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This might be silly question, but here it goes:

Suppose I have an integral $\int_a^b f(x) dx$, and by the F.T.C we have as $F(b) - F(a)$ as a result.

Now suppose that instead of $F(b) - F(a)$, I want to find the value $F(b) + F(a)$.

Of course, one could trivially say just separate them, flip the negative for $F(a)$ and then add it to $F(b)$.

However, imagine a scenario where I cannot separate them, nor change the bounds on the integral. I can only manipulate the integrand itself.

That is, is there a function $g(x)$ that we can devise such that $\int_a^b g(x) f(x) dx = F(b) + F(a)$, given $\int_a^b f(x) dx = F(b) - F(a)$?

Edit: The solution $\int_a^b g(x) f(x) dx = F(b) + F(a) + C$ where $C$ is some constant, is also acceptable.

To give an example that makes this more clear: if we had a 2 variable function $f(x, t)$, $\int_a^b f(x, t) dx = F(b,t) - F(a,t)$, then $\int_a^b g(x) f(x, t) dx = F(b,t) + F(a,t) + C$ suffices, but $\int_a^b g(x) f(x, t) dx = F(b,t) + F(a,t) + O(t)$ would not, as it is no longer a constant but also dependent on $t$.

Here is a nice example: For $n \in \mathbb{N}$, $x,c > 0$

$\Re[\Gamma[-\frac{n}{2}, -x] = \Re[\Gamma[-\frac{n}{2}, -x -c]] = \Gamma[-\frac{n}{2}]$, source: Incomplete Gamma function of negative arguments

Thank you in advance for any feedback and help!

(Also, I'd be happy to explain a data streaming setting where this applies, but didn't want to overcrowd this post with irrelevant details. I'm happy to provide context though if requested.)

BBadman
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  • This cannot be, because $\int_a^bh=-\int_b^ah.$ – Anne Bauval Apr 29 '23 at 05:47
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    Since the value of $F(b)+F(a)$ depends on the arbitrary constant $+C$, the sum can be any value, regardless of what you do with $f(x)$. – coiso Apr 29 '23 at 05:48
  • No. Suppose you have $F_0(x)$ such that $\frac{d}{dx}F_0(x)=f(x)$, and $F_c(x)=F_0(x)+c$ for $c\not=0$ which means $\frac{d}{dx}F_c(x)=f(x)$ too. You get $\int_a^b f(x) ,dx=F_0(b)-F_0(a)=F_c(b)-F_c(a)$ but you could not have both $\int_a^b g(x) f(x) dx = F_0(b) + F_0(a)$ and $\int_a^b g(x) f(x) dx = F_c(b) + F_c(a)= F_0(b) + F_0(a)+2c$ since they differ by $2c$ – Henry Apr 29 '23 at 05:48

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Remember, $F$ is only determined up to a constant, so the value you want isn't well defined.

As a concrete example, suppose $f(x) =2x$. Then $F(x)$ might be equal to $x^2$, and, of course the FTC would work, and $F(b)+F(a)$ would equal $b^2+a^2$. But $F(x)$ could also be $x^2+1$, and the FTC would still be satisfied, but $F(b)+F(a)$ would equal $b^2+a^2+2$.

One way to get rid of the arbitrary constant is to pick some $x_0$, and define

$$F(x)=\int_{x_0}^x f(t) dt$$

If you are dealing with a situation where these integrals have physical meanings, perhaps there is a sensible value for $x_0$ laying around? You mentioned that this is for a data streaming application - is there a "stream start time" that would work for $x_0$? I confess that with just the information you've given, it's hard for me to imagine a situation where $F(a)+F(b)$ is meaningful.

JonathanZ
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  • Thanks for the response! Good point! You're right, the way I presented it was ill-posed. But actually, for our purposes, a constant factor is completely fine. Going off your example, supposing we had a function $f(x) = 2x$, then we can think of all functions $F = { x^2 + c| c \in \mathbb{R} }$, as forming a sort of equivalence class of solutions, and say any solution that falls in this set is valid. So $a^2 + b^2 + 10$ can be thought of as just as the same solution $a^2 + b^2$, for instance (essentially just discounting constant factors). Does that make sense? – BBadman Apr 29 '23 at 07:50
  • I've tried to write several attempts at giving a good example, but ultimately thought it might just be better to leave a reference that covers a similar idea well: http://conferences.sigcomm.org/imc/2007/papers/imc111.pdf (Sec. 2). One thing they do in this paper is devise a way to compute $x_1 \ln(x_1) + ... + x_n \ln(x_n)$, only effectively using $x_1^p + ... + x_n^p$, for $p \in [0,2]$. This idea would be similarly motivated, in that we can still recover $a^2 + b^2$ given $a^2 + b^2 + C$, since the extra constant could be removed with a linear operation. – BBadman Apr 29 '23 at 08:15
  • Sorry for all the comments, ran out of space, but ya please let me know if I there any questions, would be happy to try to clarify. Best, – BBadman Apr 29 '23 at 08:17
  • So I glanced at that reference, and that only yields an approximation, and only applies to those particular functions. Whereas you've asked about an arbitrary $f$. My guess is that the answer to the general question is "no". It feels like there is also some work to be done refining just exactly what it is you want, and math.se isn't really set-up to go through the multiple cycles to do that. – JonathanZ Apr 29 '23 at 17:29
  • If someone who is working on exactly what you're interested in stumbled into this question, they might be able to help you. I think that's very unlikely. But good luck with your project! – JonathanZ Apr 29 '23 at 17:30
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    Understood, fair enough. Thanks! – BBadman Apr 29 '23 at 21:13