}dx=-\log(s)-\gamma-\log \log 2+O(s\log(s)).$$ Let $\Lambda(n)$ be the Von Mangoldt Lambda function, and $\gamma_0$ the Euler-Mascheroni Constant. Then we have the expansion of the similar sum $$\sum_{n\leq x}\frac{\Lambda(n)}{n\log n}=\log\log x+\gamma_{0}+O\left(\frac{1}{\log x}\right),$$ which appears in the proof of theorem 2.7 in Montgomery and Vaughn. Let $$S(x)=\sum_{2\leq k\leq x}\frac{1}{k\log k}- \sum_{n\leq x}\frac{\Lambda(n)}{n\log n},$$ and examine $I=\delta \int_1^\infty S(x)x^{-\delta -1}dx$ as $\delta\rightarrow 0$. As $S(x)=(K-\gamma_0)+O(1/\log x)$, it follows that $I=K-\gamma_0+O(\delta \log (1/\delta)$. Then, since $$\sum_{n=1}^{\infty}a_{n}n^{-s}=s\int_{1}^{\infty}A(x)x^{-s-1}dx$$ (Theorem 1.3 of Montgomery and Vaughn) we see that $$\sum_{n=2}^{\infty}\frac{n^{-\delta}}{n\log n}-\log \zeta(\delta+1)=O(\delta\log(1/\delta)$$ as $\delta\rightarrow 0$, and so $$\sum_{n=2}^\infty \frac{n^{-\delta-1}}{\log n}=-\log \delta+(K-\gamma)+O(\delta\log(1/\delta).$$ Now, writing the left hand side as a Riemann Stieltjes integral and simplifying the expression for $K$ combined with the resulting terms allows us to conclude that $$\int_{2}^\infty \frac{x^{-\delta-1}}{\log x}dx=-\log \delta -\gamma-\log \log 2+O\left(\delta\log(1/\delta)\right).$$