Proving two functions are equal under an integral

[dreNL]

I have the following problem.
If
\int_0^\infty f(s)ds=\int_0^\infty g(s)ds
What are sufficient conditions such that f(s)=g(s)?

I know that two functions f(s),g(s) are equal if their domain, call it S, is equal and if f(s)=g(s) for all s\in S but I can't figure this one out.

The full problem is actually

-\int_0^\infty\frac{d}{ds}\left(f(\vec{r}+s\hat{\Omega},E,\hat{\Omega})e^{-\Sigma_ts}\right)ds
=\int_0^\infty\Sigma_te^{-\Sigma_ts}g(\vec{r}+s\hat{\Omega},E,\hat{\Omega})ds
and therefore hopefully
-\frac{d}{ds}\left(f(\vec{r}+s\hat{\Omega},E,\hat{\Omega})e^{-\Sigma_ts}\right)=\Sigma_te^{-\Sigma_ts}g(\vec{r}+s\hat{\Omega},E,\hat{\Omega})

Please help me! I'm finishing soon with my work and proving this (or at least having sufficient conditions) would be very welcome!

 

[HallsofIvy]

I have the following problem.
If
\int_0^\infty f(s)ds=\int_0^\infty g(s)ds
What are sufficient conditions such that f(s)=g(s)?l

Just knowing that the integrals over a specified interval are 0 tells you essentially nothing about f and g themselves. The only "sufficient" condition is that f(x)= g(x)!

I know that two functions f(s),g(s) are equal if their domain, call it S, is equal and if f(s)=g(s) for all s\in S but I can't figure this one out.

The full problem is actually

-\int_0^\infty\frac{d}{ds}\left(f(\vec{r}+s\hat{\Omega},E,\hat{\Omega})e^{-\Sigma_ts}\right)ds
=\int_0^\infty\Sigma_te^{-\Sigma_ts}g(\vec{r}+s\hat{\Omega},E,\hat{\Omega})ds
and therefore hopefully
-\frac{d}{ds}\left(f(\vec{r}+s\hat{\Omega},E,\hat{\Omega})e^{-\Sigma_ts}\right)=\Sigma_te^{-\Sigma_ts}g(\vec{r}+s\hat{\Omega},E,\hat{\Omega})

Please help me! I'm finishing soon with my work and proving this (or at least having sufficient conditions) would be very welcome!

 

[dreNL]

Thank you HallsofIvy,

How about if I restate the problem as:
\int_0^\infty e^{-\Sigma_t s}f(s)ds=\int_0^\infty e^{-\Sigma_t s}g(s)ds
And it should hold for any \Sigma_t \in [0,\infty)

for very large \Sigma_t\to\infty, the only contribution is just right from zero and because the integrations over a very small interval return the same value I can imagine f(s)=g(s) there. For slightly less large \Sigma_t the integrations still return the same value and so forth.
This is by no means mathematical proof (as I am not a mathematician). From a physical point of view it makes some sense..
Actually, does it help if I tell you the functions f(s),g(s) I am considering are all well behaved (finite number of discontinuities)?

As you told me I need to proove that f(s)=g(s) for every any s\in [0,\infty). Is there now any hope of stating conditions for f(s),g(s) so that I can use the above to make it plausible that f(s)=g(s) or even give proof?
Thanks in advance

 

[chiro]

Thank you HallsofIvy,

How about if I restate the problem as:
\int_0^\infty e^{-\Sigma_t s}f(s)ds=\int_0^\infty e^{-\Sigma_t s}g(s)ds
And it should hold for any \Sigma_t \in [0,\infty)

for very large \Sigma_t\to\infty, the only contribution is just right from zero and because the integrations over a very small interval return the same value I can imagine f(s)=g(s) there. For slightly less large \Sigma_t the integrations still return the same value and so forth.
This is by no means mathematical proof (as I am not a mathematician). From a physical point of view it makes some sense..
Actually, does it help if I tell you the functions f(s),g(s) I am considering are all well behaved (finite number of discontinuities)?

As you told me I need to proove that f(s)=g(s) for every any s\in [0,\infty). Is there now any hope of stating conditions for f(s),g(s) so that I can use the above to make it plausible that f(s)=g(s) or even give proof?
Thanks in advance

Could you possibly use the FTC to show that if these functions had antiderivate expressions then based on these expressions the only possible way for these representations to equate is for the appropriate antiderivative expressions be equal and hence the expression f(s) = g(s)?