Markiyan Hirnyk

Markiyan Hirnyk

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11 years, 61 days

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These are questions asked by Markiyan Hirnyk

How to find the integral of (x+y)/(x+y+z) over the part of the unit ball  centered at the origin which lies in the positive octant { x>=0 , y>=0, z>=0 } ? Numeric calculations suggest Pi/9.

How to find the double integral of sin(x^2)*cos(y^2) over the disk of radius R which is centered at the origin? 

Here is my try 

restart; evalf(VectorCalculus:-int(sin(x^2)*cos(y^2), [x, y] = Circle(`<,>`(0, 0), 1), inert), 15);
                       0.722091449378409
identify(%);
                       0.722091449378409

 

It is suggested  

hypergeom([1/3, 2/3], [3/2], (27/4)*z^2*(1-z)) = 1/z

if z > 1. Here is my try to prove that with Maple:


 

a := `assuming`([convert(hypergeom([1/3, 2/3], [3/2], (27/4)*z^2*(1-z)), elementary)], [z > 1])

-(1/((1/2)*(27*z^3-27*z^2+4)^(1/2)+(3/2)*z*(3*z-3)^(1/2))^(1/3)-1/((1/2)*(27*z^3-27*z^2+4)^(1/2)-(3/2)*z*(3*z-3)^(1/2))^(1/3))/(z*(3*z-3)^(1/2))

(1)

b := `assuming`([simplify(a, symbolic)], [z >= 1])

2*(-(12*(3*z+1)^(1/2)*z-12*z*(3*z-3)^(1/2)-8*(3*z+1)^(1/2))^(1/3)+(12*(3*z+1)^(1/2)*z+12*z*(3*z-3)^(1/2)-8*(3*z+1)^(1/2))^(1/3))/((3*z-3)^(1/2)*(12*(3*z+1)^(1/2)*z+12*z*(3*z-3)^(1/2)-8*(3*z+1)^(1/2))^(1/3)*(12*(3*z+1)^(1/2)*z-12*z*(3*z-3)^(1/2)-8*(3*z+1)^(1/2))^(1/3)*z)

(2)

plot(1/b, z = 1 .. 10)

 

simplify(diff(1/b, z), symbolic)

-48*(((3*z-2)*(3*z+1)^(1/2)+z*(3*z-3)^(1/2))*((12*z-8)*(3*z+1)^(1/2)-12*z*(3*z-3)^(1/2))^(1/3)+((12*z-8)*(3*z+1)^(1/2)+12*z*(3*z-3)^(1/2))^(1/3)*((-3*z+2)*(3*z+1)^(1/2)+z*(3*z-3)^(1/2)))/((3*z+1)^(1/2)*(3*z-3)^(1/2)*((12*z-8)*(3*z+1)^(1/2)+12*z*(3*z-3)^(1/2))^(2/3)*((12*z-8)*(3*z+1)^(1/2)-12*z*(3*z-3)^(1/2))^(2/3)*(((12*z-8)*(3*z+1)^(1/2)-12*z*(3*z-3)^(1/2))^(1/3)-((12*z-8)*(3*z+1)^(1/2)+12*z*(3*z-3)^(1/2))^(1/3))^2)

(3)

``


 

Download simplification.mw

Can somebody execute this code on a powerful comp and report the result in MaplePrimes? That would be kind of her/him.


 

RealDomain:-solve(a^2+b^2+c^2+a*b+a*c+b*c-  (a+b-c)*sqrt(2*a*b+a*c+b*c)-(a+c-b)*sqrt(a*b+2*a*c+b*c)-(b+c-a)*sqrt(a*b+a*c+2*b*c));

Error, (in assuming) when calling '`resultant/modular2`'. Received: 'Maple was unable to allocate enough memory to complete this computation.  Please see ?alloc'

 

``


 

Download want_for_execution.mw

Let us consider the improper integral

int((abs(sin(2*x))-abs(sin(x)))/x, x = 0 .. infinity);

Si(Pi)-Si((1/2)*Pi)+sum(-(-1)^_k*Si(Pi*_k)+signum(sin((1/2)*Pi*_k))*Si((1/2)*Pi*_k)+Si(Pi*_k+Pi)*(-1)^_k-signum(cos((1/2)*Pi*_k))*Si((1/2)*Pi*_k+(1/2)*Pi), _k = 1 .. infinity)
                    

Mathematica 11 produces a similar expression and a warning

Integrate::isub: Warning: infinite subdivision of the integration domain has been used in computation of the definite integral \!\(\*SubsuperscriptBox[\(\[Integral]\), \(0\), \(\[Infinity]\)]\(\*FractionBox[\(\(-Abs[Sin[x]]\) + Abs[Sin[2\ x]]\), \(x\)] \[DifferentialD]x\)\). If the integral is not absolutely convergent, the result may be incorrect.

Up to Pedro Tamaroff http://math.stackexchange.com/questions/61828/proof-of-frullanis-theorem , the answer is 2/Pi*ln(2) because of 

J := int(abs(sin(2*x))-abs(sin(x)), x = 0 .. T) assuming T>2;
-1/2-signum(sin(T))*signum(cos(T))*cos(T)^2+(1/2)*signum(sin(T))*signum(cos(T))+cos(T)*signum(sin(T))+floor(2*T/Pi)

B := limit(J/T, T = infinity);
                               2 /Pi

K := x*(int((abs(sin(2*t))-abs(sin(t)))/t^2, t = x .. 1)) assuming x>0,x<1;

     2*sin(x)*cos(x)-2*Ci(2*x)*x+Ci(x)*x+sin(1)*x-sin(2)*x+2*Ci(2)*x-Ci(1)*x-sin(x)

                         
A := limit(K, x = 0, right);
                               0

Its numeric calculation results 

evalf(Int((abs(sin(2*x))-abs(sin(x)))/x, x = 0 .. infinity));
                        Float(undefined)

which seems not to be true.

The question is: how to obtain the reliable results for it with Maple, both symbolic and numeric? 

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