Cancelling across integrals

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

No.

BvU said:

No.

Thank you. Is there any chance you can help me understand?

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.

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?

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.

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."

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.

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)##?

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!

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?

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)##.

renormalize, thanks for your reply. In your example, can we not conclude that ##(x-1)y=(x-1)a##?

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.

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##.

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

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.

Hill said:

No to both questions. Plenty of counterexamples have been given above.

Thank you Hill.