Equality of integrals VS equality of integrands

Does $$\int_{t=0}^{\infty}f(t)dt=\int_{t=0}^{\infty}g(t)dt$$ imply $$f(t)=g(t)$$ ?

PeroK
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Does $$\int_{t=0}^{\infty}f(t)dt=\int_{t=0}^{\infty}g(t)dt$$ imply $$f(t)=g(t)$$ ?
How could that possibly be true?

etotheipi
All you need is that the area under the curve from ##0## to ##\infty## is the same. There are a lot of different functions that satisfy this. E.g.

$$f(x) = \begin{cases} 0, & x< 0 \\ 1, & 0\leq x\leq 1 \\ 0, & 1< x \end{cases}$$ and $$g(x) = \begin{cases} 0, & x< 1 \\ 1, & 1\leq x\leq 2 \\ 0, & 2< x \end{cases}$$

• Delta2, member 587159, Ahmed Mehedi and 1 other person
@PeroK @dRic2
How could that possibly be true?
Is it possible to prove the equality using calculus of variation .... ? Or may be a special case of it ... ? I am not sure though ...

member 587159
This is false for many different reasons.

For example @etotheipi gave two different functions.

However, you can also do the following: Consider an integrable function ##f## and change its value in one (or finitely many) points. Let the so obtained be function be ##g##. Then clearly ##f\neq g## yet ##\int f = \int g##.

$$R[0,\infty[ \to \mathbb{R}: f \mapsto \int_{0}^\infty f$$
is injective. This is a linear functional on the space of Riemann-integrable functions on ##[0, \infty[##, and a linear functional on an infinite dimensional vector space is never injective.

Note that some partial results do hold:

(1) If ##f \geq 0## and ##\int f = 0## then ##f = 0## almost everywhere.

(2) If ##f \geq 0## and ##f## is continuous with ##\int f =0##, then ##f=0## (everywhere).

• PeroK, etotheipi and Ahmed Mehedi
PeroK
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@PeroK @dRic2

Is it possible to prove the equality using calculus of variation .... ? Or may be a special case of it ... ? I am not sure though ...
Are you really asking that if:
$$\int_{t=0}^{\infty}f(t)dt= 0$$
Then ##f(t) = 0## everywhere?

• etotheipi
etotheipi
There are some specific and limited circumstances under which you can equate certain integrands (I don't know whether it's totally rigorous, so perhaps @PeroK or @Math_QED can advise...). For instance, consider the following statement of Gauss' Law: $$\frac{Q}{\epsilon_0} = \int_V \frac{\rho}{\epsilon_0} \, dV = \oint_S \vec{E} \cdot d\vec{S} = \int_V \nabla \cdot \vec{E} \, dV$$ $$\int_V \frac{\rho}{\epsilon_0} \, dV = \int_V \nabla \cdot \vec{E} \, dV$$ Since this holds for any domain ##V##, you may deduce ##\nabla \cdot \vec{E} = \frac{\rho}{\epsilon_0}##.

I would suspect that if ##\int_{a}^{b}f(t)dt=\int_{a}^{b}g(t)dt## for all possible ##a, b##, then you would be able to say ##f(t) = g(t)##. But that's quite different to having fixed limits.

• Delta2, Ahmed Mehedi and PeroK
PeroK
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I would suspect that if ##\int_{a}^{b}f(t)dt=\int_{a}^{b}g(t)dt## for all possible ##a, b##, then you would be able to say ##f(t) = g(t)##. But that's quite different to having fixed limits.

If two continuous functions different at a single point, then they differ on an interval. Moreover, if ##f(x_0) > g(x_0)##, then ##f(x) > g(x)## on some interval containing ##x_0##.

This is needed to extract the Euler-Lagrange equations from the calculus of variations.

• • hutchphd and etotheipi
member 587159
I would suspect that if ##\int_{a}^{b}f(t)dt=\int_{a}^{b}g(t)dt## for all possible ##a, b##, then you would be able to say ##f(t) = g(t)##. But that's quite different to having fixed limits.

