Re: a definite integral

*To*: mathgroup at smc.vnet.net*Subject*: [mg83846] Re: a definite integral*From*: "David W.Cantrell" <DWCantrell at sigmaxi.net>*Date*: Sun, 2 Dec 2007 04:10:45 -0500 (EST)*References*: <fire23$rjr$1@smc.vnet.net>

"Arnold Knopfmacher" <Arnold.Knopfmacher at wits.ac.za> wrote: > Given the function > > f=(1 - z - u*z - Sqrt[-4*z + (1 + z - u*z)^2])/ > (2*u*z*Sqrt[-4*z + (1 + z - u*z)^2]) > > I wish to compute > g[z]=Integrate[f,{u,0,1}] > > (Here z and u are independent variables, that are real and positive and > the function inside of Sqrt[ ] is nonnegative.) If it is to be nonnegative for all u in the interval [0, 1], then we must have z <= 1/4. > Mathematica 5.2 can only evaluate this if I ask for the indefinite > integral Specifically In[8]:= indef[u_, z_] = Integrate[(1 - z - u*z - Sqrt[-4*z + (1 + z - u*z)^2])/ (2*u*z* Sqrt[-4*z + (1 + z - u*z)^2]) , u] Out[8]= (-Log[(-u)*z] + Log[1 + z - u*z + Sqrt[-4*z + (1 + z - u*z)^2]] + Log[-((2*(1 - 3*z - u*z + z*(1 + z - u*z)))/(u*(-1 + z)^2*z^2)) - (2*Sqrt[-4*z + (1 + z - u*z)^2])/(u*(-1 + z)*z^2)])/(2*z) AFAIK, that result is correct. > but this then gives errors when u=0 or u=1 is substituted. I do not get an error at u=1: In[9]:= indef[1, z] Out[9]= (Log[1 + Sqrt[1 - 4*z]] + Log[-((2*(1 - 3*z))/((-1 + z)^2*z^2)) - (2*Sqrt[1 - 4*z])/((-1 + z)*z^2)] - Log[-z])/(2*z) OTOH, as you indicated, there is a problem at u=0: In[10]:= indef[0, z] a warning... Out[10]= Indeterminate but that is not surprising. It is unfortunate, however, that version 5.2 cannot correctly calculate the limit In[11]:= Limit[indef[u, z], u -> 0] Out[11]= Infinity/z and that is a bug. (I wonder if version 6 does better.) Instead, the limit above should have given 1/(2 z) Log[8/(1 - z)^4] or something equivalent. I do not know how to get Mathematica to give that result. But subtracting it from Out[9], we get your desired g[z], which simplifies to Log[((1 - z)^2*(1 - Sqrt[1 - 4*z] - 2*z))/(2*z^2)]/(2*z) > I believe it is because > Mathematica is choosing the wrong branches of the square root or of the > Log function when it computes the indefinite integral. I doubt that is the explanation in this case. > The function f is integrable, as can be obseved by first finding say > Series[f,{z,0,10}] and then integrating term by term to find the > (correct) initial power series terms of g[z]. Yes, and the terms obtained in that way agree with In[16]:= Series[Log[((1 - z)^2*(1 - Sqrt[1 - 4*z] - 2*z))/(2*z^2)]/(2*z), {z, 0, 10}] Out[16] = z + 3*z^2 + (17*z^3)/2 + 25*z^4 + (461*z^5)/6 + 245*z^6 + (3217*z^7)/4 + 2701*z^8 + (92377*z^9)/10 + 32065*z^10 + O[z]^11 David W. Cantrell