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I'm wondering if this is possible for the general case. In other words, I'd like to take $$\int_a^b{g(x)dx} + \int_c^d{h(x)dx} = \int_e^f{j(x)dx}$$ and determine $e$, $f$, and $j(x)$ from the other (known) formulas and integrals. I'm wondering what restrictions, limitations, and problems arise.

If this is not possible in the general case, I'm wondering what specific cases this would be valid for, and also how it could be done. It's a curiosity of mine for now, but I can think of some possible problems and applications to apply it to.

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Where the last upper limit $f$ is the function $f$ in the first integral? – GEdgar Jul 5 '12 at 21:10
@GEdgar: I edited the integrals so they aren't confused with variables. – Matt Groff Jul 5 '12 at 21:14
Do a substitution in the second integral in order to get an integral from $a$ to $b$. – Davide Giraudo Jul 5 '12 at 21:18
@DavideGiraudo: I just came up with that idea... I'm going to try to carry it out for everyone... I hope that will save you all some work. I figured that the same thing should be able to work for two nested integrals with two different variables with slightly more tinkering. – Matt Groff Jul 5 '12 at 21:34
Well in principle yes: let $j(x)=1$. Then let $e=0$, and let $f=\int_a^bg(x)\,dx + \int_c^d h(x)\,dx$. You get a nice equality, though not at all practical as a way to evaluate integrals. – Arturo Magidin Jul 6 '12 at 0:41
up vote 3 down vote accepted

Certainly. In fact $e$ and $f$ can be anything you want, as long as they are not equal. An affine transformation is one way to do it. Namely if $$j(x)=\frac{b-a}{f-e}g\left(\frac{b-a}{f-e}(x-e)+a\right) +\frac{d-c}{f-e}h\left(\frac{d-c}{f-e}(x-e)+c\right),$$ then $$\int_a^bg(u)du+\int_c^dh(v)dv=\int_e^fj(x)dx.$$

This transformation follows from the change of variables $$u=\frac{b-a}{f-e}(x-e)+a,\qquad v=\frac{d-c}{f-e}(x-e)+c.$$

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Could you show me where you got this from? I'd really enjoy seeing the steps to this, but I don't want to cause you unnecessary work. – Matt Groff Jul 6 '12 at 17:28
@MattGroff: try working out the change of variables I indicated in an edit to my answer. – JL344 Jul 7 '12 at 2:33
Thanks! I get $x=\frac{b-a}{f-e}(u-a)+e$ and $dx=\frac{b-a}{f-e}du$. Now, if I switch the $(x-e)$ in the definition of $u$ to $(x-f)$, I get $b-f$ for the upper limit of integration, instead of $f$. Shouldn't we be getting $f$ for the upper limit of integration? – Matt Groff Jul 7 '12 at 4:07
I think you might have those fractions upside-down. Shouldn't you get $x=\frac{f-e}{b-a}(u-a)+e$ and $dx=\frac{f-e}{b-a}du$? But you don't really need to do that anyways. Just express $u$, $du$, $v$, and $dv$ in terms of $x$ and $dx$, and plug those substitutions into the first two integrals. – JL344 Jul 7 '12 at 4:20
You're right, I mixed up the fraction. $x$ and $dx$ checked out, but I'm having trouble with the limits of integration. I'm trying to check my work. I'm also working on how to derive all of this, too. – Matt Groff Jul 7 '12 at 4:53

Assume $a< b$ and $c< d$. Let
$$ e \le \min\{a,c\}, \ f \ge \max\{b,d\}, $$ and set $j=\tilde{g}+\tilde{h}$ where $$ \tilde{g}:=g.1_{[a,b]},\tilde{h}:=h.1_{[c,d]}: [e,f] \to \mathbb{R}. $$ Then $$ \int_a^b g+\int_c^d h=\int_e^f(\tilde{g}+\tilde{h})=\int_e^f j. $$

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Could you please explain the $g.1_{[e,f]}$ and $h.1_{[e,f]}$ terms more? I'm not quite sure I follow what you're trying to do. I ran into problems trying to carry out the general method in detail, and I'm hoping that you can shed some light on the subject. Let me know if (and how) I can help... – Matt Groff Jul 5 '12 at 23:21
@MattGroff If $X$ is a set, the function $1_X$ is the indicator function, and is designed to equal $1$ if an argument is a member of $X$, and $0$ otherwise. Mercy is using a period for multiplication, though I think it would be better as $\cdot$ (which is \cdot in $\LaTeX$). – anon Jul 6 '12 at 1:27

Here's a method that should allow one a large degree of freedom as well as allowing Rieman Integration (instead of Lesbegue Integration or some other method):

Let $\tilde{g}$ be such that: $$\int_a^b{g(x)dx} = \int_e^f{\tilde{g}(x)dx}$$ ...and $\tilde{h}$ follows similarly. Then they can be both added inside a single integral.

The first method that comes to mind is to let $\tilde{g}(x) = \dot{g}\cdot g(x)dx$, where $\dot{g}$, a constant, is the ratio of the old and new integrations. A similar method that comes to mind is to actually let $\dot{g}$ be a function.

Another method that I'm exploring, and is somwhat questionable, is to attempt to use $e$ and $f$ as functions, possibly even of $x$, although this may be undefined or just plain wrong.

I'll add ideas to this as I hopefully come up with better methods.

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$\tilde{g}(x)$ is a "real", and $\dot{g}\cdot g(x)dx$ isn't, so what do you mean by $\tilde{g}(x)=\dot{g}\cdot g(x)dx$? – Mercy King Jul 6 '12 at 0:26
@Mercy: Sorry for the slow reply. I was trying to denote a multiplier by $\dot{g}$. I realize that that $\dot{g}$ sometimes means other things, but it is supposed to evaluate to a real number in this case. I believe that JL$_{344}$ did something like what I was thinking, especially with the $\frac{b-a}{f-e}$ in front of $g$. – Matt Groff Jul 6 '12 at 4:49

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