Re: a definite integral

*To*: mathgroup at smc.vnet.net*Subject*: [mg83917] Re: a definite integral*From*: m.r at inbox.ru*Date*: Tue, 4 Dec 2007 04:28:36 -0500 (EST)*References*: <fire23$rjr$1@smc.vnet.net> <fittd2$7ph$1@smc.vnet.net>

On Dec 2, 3:20 am, "David W.Cantrell" <DWCantr... at sigmaxi.net> wrote: > "Arnold Knopfmacher" <Arnold.Knopfmac... 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 Actually Limit[indef[u, z], u -> 0] should not give 1/(2 z) Log[8/(1 - z)^4]. This is correct for 0 < z < 1 but for 1 < z we do get a logarithmic term. In version 6: In[1]:= tmp = (1/(2 z))(-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)]) In[2]:= Limit[tmp, u -> 0, Assumptions -> 0 < z < 1] Out[2]= Log[8/(-1 + z)^4]/(2 z) In[3]:= Assuming[1 < z, Series[tmp, {u, 0, 0}] // Simplify] Out[3]= SeriesData[u, 0, {(-2 Log[u] + Log[8/z^2])/(2 z)}, 0, 1, 1] Maxim Rytin m.r at inbox.ru