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Posted on 2008-03-17 16:04 By William Fish ( Maple Rating 2

161)

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I would like Maple to assist me with the following definite double integral:

int(int(x/(x^2+y^2+z^2)^(3/2), y = -b .. b), z = c .. a+c)

so far, I have failed. Can anybody help?

Here is my worksheet:

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Double trouble

Comment on 2008-03-17 19:21 from Robert Israel ( Maple Rating 4

1244)

> f:= x/(x^2+y^2+z^2)^(3/2);
   F1:= int(f,z=-b..b) assuming positive;

= 2/(x^2+y^2+b^2)^(1/2)*b/(x^2+y^2)*x

So far so good

> Q:=int(F1,y=c..a+c) assuming positive;

= -I*arctanh((b^2+c*x*I+x^2)/b/(b^2+c^2+x^2)^(1/2))-I*arctanh((-b^2+c*x*I-x^2)/b/(b^2+c^2+x^2)^(1/2))+arctanh((b^2+a*x*I+c*x*I+x^2)/b/(2*a*c+c^2+x^2+a^2+b^2)^(1/2))*I+arctanh((-b^2+a*x*I+c*x*I-x^2)/b/(2*a*c+c^2+x^2+a^2+b^2)^(1/2))*I

It's curious that Maple finds an integral that appears complex, even though (under the assumption that x, a, b, c are all positive) the integral should obviously be real.

> F2:= simplify(evalc(int(F1,y))) assuming positive;

= -arctan(x*y/(x^2+b^2+b*(x^2+y^2+b^2)^(1/2)))+1/2*arctan(-x*y,b*(x^2+y^2+b^2)^(1/2)-x^2-b^2)-1/2*arctan(x*y,b*(x^2+y^2+b^2)^(1/2)-x^2-b^2)

> Q := eval(F2,y=a+c) - eval(F2, y=c);

= -arctan(x*(a+c)/(x^2+b^2+b*(x^2+(a+c)^2+b^2)^(1/2)))+1/2*arctan(-x*(a+c),b*(x^2+(a+c)^2+b^2)^(1/2)-x^2-b^2)-1/2*arctan(x*(a+c),b*(x^2+(a+c)^2+b^2)^(1/2)-x^2-b^2)+arctan(x*c/(x^2+b^2+b*(b^2+c^2+x^2)^(1/2)))-1/2*arctan(-x*c,b*(b^2+c^2+x^2)^(1/2)-x^2-b^2)+1/2*arctan(x*c,b*(b^2+c^2+x^2)^(1/2)-x^2-b^2)

Minor point

Comment on 2008-03-18 04:03 from John Fredsted ( Maple Rating 4

566)

To be strictly true to the original integral, shouldn't the integral F1 be done with respect to y, and the integral Q with respect to z?

Note added: The result is, of course, the same because the integrand is symmetric in y and z. Realized after posting. Sometimes I post, and then I think. I apologize.

Assuming

Comment on 2008-03-19 10:12 from William Fish ( Maple Rating 2

161)

Robert,

As you know, I tried

assume(-b <= y, y <= b, c <= z, z <= a+c)

and it didn't help. Is this because it gives Maple no new information because the definite integrals already provided this information?

assuming positive

worked wonders. I tried this too

assuming real,positive

and it seemed to do good things also. Here is a worksheet:

View 4937_Page92JohnFredsted.mw on MapleNet or Download 4937_Page92JohnFredsted.mw
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Can you help me understand what assuming does here?

Here is a quote from page 92 and 93 of When Least Is Best by Paul J. Nahin:

"...set the derivative...equal to zero and solve...proves to be an astonishingly ugly business!"

Assuming

Comment on 2008-03-19 11:52 from Robert Israel ( Maple Rating 4

1244)

You made no assumption on x. Basically I think the problem that Maple needs to worry about is that the antiderivative may have a discontinuity (see e.g. the thread Maple Integration Error). Certainly there is likely to be trouble if the integrand itself went to infinity, which could happen if x was imaginary.

Nahin's Answer

Comment on 2008-03-17 21:29 from William Fish ( Maple Rating 2

161)

Robert Israel,

Here is Nahin's answer:

arcsin(((b^2-x^2)*(x^2+(a+c)^2)-2*x^2*b^2)/((b^2+x^2)*(x^2+(a+c)^2)))-arcsin(((b^2-x^2)*(x^2+c^2)-2*x^2*b^2)/((b^2+x^2)*(x^2+c^2)))

As far as I can tell, your answer is the same as Nahin's. Nahin says "a good table of integrals is the "method" I used!". Your method is clear. I can follow it. Someday I like to think that I will be able to do it. You have no idea how entertaining I find your mathematics. Thank you.

Here is my worksheet:

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Assume

Comment on 2008-03-20 16:48 from William Fish ( Maple Rating 2

161)

I tried assuming x is between 0 and 20. I tried making assumtions about a, b & c. None of this helped. Assuming positive worked wonders. What exactly was assumed to be positive? Assuming real seemed to have the same affect as assuming positive.

evalc(...) assuming positive;

worked like magic (Arthur C. Clarke's 3rd law).

I read Maple Int/int error? CASs are complex and imperfect.

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