I Is g(x) Equal to g(a) If Their Integrals Are Equivalent?

AI Thread Summary
The discussion centers on whether the equality of integrals involving functions g(x) and constants g(a) allows for the conclusion that g(x) equals g(a). Participants clarify that while integrals can be equal, it does not imply that the functions themselves are equal, as counterexamples demonstrate. The conversation also touches on the conditions under which integrals can be manipulated, emphasizing that assumptions about the behavior of f(x) are crucial. Ultimately, the consensus is that without additional constraints, one cannot cancel terms or equate functions based solely on integral equality. The thread concludes with a strong emphasis on the need for caution in interpreting integral relationships.
Steve Zissou
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TL;DR Summary
wondering if integrand terms can be cancelled
Howdy all,

Let's say we have, in general an expression:

$$ \int f(x) g(x) dx $$

But in through some machinations, we have, for parameter ##a##,

$$ \int f(x) g(x) dx = \int f(x) g(a) dx $$

...can we conclude that ## g(x) = g(a) ## ????

Thanks
 
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No.
 
BvU said:
No.
Thank you. Is there any chance you can help me understand?
 
let f(x)=1
let g(x)=x
integrate from 0 to 1
then a=½ satisfies your relation
 
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Frabjous said:
let f(x)=1
let g(x)=x
integrate from 0 to 1
then a=½ satisfies your relation
Thank you Frabjous.
How about this? Let's say we have
$$ \int f(x) g(a) dx = \int f(x) g(b) dx $$
...where a and b are parameters. Can we say that ## g(a)=g(b) ##??
...or even that ## a = b ##?
Thanks again
 
Steve Zissou said:
Thank you Frabjous.
How about this? Let's say we have
$$ \int f(x) g(a) dx = \int f(x) g(b) dx $$
...where a and b are parameters. Can we say that ## g(a)=g(b) ##??
Assuming that ∫f(x) is well behaved, yes. g(a) and g(b) are constants that can be pulled out of the integral.

Edit: By well behaved, I mean not equal to ±∞ or 0.
 
Last edited:
Steve Zissou said:
TL;DR Summary: wondering if integrand terms can be cancelled

Howdy all,

Let's say we have, in general an expression:

$$ \int f(x) g(x) dx $$

But in through some machinations, we have, for parameter ##a##,

$$ \int f(x) g(x) dx = \int f(x) g(a) dx $$

...can we conclude that ## g(x) = g(a) ## ????

Thanks
Are these indefinite integrals?
 
:smile:

On the left, ##g## is a function of ##x##. On the right ##g(a)## is a number. ##g(x) = g(a)## would mean ##g(x)## is a constant.

Steve Zissou said:
How about this? Let's say we have
$$ \int f(x) g(a) dx = \int f(x) g(b) dx $$
...where a and b are parameters. Can we say that ## g(a)=g(b) ##??
...or even that ## a = b ##?
Thanks again
No. If ##\int f(x) = 0 ## nothing can be said about ##a## and ##b##

##\ ##
 
PeroK said:
Are these indefinite integrals?
I was hoping to be able to make definite integrals here.
 
  • #10
I'm very sorry guys, allow me to make my examples firmer. I don't want to waste your time. I'll be back with a "better question."
 
  • #11
Steve Zissou said:
I was hoping to be able to make definite integrals here.
There's a general mathematical idea that ##A = B## if and only if ##A - B = 0##. You can apply that to your questions. Also, ##g(a)## and ##g(b)## are just numbers. For example:
$$\int_0^1f(x)g(x) \ dx = \int_0^1kf(x) \ \Rightarrow \ \int_0^1f(x)(g(x)-k) \ dx = 0$$But, lots of definite integrals are zero. So, for any function ##f##, there will be lots of functions ##g## where that integral is zero. In fact, if ##h## is some function where ##\int_0^1 h(x) \ dx = 0##, then ##g(x) = k + \frac{h(x)}{f(x)}## will be a counteraxample to your claim.
 
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  • #12
Thank you all for looking at this. I wish I had a more concrete example but I'm just trying to understand a certain concept here. So please allow me to rephrase my original question.
How about this? Let's say ## f = f(x,y) ## and also ## g = g(x,y) ##.
$$\int_{0}^{\infty}f(x,y)g(x,y)dx =\int_{0}^{\infty}f(x,y)g(x,a)dx$$
where ##a## is some number.
Can we conclude that ##g(x,y)=g(x,a)##?
 
  • #13
Steve Zissou said:
Thank you all for looking at this. I wish I had a more concrete example but I'm just trying to understand a certain concept here. So please allow me to rephrase my original question.
How about this? Let's say ## f = f(x,y) ## and also ## g = g(x,y) ##.
$$\int_{0}^{\infty}f(x,y)g(x,y)dx =\int_{0}^{\infty}f(x,y)g(x,a)dx$$
where ##a## is some number.
Can we conclude that ##g(x,y)=g(x,a)##?
Are you just making this stuff up? What is the point of all, one wonders!
 
  • #14
Steve Zissou said:
g(x,y)=g(x,a)
Since a is some number, you do realize that this means g(x,y)=g(x)? What concept are you actually trying to understand?
 
  • #15
Steve Zissou said:
Let's say ## f = f(x,y) ## and also ## g = g(x,y) ##.
$$\int_{0}^{\infty}f(x,y)g(x,y)dx =\int_{0}^{\infty}f(x,y)g(x,a)dx$$
where ##a## is some number.
Can we conclude that ##g(x,y)=g(x,a)##?
Nope! A trivial counterexample: ##f\left(x,y\right)=e^{-x},\;g\left(x,y\right)=\left(x-1\right)y##. Both of your integrals equal zero, but ##g\left(x,y\right)\neq g\left(x,a\right)##.
 
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  • #16
renormalize, thanks for your reply. In your example, can we not conclude that ##(x-1)y=(x-1)a##?
 
  • #17
I'm imagining that the two integrals are Riemann sums in a limit. So I'm seeing the ##f(x,y)##'s all cancelling out as additive terms.
 
  • #18
Steve Zissou said:
renormalize, thanks for your reply. In your example, can we not conclude that ##(x-1)y=(x-1)a##?
No, the left integral vanishes for any arbitrary ##y##. Nothing forces ##y=a##.
 
  • #19
Thank you to those who have shared their insights here.
Let me rephrase one more time.
$$\int_{0}^{\infty}f(x)g(x)dx=\int_{0}^{\infty}f(x)h(x)dx$$
1) Can we "cancel" the ##f(x)## terms on each side, thereby now having
$$\int_{0}^{\infty}g(x)dx=\int_{0}^{\infty}h(x)dx$$
2) Can we also now say
$$g(x)=h(x)$$
Thanks
 
  • #20
Steve Zissou said:
Thank you to those who have shared their insights here.
Let me rephrase one more time.
$$\int_{0}^{\infty}f(x)g(x)dx=\int_{0}^{\infty}f(x)h(x)dx$$
1) Can we "cancel" the ##f(x)## terms on each side, thereby now having
$$\int_{0}^{\infty}g(x)dx=\int_{0}^{\infty}h(x)dx$$
2) Can we also now say
$$g(x)=h(x)$$
Thanks
No to both questions. Plenty of counterexamples have been given above.
 
  • #21
Hill said:
No to both questions. Plenty of counterexamples have been given above.
Thank you Hill.
 
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