Integral (-∞ to ∞) of Zero and ∞/∞

What mean in physical sense undefined, I must know is alpha induced zero, when circulation Γ=constant, for infinity wing and finite wing?Undefined means that the expression has no definite value. It could be anything. In this case, it means that the limit of the expression can vary depending on how you approach the point where it is undefined.To find the limit in this case, you need to use the definition of a limit, which involves evaluating the expression as the variable approaches the undefined point from both sides. However, as I mentioned before, the fact that your expression has a derivative of ##\Gamma## in it suggests that ##\Gamma## may not actually be constant, so you may need to re-evaluate your approach to this problem
  • #1
Jurgen M
Homework Statement
I have problem to solve this tasks.
Relevant Equations
Calculus, infinity
Are both integral on picture below equal zero?
I think both are zero, area of zero section under function must be zero.
INTEGRAL.png


If M=, b= , what is reslut?
Logically ∞/∞ will be 1..but...
BESKON.png
 
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  • #2
Jurgen M said:
Homework Statement:: I have problem to solve this tasks.
Relevant Equations:: Calculus, infinity

Are both integral on picture below equal zero?
I think both are zero, area of zero section under function must be zero.
View attachment 317947
Where did these integrals come from? Are they from a textbook? I don't understand why someone would write an expression as ##\frac 0 {y_0 - y}##. Clearly that equals zero for all y such that ##y \ne y_0##.
Jurgen M said:
If M=, b= , what is reslut?
Logically ∞/∞ will be 1..but...
View attachment 317948
##[\frac \infty \infty]## is one of several types of indeterminate expressions, which means it can come out as pretty much any number. To evaluate such an expression, you need to use limits.

For example, here are three similar expressions, all with different values:
  1. ##\lim_{x \to \infty} \frac {x^2}x = \infty##
  2. ##\lim_{x \to \infty} \frac {x}x = 1##
  3. ##\lim_{x \to \infty} \frac x {x^2} = 0##
 
  • #3
Mark44 said:
Where did these integrals come from? Are they from a textbook? I don't understand why someone would write an expression as ##\frac 0 {y_0 - y}##. Clearly that equals zero for all y such that ##y \ne y_0##.

##[\frac \infty \infty]## is one of several types of indeterminate expressions, which means it can come out as pretty much any number. To evaluate such an expression, you need to use limits.

For example, here are three similar expressions, all with different values:
  1. ##\lim_{x \to \infty} \frac {x^2}x = \infty##
  2. ##\lim_{x \to \infty} \frac {x}x = 1##
  3. ##\lim_{x \to \infty} \frac x {x^2} = 0##
Yes from textbook,in denominator was derviaton of constant,so I put it just zero instead..So both integrals are zero?For second case, what is result if use limes?
 
  • #4
In the absence of any functional relationship between the numerator and denominator, I don't think there can be a unique result. [tex]
\begin{split}
\lim_{x \to 0} \lim_{y \to 0} \frac{y}{x} &= 0 \\
\lim_{y \to 0} \lim_{x \to 0} \frac{y}{x} &= ??? \end{split}[/tex]
 
  • #5
Jurgen M said:
Yes from textbook,in denominator was derviaton of constant,so I put it just zero instead..So both integrals are zero?
Do you mean numerator, not denominator? To answer your question I would need to see what the fraction was before you differentiated it.
Jurgen M said:
For second case, what is result if use limes?
Here is a lime:
limes.jpg

The word you're looking for is limit.

The expression you showed is $$\frac{M^2}{1/2 \rho \pi V^2 b^2}$$

In your image you have "ro" which I assume you mean the Greek letter lowercase rho (##\rho##). Also, 3.14 is a very crude approximation for ##\pi##.
It's impossible to say what the limit here is without knowing whether there is a relationship between M and b.
 
  • #6
Jurgen M said:
Yes from textbook,in denominator was derviaton of constant,so I put it just zero instead..So both integrals are zero?
Did you mean derivative of a constant? I was confused because I thought you meant to write "derivation of a constant" which makes no sense. If you did mean to type derivation, you should know the word doesn't mean what you apparently think it does.
 
