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Is there a simpler way to integrate this?

aruwin

Member
Jul 4, 2012
121
Hi, I was wondering if there's a simpler way to integrate this? I got the answer by expanding one by one but that's such a long process! I got the answer 104/5 which is correct,though.


∫∫∫D (x + y + z)^4 dxdydz
D = {(x,y,z) | -1≤ x ≤1, -1≤y≤1, -1≤z≤ }
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,725
Hi, I was wondering if there's a simpler way to integrate this? I got the answer by expanding one by one but that's such a long process! I got the answer 104/5 which is correct,though.


∫∫∫D (x + y + z)^4 dxdydz
D = {(x,y,z) | -1≤ x ≤1, -1≤y≤1, -1≤z≤1 (?) }
I don't see any quicker method than to integrate one variable at a time: $$\begin{aligned}\int_{-1}^1 \int_{-1}^1 \int_{-1}^1 (x+y+z)^4dx\,dy\,dz &= \int_{-1}^1 \int_{-1}^1 \Bigl[ \tfrac15(x+y+z)^5\Bigr]_{x=-1}^1dy\,dz \\ &= \int_{-1}^1 \int_{-1}^1 \tfrac15\bigl((1+y+z)^5 - (-1+y+z)^5\bigr)\,dy\,dz \\ &= \int_{-1}^1 \tfrac15 \Bigl[\tfrac16(1+y+z)^6 -\tfrac16(-1+y+z)^6\Bigr]_{y=-1}^1dz \\ &= \int_{-1}^1 \tfrac1{30}\bigl((2+z)^6 -2z^6 + (-2+z)^6\bigr)\,dz \\ &= \tfrac1{30}\Bigr[\tfrac17(2+z)^7 -\tfrac27z^7 + \tfrac17(-2+z)^7\Bigr]_{-1}^1 \\ &= \tfrac1{210}(3^7 - 1 - 4 - 1 + 3^7) = \tfrac{4368}{210}, \end{aligned}$$ which cancels down to $\tfrac{104}5$.
 

aruwin

Member
Jul 4, 2012
121
I don't see any quicker method than to integrate one variable at a time: $$\begin{aligned}\int_{-1}^1 \int_{-1}^1 \int_{-1}^1 (x+y+z)^4dx\,dy\,dz &= \int_{-1}^1 \int_{-1}^1 \Bigl[ \tfrac15(x+y+z)^5\Bigr]_{x=-1}^1dy\,dz \\ &= \int_{-1}^1 \int_{-1}^1 \tfrac15\bigl((1+y+z)^5 - (-1+y+z)^5\bigr)\,dy\,dz \\ &= \int_{-1}^1 \tfrac15 \Bigl[\tfrac16(1+y+z)^6 -\tfrac16(-1+y+z)^6\Bigr]_{y=-1}^1dz \\ &= \int_{-1}^1 \tfrac1{30}\bigl((2+z)^6 -2z^6 + (-2+z)^6\bigr)\,dz \\ &= \tfrac1{30}\Bigr[\tfrac17(2+z)^7 -\tfrac27z^7 + \tfrac17(-2+z)^7\Bigr]_{-1}^1 \\ &= \tfrac1{210}(3^7 - 1 - 4 - 1 + 3^7) = \tfrac{4368}{210}, \end{aligned}$$ which cancels down to $\tfrac{104}5$.
Yeap,that's how I got it :)
Ok, what about this one right here? This seems a little more complicated because there's fraction which is hard to decompose. Can you help me out with this one?

J = ∫∫D [(x-y)/(x+y)^3] dxdy

D = {(x,y) | 0 ≤x≤1, 0≤y≤1}
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,725
Ok, what about this one right here? This seems a little more complicated because there's fraction which is hard to decompose. Can you help me out with this one?

J = ∫∫D [(x-y)/(x+y)^3] dxdy

D = {(x,y) | 0 ≤x≤1, 0≤y≤1}
I would write the numerator as $x-y = (x+y)-2y$. Then the fraction becomes $(x+y)^{-2} -2y(x+y)^{-3}$, which you can easily integrate with respect to $x$.

[But before both of us get into trouble with the Moderators, you should in future stick to the forum rule that tells you to start a new thread if you want to ask a new question.]
 

aruwin

Member
Jul 4, 2012
121
I would write the numerator as $x-y = (x+y)-2y$. Then the fraction becomes $(x+y)^{-2} -2y(x+y)^{-3}$, which you can easily integrate with respect to $x$.

[But before both of us get into trouble with the Moderators, you should in future stick to the forum rule that tells you to start a new thread if you want to ask a new question.]
Sorry, I thought that because this is basically about multiple integrals, maybe I could just continue it here.
Ok, I have decomposed the fractions successfully and I got the answer -1/2 which is wrong. The answer given to me is undefined (infinity). Is there any explanation to this?
 
Last edited:

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,725
Ok, I have decomposed the fractions successfully and I got the answer -1/2 which is wrong. The answer given to me is undefined (infinity). Is there any explanation to this?
Ah, yes, I didn't notice it before, but this is a tricky question. If you follow my previous hint then you get $$\begin{aligned} \int_0^1 \int_0^1 \frac{x-y}{(x+y)^3}\,dx\,dy &= \int_0^1 \int_0^1 \bigl((x+y)^{-2} - 2y(x+y)^{-3}\bigr)\,dx\,dy \\ &= \int_0^1 \Bigl[-(x+y)^{-1} + y(x+y)^{-2}\Bigr]_{x=0}^1dy \\ &= \int_0^1 \bigl(-(y+1)^{-1} + \rlap{\color{red}/}y^{-1} + y(y+1)^{-2} - \rlap{\color{red}/}y^{-1}\bigr)\,dy. \end{aligned}$$

If you now cancel those $y^{-1}$s and integrate, then you get the answer -1/2 (which is wrong). However, some alarm bells should have started to sound when you did the cancellation, because the y-integral goes from 0 to 1, and the function $y^{-1}$ becomes infinite at the left end of that interval.

Another cause for concern is that if you use the same method to do the y-integral first, before the x-integral, then you would end up with the result +1/2 rather than -1/2. (You don't actually have to do the calculation to see that. Just observe that the function $(x-y)/(x+y)^3$ is anti-symmetric: it changes sign when you interchange $x$ and $y$.)

What all this shows is that this integral is an improper integral. The function becomes infinite at the origin, which is at one corner of the domain of integration. As with functions of a single variable, you have to be very cautious with such integrals, and ensure that the improper integral converges. In this case, it does not.
 

aruwin

Member
Jul 4, 2012
121
If you now cancel those $y^{-1}$s and integrate, then you get the answer -1/2 (which is wrong). However, some alarm bells should have started to sound when you did the cancellation, because the y-integral goes from 0 to 1, and the function $y^{-1}$ becomes infinite at the left end of that interval.
How do we determine that this integral is improper and how do you know that the function $y^{-1}$ becomes infinite at the left end of that interval?
 

Fantini

"Read Euler, read Euler." - Laplace
MHB Math Helper
Feb 29, 2012
342
I would risk saying it's a bit of practice. The function $1/y$ has a discontinuity at the origin, thus if zero is one of the limits of integration then we'd have an improper integral. Improper integrals tend to happen when we want to integrate the function where it isn't defined, leading to questions if it converges.

We know $y^{-1}$ becomes infinite at the left end of the interval because $$\lim_{y \to 0^+} \frac{1}{y} = + \infty.$$

Cheers.