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Specifically, for a double integral $$\int_a^b \int_{g_1(x)}^{g_2(x)} f(x,y) \, dy \, dx$$ how would you change the order of integration without having to sketch it out? I came across this while researching which talks about the use of the Heaviside function, however I am unsure how to apply this process to all double integrals.

Thanks!

Michael Hardy
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Richard Robinson
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2 Answers2

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I consider it similar to reversing the order of summation in a double sum.

I'm going to try to think this through logically.

In this case, $\int_a^b \int_{g_1(x)}^{g_2(x)} f(x,y) \, dy \, dx$, $g_1(x) \le y \le g_2(x)$. Therefore, assuming that $g_1$ and $g_2$ are strictly monotonic increasing and therefore have an inverse, and also satisfy $g_1(x) \le g_2(x)$, $x \le g_1^{(-1)}(y)$ and $x \ge g_2^{(-1)}(y)$ so the new inner integral will go from $g_2^{(-1)}(y)$ to $g_1^{(-1)}(y)$.

Since $a \le x \le b$, $y \le g_2(b)$ and $y \ge g_1(a)$ so the outer integral would go from $g_1(a)$ to $g_2(b)$.

So the integral would be $\int_{g_1(a)}^{g_2(b)} \int_{g_2^{(-1)}(y)}^{g_1^{(-1)}(y)} f(x, y) \,dx\,dy$.

Michael Hardy
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marty cohen
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  • So for example, how would you reverse $\int_0^3 \int_{x^2}^{12-x} f(x,y) , dy , dx$ ? – Richard Robinson Jun 20 '18 at 00:29
  • @RichardRobinson Your $g_1, g_2$ aren't monotonically increasing. So first step might be to try and rearrange this so that you can apply this tool. – Mason Jun 20 '18 at 00:45
  • @Mason ahh I didn't realize that before; I haven't learned how to rearrange it such that they are indeed monotonically increasing – how might I do so? (thank you very much, I'm only in my second week of multivariate calc) – Richard Robinson Jun 20 '18 at 01:20
  • I am hesitant to get to involved I only glanced through this post. What I mean is you can use: $\int_a^b=\int_a^c+\int_c^b$. – Mason Jun 20 '18 at 01:34
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\begin{align} & \int_a^b \int_{g_1(x)}^{g_2(x)} f(x,y) \, dy \, dx \\[10pt] = {} & \iint\limits_{\begin{array}{c} a\,\le\,x\,\le\,b \\g_1(x) \,\le\,y\,\le\,g_2(x) \end{array}} f(x,y) \, d(x,y) \\[10pt] = {} & \int_{\min g_1}^{\max g_2} \left(\,\, \int\limits_{x\,\in\,g_1^{-1}(-\infty,y] \,\cap\,g_2^{-1}[y,+\infty)} f(x,y) \, dx \right) \,dy \end{align} where $\displaystyle g_1^{-1}(-\infty,y] = \{ x : g_1(x) \le y \},$ and similarly for the other set. (This does not mean that $g_1$ or $g_2$ has an inverse function; it just means one can take the inverse image of a set under a function.)

Without more information about $g_1$ and $g_2,$ I don't think one can be more specific.

Michael Hardy
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  • Thanks! Assuming all the normal circumstances of $g_1$ and $g_2$, is there any more simple, easy to follow formula? (I'm only in first year of Uni, thanks) – Richard Robinson Jun 19 '18 at 23:58
  • @RichardRobinson : Here's a simple example resembling some that sometimes arise: $$ \int_0^1 \left( \int_x^1 f(x,y) , dy\right) , dx = \iint\limits_{0,\le, x,\le y,\le 1} f(x,y) , d(x,y) = \int_0^1 \left( \int_0^y f(x,y), dx \right) , dy.$$ – Michael Hardy Jun 20 '18 at 06:14