Question: An integral of ln(t)^2/((t^2+1)*(t-z))

f := (z, t) -> ln(t)^2/((t^2+1)*(t-z));

int(f(z, t), t = 0 .. infinity) assuming Im(z) > 0;
       int(ln(t)^2/((t^2+1)*(t-z)), t = 0 .. infinity)

int(f(a + I*b, t), t = 0 .. infinity) assuming a::real, b > 0; # 0*infinity
       -sqrt(a^2+b^2-2*b+1)*signum(I*arctan(b, a)-I*arctan(-b, -a)-I*Pi)*
       infinity/((I*b-I+a)*(I*b+I+a))

int(f(z, t), t = 0 .. infinity) assuming Re(z) > 0, Im(z) > 0;
       int(ln(t)^2/((t^2+1)*(t-z)), t = 0 .. infinity)

int(f(a + I*b, t), t = 0 .. infinity) assuming a > 0, b > 0;
      -((3*I)*Pi^3*b+3*Pi^3*a-(16*I)*Pi^2*arctan(b/a)-(6*I)*Pi*ln(a^2+b^2)^2+(24*I)*
      Pi*arctan(b/a)^2+(6*I)*ln(a^2+b^2)^2*arctan(b/a)-(8*I)*arctan(b/a)^3-8*Pi^2*
      ln(a^2+b^2)+24*Pi*ln(a^2+b^2)*arctan(b/a)+ln(a^2+b^2)^3-12*ln(a^2+b^2)*
      arctan(b/a)^2)/(24*(-b^2+(2*I)*a*b+a^2+1))

So it looks like the first three can be made to work as well (and the result in terms of z will be much neater).

 

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