Two questions regarding $\mathrm {Li}$ from "Edwards"

Question

I would appreciate help understanding a relation in Edwards's "Riemann's Zeta Function."

On page 30 he has:

$$\int_{C^{+}} \frac{t^{\beta - 1}}{\log t}dt = \int_{0}^{x^{\beta}}\frac{du}{\log u}= \mathrm {Li} (x^{\beta}) - i\pi$$

He states for $\beta$ positive and real, change variables $u = t^{\beta}$ which implies $\log t = \log u/\beta$ and $dt/t = du/u \beta$.

Here $C^{+}$ is a path which is a line segment from $0$ to $1 - \epsilon$ and passes over the singularity at $u = 1$ in a semi-circle in the upper half-plane and continues in a line segment from $1 + \epsilon$ to $1$.

I would appreciate help with two aspects:

-- Since most of the discussions of the logarithmic integral I have seen take the integral from $2$ rather than from $0$, how do you treat what looks like a $- \mathrm {Li}(0)$ term?

-- How do you actually get the $- i \pi$ term. I would guess it's from integrating around the half-circle above $u = 1$ in a clockwise direction. But I have tried parametrization with $u = r e^{i \theta}$. Maybe this is something I should know from complex analysis.

Thanks very much.

Answer 1

The standard notation is $\DeclareMathOperator{\li}{li} \DeclareMathOperator{\Li}{Li}$

$$\begin{align} \li x &= \int_0^x \frac{dt}{\log t}\\ \Li x &= \int_2^x \frac{dt}{\log t} = \li x - \li 2 \end{align}$$

where the $\li$ integral is to be interpreted as the Cauchy principal value

$$\li x = \lim_{\varepsilon \searrow 0} \left(\int_0^{1-\varepsilon} \frac{dt}{\log t} + \int_{1+\varepsilon}^x \frac{dt}{\log x}\right)$$

for $x > 1$ (and similarly for the $\Li$ integral if $x < 1$; neither integral is finite for $x = 1$).

Integrating over the path $C^+$ gives

$$\begin{align} \int_{C^+} \frac{t^{\beta-1}}{\log t}\, dt &= \int_0^{1-\varepsilon} \frac{t^{\beta-1}}{\log t}\,dt + \int_{1+\varepsilon}^x \frac{t^{\beta-1}}{\log t}\,dt + \underbrace{\int_0^\pi \frac{(1+\varepsilon e^{i(\pi-\varphi)})^{\beta-1}}{\log (1+\varepsilon e^{i(\pi-\varphi)})} (-i\varepsilon e^{i(\pi-\varphi)})\,d\varphi}_{\rho(\varepsilon)}\\ &= \int_0^{(1-\varepsilon)^{\beta}} \frac{du}{\log u} + \int_{(1+\varepsilon)^{\beta}}^{x^\beta} \frac{du}{\log u} + \rho(\varepsilon). \end{align}$$

Even though $(1-\varepsilon)^\beta$ and $(1+\varepsilon)^\beta$ aren't (for $\beta \neq 1$) quite symmetric about $1$, we have

$$\lim_{\varepsilon\searrow 0} \left(\int_0^{(1-\varepsilon)^{\beta}} \frac{du}{\log u} + \int_{(1+\varepsilon)^{\beta}}^{x^\beta} \frac{du}{\log u}\right) = \li(x^\beta),$$

since the asymmetry is of order $O(\varepsilon^2)$.

By a Laurent expansion of $\dfrac{z^{\beta-1}}{\log z}$ about $1$ (or by multiple other means), one can easily see that

$$\lim_{\varepsilon\searrow 0} \rho(\varepsilon) = -\pi i \operatorname{Res}_{z=1} \frac{z^{\beta-1}}{\log z} = -\pi i.$$

Since the integral over $C^+$ does not depend on $\varepsilon > 0$ by Cauchy's integral theorem, we have

$$\int_{C^+} \frac{t^{\beta-1}}{\log t}\, dt = \li(x^\beta) -\pi i.$$

Edwards seems to use non-standard notation and calls $\Li$ what usually is called $\li$.

Answer 2

Here is a try for part two, the $- \pi i$ term:

Change variable from $u$ to $e^{t}$. Then $\log u = t$ and $du = e^{t}dt$.

Considering the integral around a half circle, now about and above $t = 0$, in the clockwise direction, the residue of the integrand is $-i\pi\left(e^{t}\right)_{z=0} = - i \pi$.

Answer 3

Daniel Fischer provided most of the information (including Edwards' use of $\,\operatorname{Li}(x)$ instead of $\,\operatorname{li}(x)$ with the possible confusion with the polylogarithm $\operatorname{Li}_s(x)$) but let's add some details and generalities.