  • #7
Jurgen M said:
I think both are zero, area of zero section under function must be zero.
You need to be a little careful. The integrands are zero when ##y\ne y_0##, but what about when ##y=y_0##? That point is clearly in the interval of the first integral. It may or may not be in the second integral. The quotient ##0/0## is undefined, so you have to figure out how to handle that complication before you can give a definite answer.
 
  • #8
Mark44 said:
Do you mean numerator, not denominator? To answer your question I would need to see what the fraction was before you differentiated it.

The word you're looking for is limit.

The expression you showed is $$\frac{M^2}{1/2 \rho \pi V^2 b^2}$$

In your image you have "ro" which I assume you mean the Greek letter lowercase rho (##\rho##). Also, 3.14 is a very crude approximation for ##\pi##.
It's impossible to say what the limit here is without knowing whether there is a relationship between M and b.
Yes I mean on numerator.
Γ is const, so dΓ/dy=0

here is full integrals
final integ.png

Mark44 said:
The expression you showed is $$\frac{M^2}{1/2 \rho \pi V^2 b^2}$$

In your image you have "ro" which I assume you mean the Greek letter lowercase rho (##\rho##). Also, 3.14 is a very crude approximation for ##\pi##.
It's impossible to say what the limit here is without knowing whether there is a relationship between M and b.
ro is density
b is wingspan, M is lift force, when wingspan go to infinity it is logical that M(lift force) go to infinity as well..
So I put ∞ in numerator instead M..
 
  • #9
vela said:
You need to be a little careful. The integrands are zero when ##y\ne y_0##, but what about when ##y=y_0##? That point is clearly in the interval of the first integral. It may or may not be in the second integral. The quotient ##0/0## is undefined, so you have to figure out how to handle that complication before you can give a definite answer.
What mean in physical sense undefined, I must know is alpha induced zero, when circulation Γ=constant, for infinity wing and finite wing?

final integ.png
 
  • #10
I would start with one of your first integrals,
$$\frac 1 {4\pi V}\int_{-b}^b \frac{\frac{d\Gamma}{dy}}{y_0 - y}~dy$$
First, is ##y_0## in the interval [-b, b]? If so, the integral is undefined.
Second, you say that ##\Gamma## (uppercase Greek letter gamma), the circulation, is constant. Since the integral has the derivative of ##\Gamma## in it, that suggests that ##\Gamma## might not be constant.

It would be helpful if you showed where this integral comes from, by posting a photo of the textbook where this integral is discussed.

Jurgen M said:
ro is density
As already mentioned, "ro" should be "rho", the lowercase Greek letter ##\rho##. Density is often represented by ##\rho##.
Jurgen M said:
b is wingspan, M is lift force, when wingspan go to infinity it is logical that M(lift force) go to infinity as well..
So I put ∞ in numerator instead M..
That makes no sense. You can't just plug in ##\infty## into an expression. In addition, if your logic is correct, you have b being infinitely large as well, so the expression you're trying to evaluate is the indeterminate form ##[\frac \infty \infty]##. I showed three examples earlier in this thread of this type of indeterminate limit, all with different values.
 
  • #11
In the case of an infinite wingspan, it makes more sense to talk of lift per unit length along the wing, rather than the total lift which is indeed infinite.

Effectively you have reduced the spatial dimension by 1, and whatever results you have for a finite wingspan might not be applicable.
 
  • #12
Mark44 said:
I would start with one of your first integrals,
$$\frac 1 {4\pi V}\int_{-b}^b \frac{\frac{d\Gamma}{dy}}{y_0 - y}~dy$$
First, is ##y_0## in the interval [-b, b]? If so, the integral is undefined.
Second, you say that ##\Gamma## (uppercase Greek letter gamma), the circulation, is constant. Since the integral has the derivative of ##\Gamma## in it, that suggests that ##\Gamma## might not be constant.

It would be helpful if you showed where this integral comes from, by posting a photo of the textbook where this integral is discussed.As already mentioned, "ro" should be "rho", the lowercase Greek letter ##\rho##. Density is often represented by ##\rho##.
That makes no sense. You can't just plug in ##\infty## into an expression. In addition, if your logic is correct, you have b being infinitely large as well, so the expression you're trying to evaluate is the indeterminate form ##[\frac \infty \infty]##. I showed three examples earlier in this thread of this type of indeterminate limit, all with different values.
Yes for wing(3D) gama is not constant, but in book write for airfoil(2D) is constant so alpha induced=0, because airfoil dont have induced drag....
It not clear why book say it is zero if integral is undefined like you say.

