$$ \begin{array}{l}{\text { (3) Let } f \text { be an integrable function. Express the integral }} \\ {\qquad \int_{0}^{1} \int_{y}^{2 y} \int_{0}^{x+y} f(x, y, z)\, d z\, d x \,d y} \\ {\text { as a sum of iterated integrals in the order } d x d y d z \text { . }}\end{array} $$

For this to be in order $dx\, dy\, dz$, we need $f$ to be a continuous function and partial derivative need to exist but no condition is given as such so how to tackle such kind of function?

  • $\begingroup$ No? If $f\in L^1(\mathbb{R}^3)$ (or $L^1([0,1]\times [0,2]\times [0,3])$ as seems to be your case), then the interchangability of integration order is part of the Fubini theorem. $\endgroup$ – WoolierThanThou Oct 10 '19 at 6:52
  • $\begingroup$ @WoolierThanThou so what should be answer can you explain please? $\endgroup$ – maths student Oct 10 '19 at 7:22

The answer is $\int_0^{3}\int_0^{1} \int_{\max \{z-y,y\}}^{2y} f(x,y,z)\,dx\,dy\,dz$. Justification is by Fubini's Theoorem. [ Note that the inequalities $0<z<x+y, y<x<2y, 0<y<1$ are equivalent to the inequalities $\max \{z-y,y\} <x <2y, 0<y<1$ and $0<z<3$].

  • $\begingroup$ Isn't it the case that x range from 0 to 2 $\endgroup$ – maths student Oct 10 '19 at 7:31
  • $\begingroup$ $x$ can on;y go up to $2y$. Of course, $2y <2$ but the limit of integration w.r..t $x$ has to go up to $2y$ not $2$. @mathsstudent $\endgroup$ – Kavi Rama Murthy Oct 10 '19 at 7:35
  • $\begingroup$ Then for z why the limit of integration from 0 to 3? $\endgroup$ – maths student Oct 10 '19 at 7:37
  • $\begingroup$ @mathsstudent Once you have taken care of all the restrictions on $x$ and $y$ you have to integrate w.r.t. $z$ from its minimum value to its maximum value. [No other variable can appear in the final integral]. The maximum value of $z$ is $3$ because the maximum value of $y$ is $2$ and the maximum value of $x$ is $1$. $\endgroup$ – Kavi Rama Murthy Oct 10 '19 at 7:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.