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Question: Calculating Integrals via different ways

Question:Calculating Integrals via different ways

digerdiga 385 Maple 17
integration
January 17 2014
1

Mapleprimes_Integral.mw

I have a question regarding following problem:

assume(a > 0, a < 1, t > 0, Z0 > 0, z > 0)

f1 := proc (z) options operator, arrow; 1/z end proc

proc (z) options operator, arrow; 1/z end proc

 (1)

I_1 := int(f1(z)*ln((a*z+1)/(1+z/a)), z = 0 .. Z0); 1; MultiSeries:-asympt(%, Z0, 3)

2*ln(Z0)*ln(a)+(a^2-1)/(a*Z0)-(1/4)*(a^4-1)/(a^2*Z0^2)+O(1/Z0^3)

 (2)

Using the representation which should hold for all a>0 and z>0

int(z*exp(t)*(a^2-1)/((exp(t)+a*z)*(exp(t)*a+z)), t = 0 .. infinity); 1; combine(%)

ln((a*z+1)*a/(z+a))

 (3)

I'm calculating the result the other way around

int(z*exp(t)*(a^2-1)*f1(z)/((exp(t)+a*z)*(exp(t)*a+z)), z = 0 .. Z0); 1; I_2 := int(%, t = 0 .. infinity); 1; MultiSeries:-asympt(%, Z0, 3)

2*ln(Z0)*ln(a)+(a^2-1)/(a*Z0)-(1/4)*(a^4-1)/(a^2*Z0^2)+O(1/Z0^3)

 (4)

plot(eval([I_1, I_2], a = 1/2), Z0 = 0 .. 10)

  

So the results are the same.

But if I calculate this with another function

f2 := proc (z) options operator, arrow; 1/(z*(z+a)) end proc

proc (z) options operator, arrow; 1/(z*(z+a)) end proc

 (5)

I_3 := int(f2(z)*ln((a*z+1)/(1+z/a)), z = 0 .. Z0); 1; MultiSeries:-asympt(%, Z0, 3)

-dilog(a^2)/a-2*ln(a)/Z0+(1/2)*(2*ln(a)*a^2+a^2-1)/(a*Z0^2)+O(1/Z0^3)

 (6)

int(z*exp(t)*(a^2-1)*f2(z)/((exp(t)+z*a)*(exp(t)*a+z)), z = 0 .. Z0); 1; I_4 := IntegrationTools:-Change(int(%, t = 0 .. infinity), t = ln(z)); 1; MultiSeries:-asympt(%, Z0, 3); 1; simplify(convert(convert(MultiSeries:-series(I_4, Z0, 1), polynom), polynom))

ln(a)*(ln(Z0)+ln(a))/a

 (7)

I get another result :-/ The Integral doesn't even vanish in the limit Z0 -> 0

Though if I take the limit prior:

int(z*exp(t)*(a^2-1)*f2(z)/((exp(t)+z*a)*(exp(t)*a+z)), z = 0 .. infinity);

-dilog(a^2)/a

 (8)

the result is correct. What is the problem here?



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