What indeterminate integral mean in reality, not possible?

Also what is difference in undfined and indeterminate?
 
  • #13
Jurgen M said:
Yes for wing(3D) gama is not constant, but in book write for airfoil(2D) is constant so alpha induced=0, because airfoil dont have induced drag....
It not clear why book say it is zero if integral is undefined like you say.
I'm not up on airfoil mathematics, so I can't say anything meaningful about gamma (##\Gamma##), alpha (##\alpha##), or induce drag. I'm curious as to how the integral you showed was developed.

Jurgen M said:
What indeterminate integral mean in reality, not possible?
I didn't say the integral was indeterminate, but rather the expression you showed, where both M and b were infinite. Any time you have an expression of the form ##[\frac \infty \infty]## or ##[\frac 0 0 ]## (and a few others), these are indeterminate forms. They need to be evaluated using limits.

Jurgen M said:
Also what is difference in undfined and indeterminate?
Some expressions that are undefined are ##\frac 1 0## or ##\sqrt{-1}## (if you're limited to real numbers), and others that involve invalid operations. For indeterminate forms, I gave three examples back in post #2.
 
  • #14
Mark44 said:
I'm not up on airfoil mathematics, so I can't say anything meaningful about gamma (##\Gamma##), alpha (##\alpha##), or induce drag. I'm curious as to how the integral you showed was developed.

I didn't say the integral was indeterminate, but rather the expression you showed, where both M and b were infinite. Any time you have an expression of the form ##[\frac \infty \infty]## or ##[\frac 0 0 ]## (and a few others), these are indeterminate forms. They need to be evaluated using limits.

Some expressions that are undefined are ##\frac 1 0## or ##\sqrt{-1}## (if you're limited to real numbers), and others that involve invalid operations. For indeterminate forms, I gave three examples back in post #2.
If y ≠yo, integral is zero, if y=yo is undefined?
 
  • #15
Jurgen M said:
If y ≠yo, integral is zero, if y=yo is undefined?
That was already touched on by @vela in post #7.
vela said:
You need to be a little careful. The integrands are zero when ##y\ne y_0##, but what about when ##y=y_0##? That point is clearly in the interval of the first integral. It may or may not be in the second integral. The quotient ##0/0## is undefined, so you have to figure out how to handle that complication before you can give a definite answer.
By "first integral" he means the one with integration limits of ##-\infty## and ##\infty##. The other integral is the one with limits of -b and b.
 
  • #16
Mark44 said:
First, is ##y_0## in the interval [-b, b]? If so, the integral is undefined.
Why? isnt integral undefined only when y=yo?
 
  • #17
Jurgen M said:
Why? isnt integral undefined only when y=yo?
Yes, but if ##y_0 \in [-b, b]## there's a problem.
 

What is the meaning of "Integral (-∞ to ∞) of Zero and ∞/∞"?

The integral of (-∞ to ∞) of Zero and ∞/∞ refers to the process of finding the area under the curve of a function that approaches zero at negative and positive infinity, or a function that has an infinite limit at negative and positive infinity.

Why is this type of integral important in mathematics?

This type of integral is important in mathematics because it allows for the evaluation of functions that have infinite limits at negative and positive infinity. It also helps in solving problems related to areas, volumes, and rates of change.

What is the difference between "Zero and ∞/∞" and "∞/∞"?

The difference between "Zero and ∞/∞" and "∞/∞" is that the former refers to a function that approaches zero at negative and positive infinity, while the latter refers to a function that has an indeterminate form at negative and positive infinity.

Can the integral of "Zero and ∞/∞" be evaluated using traditional methods?

No, the integral of "Zero and ∞/∞" cannot be evaluated using traditional methods such as the fundamental theorem of calculus. It requires advanced techniques such as L'Hopital's rule or substitution methods.

Are there any real-world applications of this type of integral?

Yes, there are several real-world applications of this type of integral, such as in physics, where it is used to calculate work done by a variable force, or in economics, where it is used to calculate total profit or loss when the demand and supply curves approach zero and infinity.

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