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I came across this problem yesterday where i wanted to change variables in an integral like below. $$\iiint f\left(x_{1},x_{2},x_{3}\right) dx_{1}dx_{2}dx_{3}\tag{1}$$

so $y_1 = x_2 - x_1$ and $y_2 = x_3 - x_1$ there is no constraint available on $y_3$

Standard change of variable process would be to put $$x_1 = a_{11}y_1+a_{12}y_2 +a_{13}y_3$$ $$x_2 = a_{21}y_1+a_{22}y_2 +a_{23}y_3\tag{2}$$ $$x_3 = a_{31}y_1+a_{32}y_2 +a_{33}y_3$$

Let $\Delta = a_{11}a_{22}a_{33}+a_{12}a_{23}a_{31}+a_{13}a_{21}a_{32}-a_{13}a_{22}a_{31}-a_{12}a_{21}a_{33}-a_{11}a_{23}a_{32} > 0.\,$

Now $$\iiint f\left(x_{1},x_{2},x_{3}\right) dx_{1}dx_{2}dx_{3} =\tag{3}$$ $$=\iiint f(a_{11}y_1+a_{12}y_2 +a_{13}y_3,a_{21}y_1+a_{22}y_2 +a_{23}y_3,a_{31}y_1+a_{32}y_2 +a_{33}y_3)\Delta dy_1dy_2dy_3$$

I have two questions, Clearly the determinant has to be kept positive, for this process to work but how to pick a sensible value for $y_3$ which i can later integrate out ? Also if the original 3 variables had the range values $[0,\inf)$ how can i calculate the limits of the new variables ?

Is there a better way to do this ?

Any help would be much appreciated.

Edit: There is possibility that more context is needed here my original integral is the joint density of 3 independent exponential variables.

$$\int _{0}^{\infty }\int _{0}^{\infty }\int _{0}^{\infty } \lambda^3\mathrm e^{-\lambda(x_1+x_2+x_3)} dx_{1}dx_{2}dx_{3}\tag{4}$$

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@M. L. Thanks mate. It seems much neater now. –  Hardy Oct 5 '12 at 20:46
    
The motivation is somewhat odd since the triple integral in the Edit is simply the cube of $\int\limits_0^{+\infty}\lambda\mathrm e^{-\lambda x}\mathrm dx=1$. –  Did Oct 5 '12 at 20:48
    
@did This spanned off from the problem i was discussing with you the other day. math.stackexchange.com/questions/206953/… except i am not considering expectation here. –  Hardy Oct 5 '12 at 20:52
    
I know. And I do not understand the motivation. –  Did Oct 5 '12 at 20:55
    
@did i think i can spot one mistake in my post too I am not considering expectation, which is important, but essentially i spent a number of hours trying to transform a 3 variable integral to a 2 variable integral, i have no prior experience doing that, hence this was attempt to seek help. –  Hardy Oct 5 '12 at 20:58
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