In the book "Handbook of Mathematical Functions" by Abramowitz and Stegun, I came across this integration symbol in the first few equations defining the exponential integral functions, pictured below. Does this mean anything, or is it a printing error?
What does an integral with a horizontal bar through it mean
integrationnotation
Related Solutions
$\operatorname{erf}(x)$ is an odd function, therefore, $$ \begin{align} \int_{-\infty}^\infty(\operatorname{erf}(a+x)+\operatorname{erf}(a-x))\;\mathrm{d}x &=\lim_{L\to\infty}\;\int_{-L}^L(\operatorname{erf}(x+a)-\operatorname{erf}(x-a))\;\mathrm{d}x\\ &=\lim_{L\to\infty}\;\int_{-L+a}^{L+a}\operatorname{erf}(x)\;\mathrm{d}x-\lim_{L\to\infty}\;\int_{-L-a}^{L-a}\operatorname{erf}(x)\;\mathrm{d}x\\ &=\lim_{L\to\infty}\;\int_{L-a}^{L+a}\operatorname{erf}(x)\;\mathrm{d}x-\lim_{L\to\infty}\;\int_{-L-a}^{-L+a}\operatorname{erf}(x)\;\mathrm{d}x\\ &=4a\tag{1} \end{align} $$ since $\lim\limits_{x\to\infty}\operatorname{erf}(x)=1$ and $\lim\limits_{x\to-\infty}\operatorname{erf}(x)=-1$.
Furthermore, $$ \begin{align} \int_{-\infty}^\infty\left(\operatorname{erf}(a+x)\operatorname{erf}(a-x)+1 \right)\;\mathrm{d}x &=\int_{-\infty}^\infty(\operatorname{erf}(a+x)+1)(\operatorname{erf}(a-x)+1)\;\mathrm{d}x\\ &-\int_{-\infty}^\infty(\operatorname{erf}(a+x)+\operatorname{erf}(a-x))\;\mathrm{d}x\tag{2} \end{align} $$ To evaluate $$ \begin{align} \int_{-\infty}^\infty(\operatorname{erf}(a+x)+1)(\operatorname{erf}(a-x)+1)\;\mathrm{d}x &=\frac{4}{\pi}\int_{-\infty}^\infty\int_{-\infty}^{a+x}\int_{-\infty}^{a-x}e^{-s^2-t^2}\;\mathrm{d}s\;\mathrm{d}t\;\mathrm{d}x \end{align} $$ note that $s\le a+x$ and $t\le a-x$; i.e. $s-a\le x\le a-t$ and $s+t\le2a$. Thus, $$ \begin{align} \frac{4}{\pi}\int_{-\infty}^\infty\int_{-\infty}^{a+x}\int_{-\infty}^{a-x}e^{-s^2-t^2}\;\mathrm{d}s\;\mathrm{d}t\;\mathrm{d}x &=\frac{4}{\pi}\int\int_{s+t\le2a}\int_{s-a}^{a-t}e^{-s^2-t^2}\;\mathrm{d}x\;\mathrm{d}s\;\mathrm{d}t\\ &=\frac{4}{\pi}\int_{-\infty}^\infty\int_{-\infty}^\infty(2a-s-t)_+\;e^{-s^2-t^2}\;\mathrm{d}s\;\mathrm{d}t \end{align} $$ Change variables: $u=(s+t)/\sqrt{2}$ and $v=(s-t)/\sqrt{2}$ so that $s=(u+v)/\sqrt{2}$ and $t=(u-v)/\sqrt{2}$: $$ \begin{align} \frac{4}{\pi}\int_{-\infty}^\infty\int_{-\infty}^\infty(2a-s-t)_+\;e^{-s^2-t^2}\;\mathrm{d}s\;\mathrm{d}t &=\frac{4}{\pi}\int_{-\infty}^\infty\int_{-\infty}^\infty(2a-\sqrt{2}u)_+\;e^{-u^2-v^2}\;\mathrm{d}u\;\mathrm{d}v\\ &=\frac{4}{\sqrt{\pi}}\int_{-\infty}^{\sqrt{2}a}(2a-\sqrt{2}u)\;e^{-u^2}\;\mathrm{d}u\\ &=4a(\operatorname{erf}(\sqrt{2}a)+1)-\frac{4}{\sqrt{\pi}}\int_{-\infty}^{\sqrt{2}a}\sqrt{2}u\;e^{-u^2}\;\mathrm{d}u\\ &=4a(\operatorname{erf}(\sqrt{2}a)+1)-\frac{2\sqrt{2}}{\sqrt{\pi}}\int_{-\infty}^{\sqrt{2}a}\;e^{-u^2}\;\mathrm{d}u^2\\ &=4a(\operatorname{erf}(\sqrt{2}a)+1)+\frac{2\sqrt{2}}{\sqrt{\pi}}\;e^{-2a^2} \end{align} $$ Therefore, $$ \int_{-\infty}^\infty\left(\operatorname{erf}(a+x)+1\right)\left(\operatorname{erf}(a-x)+1\right)\;\mathrm{d}x =4a\left(\operatorname{erf}(\sqrt{2}a)+1\right)+\frac{2\sqrt{2}}{\sqrt{\pi}}\;e^{-2a^2}\tag{3} $$ Thus, the convolution of $\operatorname{erf}(x)+1$ with itself is $2x(\operatorname{erf}(x/\sqrt{2})+1)+\frac{2\sqrt{2}}{\sqrt{\pi}}e^{-x^2/2}$.
