Proving two functions are equal under an integral

• dreNL
In summary, the problem is to find sufficient conditions for f(s) and g(s) such that the integral of e^{-\Sigma_t s} over the interval [0,\infty) is equal for both functions. The only sufficient condition is that f(s) = g(s). The functions are well-behaved with a finite number of discontinuities. It may be possible to use the Fundamental Theorem of Calculus to prove that f(s) = g(s) by showing that their antiderivative expressions must be equal.
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!

dreNL said:
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!

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?

dreNL said:
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?

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)?

1. How do I prove that two functions are equal under an integral?

To prove that two functions are equal under an integral, you need to show that their integrals are equal. This means that the area under the curve of both functions must be the same.

2. Can I use any method to prove two functions are equal under an integral?

Yes, there are several methods that can be used to prove that two functions are equal under an integral. Some common methods include the fundamental theorem of calculus, substitution, and integration by parts. The method you choose will depend on the specific functions you are working with.

3. What should I do if I am unable to prove that two functions are equal under an integral?

If you are unable to prove that two functions are equal under an integral, there may be a mistake in your calculations or approach. It is important to double check your work and try a different method if necessary. You can also seek help from a math tutor or colleague for guidance.

4. Can two functions be equal under an integral but not equal everywhere?

Yes, two functions can have the same integral but be different in other areas. This is because the integral only measures the overall area under the curve, not the individual points on the curve.

5. Is it possible for two functions to have different integrals but still be equal?

No, if two functions have different integrals, they are not equal. The integral of a function is a unique value that represents the area under the curve, so if two functions have different integrals, they must be different functions.

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