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"Don't panic!"

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- Thread starter "Don't panic!"
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In summary: Sorry, that's not really how integrals work.Integrals are a way to calculate the area under a curve between two points. So if you wanted to calculate the area under a curve from -4 to 4, you would integrate from -4 to 4.If the improper integral doesn't exist, then there could exist paths for the limits of integration to take which will give you different, finite... answers.

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"Don't panic!"

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If you want a formal proof, then you'll have to start by mentioning what the definition of the integral is.

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"Don't panic!"

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micromass said:

If you want a formal proof, then you'll have to start by mentioning what the definition of the integral is.

Does this apply to Riemann integration?

I was wondering because I read that one of the main reasons that the Dirac delta function doesn't make sense is the following:

" We require that [tex]\delta (x)=\bigg\lbrace\begin{matrix}1\qquad\quad x=0\\ 0\quad\text{elsewhere}\end{matrix}[/tex] and that [tex]\int_{-\infty}^{\infty}\delta (x)=1[/tex] The problem is these to requirements cannot be satisfied simultaneously, as the integral of a function defined as above is zero".

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"Don't panic!" said:Does this apply to Riemann integration?

Yes.

I was wondering because I read that one of the main reasons that the Dirac delta function doesn't make sense is the following:

" We require that [tex]\delta (x)=\bigg\lbrace\begin{matrix}1\qquad\quad x=0\\ 0\quad\text{elsewhere}\end{matrix}[/tex] and that [tex]\int_{-\infty}^{\infty}\delta (x)=1[/tex] The problem is these to requirements cannot be satisfied simultaneously, as the integral of a function defined as above is zero".

Yes, the delta function is not a function ##\mathbb{R}\rightarrow \mathbb{R}## in the classical sense of the word. Rather, it must be defined rather differently. There are many possible definitions.

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"Don't panic!"

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micromass said:Yes, the delta function is not a function R→R\mathbb{R}\rightarrow \mathbb{R} in the classical sense of the word. Rather, it must be defined rather differently. There are many possible definitions.

How does one show that if a function is defined to have a non-zero value at a single point, but is zero elsewhere, that it's Riemann integral is zero?

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Well, what is your definition of the Riemann integral? Use that definition directly.

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"Don't panic!"

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micromass said:Well, what is your definition of the Riemann integral? Use that definition directly.

Well I tend to use the following definition, [tex]\int_{a}^{b}f(x)dx =\lim_{n\rightarrow\infty}\frac{(b-a)}{n}\sum_{i=1}^{n}f(x^{\ast}_{i})[/tex] where ##x^{\ast}_{i}\in [x_{i},x_{i+1}]##.

I can't see how I can show it from this definition unfortunately.

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"Don't panic!" said:Well I tend to use the following definition, [tex]\int_{a}^{b}f(x)dx =\lim_{n\rightarrow 0}\frac{(b-a)}{n}\sum_{i=1}^{n}f(x^{\ast}_{i})[/tex] where ##x^{\ast}_{i}\in [x_{i},x_{i+1}]##.

I can't see how I can show it from this definition unfortunately.

That definition is only useful for finite ##a## and ##b##. So let us do it in this case. Take ##a = -1##, ##b=1##. Then choose your partition

[tex] -1, -1 + \frac{2}{n}, -1 + \frac{4}{n}, ..., 1 - \frac{2}{n}, 1[/tex]

end choose your ##x^{ast}_i## arbitrary. There are essentially two different situations:

1) ##x^*_i## is never ##0##. Then ##f(x^*_i) = 0## for each ##i##. Thus ##\sum_{i=1}^n f(x^*_i)\frac{2}{n} = 0##. So ##\int_a^b f(x)dx = 0##.

2) There is some ##i## such that ##x^*_i = 0##. Then ##f(x^*_i) = c## and the other ##i## satisfy ##f(x^*_i) = 0##. Hence

[tex]\sum_{i=1}^n f(x^*_i) \frac{2}{n} = \frac{2c}{n}[/tex]

This converges to ##0## as ##n\rightarrow +\infty##. So ##\int_a^b f(x)dx = 0##.

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"Don't panic!"

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micromass said:That definition is only useful for finite aa and bb.

What is a better definition that takes into account infinite intervals?

Thanks for the demonstration for the other case (I had just thought of a similar way to do it just before I received your message, so I'm glad I understood it correctly).

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JonnyG

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It reduces to a finite sum, which micromass showed you is 0.

Look up the concept of "measure zero"

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"Don't panic!"

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- #12

JonnyG

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If you can prove, which shouldn't be too hard, that [itex] \int_{-\infty}^{\infty} f(x)d(x) [\itex] exists (meaning the limit as shown above exists and is finite),

If the improper integral doesn't exist, then there could exist paths for the limits of integration to take which will give you different, finite answers!

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JonnyG

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- #14

mathwonk

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well how much area is under that graph? isn't it obviously zero?

- #15

HallsofIvy

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Integration is a mathematical process of finding the area under a curve, which can be represented by a function. When a function is non-zero at one point, it means that the value of the function is zero everywhere except at that one point.

Integration of a function that's non-zero at one point is important because it allows us to calculate the total area under the curve, even if the function has a non-zero value at one point. This is useful in real-world applications, such as calculating the work done by a variable force.

The process of integrating a function that's non-zero at one point is the same as integrating any other function. We use techniques such as substitution, integration by parts, or partial fractions to find the antiderivative of the function, and then evaluate the integral using the limits of integration.

Yes, a function can be non-zero at more than one point and still be integrated. The only requirement for a function to be integrable is that it is continuous on the interval of integration. A function can be non-zero at multiple points and still be continuous, allowing it to be integrated.

If the function is not continuous at the point where it is non-zero, then the integral cannot be evaluated using traditional methods. In this case, we would need to use more advanced techniques, such as improper integration or the Cauchy principal value, to calculate the integral.

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