Subtract $4a$ from $(3)$ using $(1)$ and $(2)$ to get $$ \int_{-\infty}^\infty\left(\operatorname{erf}(a+x)\operatorname{erf}(a-x)+1 \right)\;\mathrm{d}x =4a\operatorname{erf}(\sqrt{2}a)+\frac{2\sqrt{2}}{\sqrt{\pi}}\;e^{-2a^2}\tag{4} $$ My guess is you want either $(3)$ or $(4)$.
Let $S$ be a set of (logical) formulae and $\psi$ be a formula. Then $S \vdash \psi$ means that $\psi$ can be derived from the formulae in $S$. Intuitively, $S$ is a list of assumptions, and $S \vdash \psi$ if we can prove $\psi$ from the assumptions in $S$.
$\vdash \psi$ is shorthand for $\varnothing \vdash \psi$. That is, $\psi$ can be derived with no assumptions, so that in some sense, $\psi$ is 'true').
More precisely, systems of logic consist of certain axioms and rules of inference (one such rule being "from $\phi$ and $\phi \to \psi$ we can infer $\psi$"). What it means for $\psi$ to be 'derivable' from a set $S$ of formulae is that in a finite number of steps you can work with (i) the formulae in $S$, (ii) the axioms of your logical system, and (iii) the rules of inference, and end up with $\psi$.
In particular, if $\vdash \psi$ then $\psi$ can be derived solely from the axioms by using the rules of inference in your logical system.
Best Answer
According to the digital copy of the text which was linked by the user Chappers, this notation is the Cauchy principle value. This use is listed in the rather extensive index of notation at the end of the text—specifically, in the section labeled Miscellaneous Notation.
The Cauchy Principal Value is a way of assigning a value to certain "improper" integrals which would otherwise be undefined. If $f : \mathbb{R} \to \mathbb{R}$ has a singularity at $c \in [a,b]$, then the Cauchy Principal value is given by
$$ -\kern-9pt\int_{a}^{b} f(x)\,\mathrm{d}x := \lim_{\varepsilon\searrow 0} \left[ \int_{a}^{c-\varepsilon} f(x)\,\mathrm{d}x + \int_{c+\varepsilon}^b f(x)\,\mathrm{d}x\right].$$
A similar definition applies if $f$ has a singularity at infinity:
$$ -\kern-9pt\int_{-\infty}^{\infty} f(x) = \lim_{R\to\infty} \int_{-R}^{R} f(x)\,\mathrm{d}x. $$
In this second case, it is easier to see how Cauchy principal value differs from the "usual" method of assigning a value to an improper integral. In the usual setting, we define $$ \int_{-\infty}^{\infty} f(x)\,\mathrm{d}x := \lim_{a\to-\infty} \int_{a}^{c} f(x)\,\mathrm{d}x + \lim_{b\to\infty} \int_{c}^{b} f(x),\mathrm{d}x, $$ where $c$ is any real number. Using this standard definition, the sine function cannot be integrated over the entire real line. However, the Cauchy principle value does exist:
$$ -\kern-9pt\int_{-\infty}^{\infty} \sin(x)\,\mathrm{d}x = \lim_{R\to\infty} \int_{-R}^{R} \sin(x) \,\mathrm{d}x = 0, $$ since sine is an odd function.
It is also well worth noting that $-\kern-7.5pt\int$ is not standardized notation for the Cauchy principal value. Most authors will, instead, use the notation $PV\kern-4pt\int$, or something similar. Also, the notation $-\kern-7.5pt\int$ is used by other authors to mean something different. For example, in his text on PDEs, Evans uses $-\kern-7.5pt\int$ to denote the average integral over a ball, i.e. $$ -\kern-9pt\int_{B(x,r)} f(y)\,\mathrm{d}y = \frac{1}{\mu(B(x,r))} \int_{B(x,r)} f(y)\,\mathrm{d}y,$$ where $B(x,r)$ denotes a ball in $n$-dimensional Euclidean space with center $x$ and radius $r$, and $\mu(B(x,r))$ denotes the $n$-dimensional volume (Lebesgue measure) of that ball.