# Improper integrals

• I
I understand what improper integrals are, but are they really integrals? The semantics are just a bit confusing.

fresh_42
Mentor
I understand what improper integrals are, but are they really integrals? The semantics are just a bit confusing.
An integral ##\int f(x)\,dx## is a solution to ##F(x)'=f(x)##.
As long as there are no boundary conditions to this differential equation, many solutions are possible. Nevertheless, they still have to solve the equation. As such they are a kind of generic solution, the set of possible flows if you like, which we call improper integral (I think; here it is call undetermined). Fixing a boundary condition means to determine a single flow of the vector field, a single solution. So in a way, generic would be the better word, but that's semantics.

FS98
An integral ##\int f(x)\,dx## is a solution to ##F(x)'=f(x)##.
As long as there are no boundary conditions to this differential equation, many solutions are possible. Nevertheless, they still have to solve the equation. As such they are a kind of generic solution, the set of possible flows if you like, which we call improper integral (I think; here it is call undetermined). Fixing a boundary condition means to determine a single flow of the vector field, a single solution. So in a way, generic would be the better word, but that's semantics.
Wouldn’t that definition of an integral rule out definite integrals as integrals?

fresh_42
Mentor

https://www.geogebra.org/m/M9SEYXSy

Do arbitrary paths through this vector field deserve the name solution or only the unique blue one, which required a deliberate choice? That's a discussion for a Wittgenstein seminar, not a matter of mathematics. The word integral refers to a solution and it serves its purpose. Nobody would benefit from a distinction here other than in the adjective. It still distinguishes all routes through this field, which do not follow a flow.

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An integral ##\int f(x)\,dx## is a solution to ##F(x)'=f(x)##.
Hi,
I do believe that is an indefinite integral.
An improper integral on the other hand is defined as an integral that has these:

1. One or both of the limits of integration are
$$\pm \infty$$

and/ or:

2. The function is not bounded over the domain of integration.

$$\forall x \in \text[a,b] \,\, \, \nexists \text{m} \in \mathbb{R} \, \, \text{s.t} \, \, |f(x)| \le \text{m}$$

To solve improper integrals, one has to use limits.
The example above is of an indefinite integral.
This is an example of an improper integral:

$$\displaystyle \int_{-\infty}^{5} \frac{1}{x} dx = \displaystyle\lim_{b \to -\infty} \displaystyle \int_{b}^{5} \frac{1}{x} dx$$

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Mark44
Mentor
which we call improper integral (I think; here it is call undetermined).
Maybe you're confusing the English terms "indefinite integral" and "improper integral."

I do believe that is an indefinite integral.
Yes, I agree, and I agree with your definition of an improper integral.

fresh_42
Mentor
Maybe you're confusing the English terms "indefinite integral" and "improper integral."
Yes, I did. I couldn't imagine or have forgotten that there is a certain name for integrals with ##\pm \infty## as boundaries. And "undetermined" as literal translation is of course basically the same word as indefinite. Thanks.

I understand what improper integrals are, but are they really integrals? The semantics are just a bit confusing.
Yes they are integrals, just that they require limits to be solved. I suggest you google Riemann sum. I think the reason they are called improper is because the summation uses limits, I am not sure as to the why they are named so.

Here is the simplest example, again:
$$\displaystyle \int_{1}^{\infty} \frac{1}{e^{x} } dx = \displaystyle\lim_{b \to \infty} \displaystyle \int_{1}^{b} \frac{1}{e^{x} } dx = [ \displaystyle\lim_{b \to \infty} \left(-\frac{1}{e^{b} } \right)] -[ - \frac{1}{e}] = \frac{1}{e}$$

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