Integral boundary notation $\int_0^{(1+)}dx~f(x)$? What does the notation for the upper boundary in the following kind of integral mean?
$$\int_0^{(1+)}dx ~f(x)$$
It can be seen here.
 A: Even though the previous answers are correct in some settings, I do not believe that is what is meant in the case you refer to.
Instead, I suggest that what is meant is a loop integral, and the notation
$$
\int_{0}^{(1+)}
$$
means that the contour they integrate over is starting at $0$, circles $1$ in positive sense (i.e. counter-clock wise) and then returns to $0$. So, why do I believe that? Well, I've seen it before. Also, look at this page about the Beta function, In particular this figure.
(The reason why one loops like this might be singularities or branch cuts.)
A: I assume that $f(x)$ is well defined for $x=0$. Maybe, $f(x)$ is "strange" when $x = 1$ (i.e. vertical asymptote, jumps, holes...).
Assume also that there exists a function $F(x)$ such that $\frac{d F(x)}{dx} = f(x)$ (i.e., $F(x)$ is a primitive of $f(x)$). Then:
$$\int_0^{1^+}dx ~f(x) = \lim_{x \to 1^+} F(x) - F(0).$$
A: Generally, the notation $a^+$ is used as a shorthand for "approaching $a$ from the right". So it suggests, at least to me, that $f(x)$ has some unconventional behaviour at $x = 1$, in particular it may be that $f(x)$ is actually a pseudo-function - for example, it may be related to the Dirac delta function $\delta(x)$ which is defined such that it is zero everywhere except at $x = 0$, and its integral over any interval containing zero is $1$.
You could represent the Dirac delta's behaviour by writing something like, for example: $\int_{-\infty}^{0^+} \delta(x) dx = 1$, which would just be shorthand for $\forall \epsilon > 0 \int_{-\infty}^{\epsilon} \delta(x) dx = 1$, or the same thing represented as a single-sided limit.