This is not quite true (it's true if you ask that ##f,g## are continuous). If they are only Riemann-integrable, best you can get is that they are equal almost everywhere.

Riemann-integral does not see the behaviour at single points, so you can take the example I made earlier in post #6 to get ##\int_a^b f = \int_a^b g## for all ##a,b## yet ##f \neq g##.

• • Delta2 and etotheipi
Are you really asking that if:
$$\int_{t=0}^{\infty}f(t)dt= 0$$
Then ##f(t) = 0## everywhere?

I am very sorry that I can not make my message clear. To be specific my question goes as follows:

Let us consider the following Lagrangian in the context of an optimization problem:

$$L=B\int_{t=0}^{\infty}e^{-\beta t}\frac{c(t)^{1-\theta}}{1-\theta}dt+\lambda \left[ k(0)+\int_{t=0}^{\infty} e^{-R(t)}e^{(n+g)t}w(t)dt - \int_{t=0}^{\infty} e^{-R(t)}e^{(n+g)t}c(t)dt \right]$$

After taking the first partial derivative of the above Lagrangian with respect to c(t) (and setting it to zero as first order optimality condition) it is written that

$$Be^{-\beta t} c(t)^{-\theta}-\lambda e^{-R(t)}e^{(n+g)t}=0$$

It seems that they take the first partial derivative of the Lagrangian with respect to c(t) and simply ignore the dt from both sides.

How the second line follows from the first line? They said it can be proved using calculus of variation. But, how can I prove it?

• PeroK
I am very sorry that I can not make my message clear. To be specific my question goes as follows:

Let us consider the following Lagrangian in the context of an optimization problem:

$$L=B\int_{t=0}^{\infty}e^{-\beta t}\frac{c(t)^{1-\theta}}{1-\theta}dt+\lambda \left[ k(0)+\int_{t=0}^{\infty} e^{-R(t)}e^{(n+g)t}w(t)dt - \int_{t=0}^{\infty} e^{-R(t)}e^{(n+g)t}c(t)dt \right]$$

After taking the first partial derivative of the above Lagrangian with respect to c(t) (and setting it to zero as first order optimality condition) it is written that

$$Be^{-\beta t} c(t)^{-\theta}-\lambda e^{-R(t)}e^{(n+g)t}=0$$

It seems that they take the first partial derivative of the Lagrangian with respect to c(t) and simply ignore the dt from both sides.

How the second line follows from the first line? They said it can be proved using calculus of variation. But, how can I prove it?

@Math_QED @etotheipi

dRic2
Gold Member
That's a functional derivative, not a partial derivative. It's an other story.

If you want a quick recipe, a functional derivative like ##\frac {\delta L} {\delta g}## where ##L =\int h[g]## can be evaluated by trowing away the integral and calculating ##\frac {\partial h[g]} {\partial g}##. Ok don't hate me for this comment.

• Ahmed Mehedi
That's a functional derivative, not a partial derivative. It's an other story.

If you want a quick recipe, a functional derivative like ##\frac {\delta L} {\delta g}## where ##L =\int h[g]## can be evaluated by trowing away the integral and calculating ##\frac {\partial h[g]} {\partial g}##. Ok don't hate me for this comment.

May be you are right! Perhaps they were talking about functional derivative and not the partial one. Can you provide me any good quick read regarding functional derivatives?

etotheipi
LOL this question took a turn...

I don't know if I can help from here on, I've only ever skimmed through the first chapter of a calculus of variations textbook.

• Ahmed Mehedi and dRic2
LOL this question took a turn...

I don't know if I can help from here on, I've only ever skimmed through the first chapter of a calculus of variations textbook.

