Vladimir K.

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18 years, 47 days

MaplePrimes Activity


These are replies submitted by Vladimir K.

@Carl Love

Thank you! I've also got the correct answer in Mathematica using epsilon-shifted contour path. 

@Carl Love I'll try to figure out, what's wrong in M14. Many Thanks

@Alejandro Jakubi 

There are some other pecularities with this integral.

Now I compute the integral 

As it is not possible directly, I split this apparently real integral in two parts. Both parts are real for n>0. Look, what's happens!!!

> Za:=simplify(int(((c-x)/x)^n,x=a..c),symbolic) assuming a>0,c>a,n>-1;
> Zb:=simplify(int(((x-c)/x)^n,x=c..b),symbolic) assuming c>0,b>c,n>-1;
>
> Z2:=(A,B,C)->subs(a=A,c=C,Za)-subs(b=B,c=C,Zb);
> INT:=simplify(Z2(A,B,(A+B)/2)) assuming A>0,B>0,C>0,C>A,B>C,n>0;
> RI:=simplify(evalc(Re(INT))) assuming A>0,B>0,C>0,C>A,B>C,n>0;
> II:=simplify(evalc(Im(INT))) assuming A>0,B>0,C>0,C>A,B>C,n>0;
> ii:=evalf(simplify(subs(A=1.,B=2.,II),trig)):
> ri:=evalf(subs(A=1,B=2,RI)):
> plot(ri,n=0..1);
> plot(ii,n=0..1);

The integral is now complex! Exception n=0,n=1/2 and n=1.

Can anybody explain this behavior? What's going wrong? 

@Alejandro Jakubi 

There are some other pecularities with this integral.

Now I compute the integral 

As it is not possible directly, I split this apparently real integral in two parts. Both parts are real for n>0. Look, what's happens!!!

> Za:=simplify(int(((c-x)/x)^n,x=a..c),symbolic) assuming a>0,c>a,n>-1;
> Zb:=simplify(int(((x-c)/x)^n,x=c..b),symbolic) assuming c>0,b>c,n>-1;
>
> Z2:=(A,B,C)->subs(a=A,c=C,Za)-subs(b=B,c=C,Zb);
> INT:=simplify(Z2(A,B,(A+B)/2)) assuming A>0,B>0,C>0,C>A,B>C,n>0;
> RI:=simplify(evalc(Re(INT))) assuming A>0,B>0,C>0,C>A,B>C,n>0;
> II:=simplify(evalc(Im(INT))) assuming A>0,B>0,C>0,C>A,B>C,n>0;
> ii:=evalf(simplify(subs(A=1.,B=2.,II),trig)):
> ri:=evalf(subs(A=1,B=2,RI)):
> plot(ri,n=0..1);
> plot(ii,n=0..1);

The integral is now complex! Exception n=0,n=1/2 and n=1.

Can anybody explain this behavior? What's going wrong? 

Thanks! The expression as incomplete beta function is compact. It could be an option for MAPLE integration also. 

Thanks! The expression as incomplete beta function is compact. It could be an option for MAPLE integration also. 

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