Limits of Integration Variable

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The discussion focuses on the correct formulation of integrals when using the Dirac delta function and the placement of variables in limits of integration. It clarifies that while H(x) can be expressed with the limit superior, there is confusion about how to properly represent it with the limit inferior. The importance of using distinct dummy variables in integrals is emphasized, suggesting that the variable in the integral should differ from the limits. Additionally, it is noted that the expressions for F(x) should not be conflated, as they represent different functions. Understanding these distinctions is crucial for accurately defining integrals and their properties.
Jhenrique
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Knowing that the limits of integration of a any function, for example:
\int_{-\infty}^{+\infty}\delta (x)dx=1
I know that's correct call your primitive through the limit superior as a variable, so
H(x)=\int_{-\infty}^{x}\delta (x)dx
But, and if I want to describe your primitive through the limit inferior as a variable? Will be so:
H(x)=\int_{-x}^{+\infty}\delta (x)dx
or:
H(x)=\int_{+x}^{+\infty}\delta (x)dx
or other?
 
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Jhenrique said:
Knowing that the limits of integration of a any function, for example:
\int_{-\infty}^{+\infty}\delta (x)dx=1
I am having a hard time understanding what you're saying and what you're asking. I can't tell if you're asking about some generic function whose integral is 1, or if your question is about the Dirac delta function (see http://en.wikipedia.org/wiki/Dirac_delta_function).

Jhenrique said:
I know that's correct call your primitive through the limit superior as a variable, so
H(x)=\int_{-\infty}^{x}\delta (x)dx
Here H is a function of x. Clearly the value of H(x) is somewhere between 0 and 1.
Jhenrique said:
But, and if I want to describe your primitive through the limit inferior as a variable? Will be so:
H(x)=\int_{-x}^{+\infty}\delta (x)dx
or:
H(x)=\int_{+x}^{+\infty}\delta (x)dx
or other?
 
I saw that when you have a definite integral of a function f(x), you can to express the primitive, F(x), placing the variable x in limit superior of integral:
F(x)=\int_{x_{0}}^{x}f(x)dx
It's known... So I ask if F(x) can be equal to this too:
F(x)=\int_{x}^{x_{1}}f(x)dx
I ask which is the correct expression to F(x) when the variable x is placed in limit inferior.
 
Jhenrique said:
I saw that when you have a definite integral of a function f(x), you can to express the primitive, F(x), placing the variable x in limit superior of integral:
F(x)=\int_{x_{0}}^{x}f(x)dx
It's known... So I ask if F(x) can be equal to this too:
F(x)=\int_{x}^{x_{1}}f(x)dx
Probably not.
$$\int_x^{x_1}f(x)dx = -\int_{x_1}^x f(x)dx$$
Does that look the same as the first one you defined as F(x) above?
Jhenrique said:
I ask which is the correct expression to F(x) when the variable x is placed in limit inferior.

Two things:
1. The integrals you wrote are functions of x. The dummy variable in the integral should be some other variable, such as t. In other words, your first integral probably should be written like this:
$$F(x) = \int_{x_0}^x f(t)dt$$
2. You should identify two different things with the same letter. In other words, the two integrals you wrote should not both be identified as F(x).
 
There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...

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