HAHA ....... I am yet to see the textbook of calculus of variation ....... Let alone chapter one ......

dRic2
Gold Member
Can you provide me any good quick read regarding functional derivatives?
Sorry, I'm not very familiar with functional derivatives. I just worked out some tricks to evaluate them in case of need. All I know comes from pag 54-56 from Lanczos' book on analytical mechanics and from 5 pages of notes by a professor of mine.

• Ahmed Mehedi
Given
$$\int_{0}^{\infty} f(t) dt = \int_{0}^{\infty} g(t) dt$$ the most we can get is :
$$\int_{0}^{\infty} f(t) dt - \int_{0}^{\infty} g(t) dt = 0$$
$$\lim_{x\to \infty} \int_{0}^{x} \left[f(t) -g(t) \right] dt = 0$$
$$\lim_{x \to \infty} \frac{d}{dx} \int_{0}^{x} \left[f(t) - g(t) \right] = 0$$
$$\lim_{x \to \infty} f(x) - g(x) = 0$$
That is to say, functions ##f## and ##g## converge to each other as they approach to infinity.

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• • Delta2, member 587159, PeroK and 2 others
Given
$$\int_{0}^{\infty} f(t) dt = \int_{0}^{\infty} g(t) dt$$ the most we can get is :
$$\int_{0}^{\infty} f(t) dt - \int_{0}^{\infty} g(t) dt = 0$$
$$\lim_{x\to \infty} \int_{0}^{x} \left[f(t) -g(t) \right] dt = 0$$
$$\lim_{x \to \infty} \frac{d}{dx} \int_{0}^{x} \left[f(t) - g(t) \right] = 0$$
$$\lim_{x \to \infty} f(x) - g(x) = 0$$
That is to say, functions ##f## and ##g## converge to each other as they approach to infinity.

A good interpretation!

• member 587159
Given
$$\int_{0}^{\infty} f(t) dt = \int_{0}^{\infty} g(t) dt$$ the most we can get is :
$$\int_{0}^{\infty} f(t) dt - \int_{0}^{\infty} g(t) dt = 0$$
$$\lim_{x\to \infty} \int_{0}^{x} \left[f(t) -g(t) \right] dt = 0$$
$$\lim_{x \to \infty} \frac{d}{dx} \int_{0}^{x} \left[f(t) - g(t) \right] = 0$$
$$\lim_{x \to \infty} f(x) - g(x) = 0$$
That is to say, functions ##f## and ##g## converge to each other as they approach to infinity.

This is false. Consider ##f=0## and $$g(x) = \begin{cases}1 \quad x \in \mathbb{N} \\ 0 \quad x \notin \mathbb{N}\end{cases}$$
Then $$\int_0^\infty f = 0 = \int_0^\infty g$$ yet $$\lim_{x \to \infty}[ f(x)-g(x)] = -\lim_{x \to \infty} g(x)$$ does not exist.

The flaw happens when you introduce the derivative.

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• etotheipi
This is false. Consider ##f=0## and $$g(x) = \begin{cases}1 \quad x \in \mathbb{N} \\ 0 \quad x \notin \mathbb{N}\end{cases}$$
Then $$\int_0^\infty f = 0 = \int_0^\infty g$$ yet $$\lim_{x \to \infty}[ f(x)-g(x)] = -\lim_{x \to \infty} g(x)$$ does not exist.

The flaw happens when you introduce the derivative.
I assumed the functions to be continuous.

member 587159
I assumed the functions to be continuous.

Even then you must justify switching the limits involved, which you didn't. I'm pretty sure the statement is even false for continuous functions.

• etotheipi
PeroK
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I assumed the functions to be continuous.
It's still false. There are continuous functions that do not converge to ##0## as ##t \rightarrow \infty## yet the integral exists.

It's a good exercise to find one.

• Infrared and member 587159
Even then you must justify switching the limits involved, which you didn't. I'm pretty sure the statement is even false for continuous functions.
We can do the switching when they are monotone.

member 587159
We can do the switching when they are monotone.

If you keep throwing in extra assumptions, eventually you will be right yes...