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I have very hard instructor for multivariate calculus. He ask if the next integral is well-defined.

$$ \iint\limits_D\,{\cos(z)\sin^3(z)\cos(y)\sin(y)\over (\cos^2(z)+\sin^2(z)\cos^2(y))(\cos^2(z)+\sin^2(z)\cos^2(y)-b)}\,dy\,dz $$

$D$ is the region $[0,\pi] \times [0,\pi]$ and $b \in (0,1]$. Is it possible to calculate integral with Mathematica or by hand for all $b$? I consider this an improper integral. For $b=1$, the free version of Wolfram Alpha says that the integral is $0$, but it is not strong enough to calculate the integral for general values of $b$.

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The integral is always zero.

To see this, preform the replacement: $$u = -\cos(y), \ v = -\cos(z) \ \rightarrow du = \sin(y)dy, \ dv = \sin(z)dz$$ The integral is then: $$I = \iint\limits_D\ {uv(1-v^2) \over (v^2+(1-v^2)u^2) (v^2+(1-v^2)u^2 - b)}\,du\,dv $$ Where $D = [-1,1]\times[-1,1]$. Note that the integral is odd in both $u$ and $v$, leading to $I = 0\ $ for all $b$.

Now, how do we know this converges? the left hand denominator is never zero, but the right side will always be zero for some pair of values $u_0,v_0$. You can easily see that the roots lie on a closed symmetrical curve, i.e. for any value $b$, if $u_0,v_0$ is a root so are $\pm u_0,\pm v_0$. This means you can divide the integral into a region outside this "ring of roots" and one inside. Due to symmetry and the oddness, each of these two integrals converges to zero, and so does the sum.

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So doesn't this merely show: the integral is zero if it converges? And your new integral maybe is $du dv$? – GEdgar Sep 7 '12 at 16:44
@GEdgar - fixed, hopefully. – nbubis Sep 7 '12 at 17:38
The issue of convergence does not go away simply by excluding the curve at which the function is undefined. Within either region the function is unbounded. In particular, restricting further to the quarter $u,v>0$ yields $-\infty$ for the region inside the "ring of roots" and $+\infty$ outside. So unless this hard multivariable instructor devised an unusual notion of convergence of the P.V.-type, the integral is not defined. – user31373 Sep 9 '12 at 3:14

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