Concerning your second question the idea is that : \begin{align} \lim_{\epsilon\to 0^+}\left[\int_{0}^{1-\epsilon}\frac{dz}{\log z}+\int_{C_{\epsilon}}\frac{dz}{\log z}+\int_{1+\epsilon}^\infty\frac{dz}{\log z}\right]&=P\int_{0}^\infty\frac{dz}{\log z}+\lim_{\epsilon\to 0^+}\int_{C_{\epsilon}}\frac{dz}{\log z}\\ &=\operatorname{li}(x) -\left(i\pi\;\underset{z=1}{\operatorname{Res}}\frac 1{\log z}\right)\\ &=\operatorname{li}(x) -i\pi\\ \end{align} where the integral representation of the logarithmic integral was replaced by $\operatorname{li}(x)$ while the integral over the little half-circle $C_\epsilon$ is of course minus the integral over the contour of radius $\epsilon$ from $\,\theta=0$ to $\theta=\phi\,$ (with $\phi=\pi$ here). At the limit $\,\epsilon\to 0$ this last integral converges to $\;\displaystyle i\,\phi\;\underset{z=1}{\operatorname{Res}}\frac 1{\log z}\,$ as proved in the third part of this answer.

Coming to your first question let's notice that the logarithmic integral is defined as a Cauchy principal value around $1$. The $-i\pi\,$ is a term (kind of adjustment) that appeared from the contour integration because of the (half)-singularity at $z=1$ and is not linked to $\;\operatorname{li}(0)=0$.

---

To make this clearer and since this way of 'cutting the pole in two parts' (i.e. the case $\phi=\pi$) appears quite often let's consider the whole contour closed by a half-circle of radius $R$ on the upper complex plane. We will distinguish these three contours $C$,

suppose that $f(z)$ is analytic in the upper half-plane (including the real line) except at simple poles and that the contribution of the upper half circle goes to $0$ as $R\to +\infty$.
We may then try to compute $\;\displaystyle I:=P\int_{-\infty}^\infty f(z)\;dz\;$ using these three contours :

a) The '$z_0$ outside' case was considered earlier. The contour integral would return at the limit $\epsilon\to 0$ and $R\to +\infty$ : $$I-\pi \,i\;\underset{z=z_0}{\operatorname{Res}}f(z)=2\,\pi\,i\sum_{z_i\;\text{inside}\;C} \;\underset{z=z_i}{\operatorname{Res}}f(z)$$ b) Here $z_0$ is inside the contour and we will get $$I+\pi \,i\;\underset{z=z_0}{\operatorname{Res}}f(z)=2\,\pi\,i\sum_{z_i\;\text{inside}\;C} \;\underset{z=z_i}{\operatorname{Res}}f(z)$$ $\quad$since $z_0$ appears in the sum at the right $\;I\,$ will be the same than in a).

c) Imagine that the residue for a pole $z_0$ on a line of the contour simply counts as a half-residue then : $$I=2\,\pi\,i\left[\sum_{z_i\;\text{inside}\;C} \;\underset{z=z_i}{\operatorname{Res}}f(z)+\frac 12\underset{z=z_0}{\operatorname{Res}}f(z)\right]$$

$\quad$in harmony with a) and b).

$\quad$We may generalize this last formula :
For every residue of simple pole $c_j$ on the contour we replace the $2\,\pi$ constant of the residue theorem by the angle $\phi_j$ between the outgoing and incoming tangent lines at $c_j$ (if these tangents exist) to get :

$$I=2\,\pi\,i\,\sum_{z_i\;\text{inside}\;C} \;\underset{z=z_i}{\operatorname{Res}}f(z)+\sum_{c_j\;\text{on}\;C} \phi_j\;i\;\underset{z=c_j}{\operatorname{Res}}f(z)$$

This third method should make things easier to remember and to compute!

---

Let's evaluate $\;I(\phi):=\lim_{\epsilon\to 0}\int_{C_\epsilon}f(z)\,dz\;$ over a partial circular contour centered at $z_0$ of angle $\phi$ (this picture and part of the discussion is extract from Ablowitz and Fokas' excellent 'Complex Variables').

If $f(z)$ is regular in a neighborhood of $z_0$ except at the simple pole $z_0$ with $\;\underset{z=z_0}{\operatorname{Res}} f(z)=C_{-1}$ then, from its Laurent expansion; $f(z)$ may be rewritten as : $$f(z)=\frac{C_{-1}}{z-z_0}+g(z)$$ with $g(z)$ analytic in the neighborhood so that : $$I(\phi)=\lim_{\epsilon\to 0}\int_{C_\epsilon}f(z)\,dz=\lim_{\epsilon\to 0}\int_{C_\epsilon}\frac{C_{-1}}{z-z_0}dz+\lim_{\epsilon\to 0}\int_{C_\epsilon}g(z)\,dz$$ The integral of $g(z)$ will disappear at the limit since $g(z)$ is analytic in the neighborhood and thus $|g(z)|<M=$constant with a path of length $\to 0$.

While we may substitute $\,z=z(\theta)=z_0+\epsilon\,e^{i(\theta+\theta_0)}\,$ in the other integral to get : $$I(\phi)=\lim_{\epsilon\to 0}\int_{C_\epsilon}\frac{C_{-1}}{z-z_0}dz=\lim_{\epsilon\to 0}\int_0^\phi\frac{C_{-1}}{\epsilon\,e^{i(\theta+\theta_0)}}i\,\epsilon\,e^{i(\theta+\theta_0)}\;d\theta$$ That is $$\boxed{\displaystyle I(\phi)=\lim_{\epsilon\to 0}\int_{C_\epsilon}f(z)\,dz=i\,\phi\;C_{-1}}$$