Re: Re: A Problem with Simplify

*To*: mathgroup at smc.vnet.net*Subject*: [mg87782] Re: [mg87760] Re: A Problem with Simplify*From*: Daniel Lichtblau <danl at wolfram.com>*Date*: Wed, 16 Apr 2008 22:33:03 -0400 (EDT)*References*: <ftkb7f$a9m$1@smc.vnet.net> <200804140943.FAA08090@smc.vnet.net> <200804161114.HAA00059@smc.vnet.net>

Alexey Popkov wrote: > [...] >>Capitalization notwithstanding, it need not do any such thing. Integrate >>should give a generic result that behaves sensibly in the limit as a->1 > >>from whatever path is specified to Limit. This is similar to the > >>situation of > > >>In[18]:= InputForm[solns = x /. Solve[a*x^2 + x == 1, x]] >>Out[18]//InputForm= {(-1 - Sqrt[1 + 4*a])/(2*a), >> (-1 + Sqrt[1 + 4*a])/(2*a)} >> >>In[20]:= Limit[solns, a->0] >>Out[20]= {-Infinity, 1} >> >>Note that a blind substitution of a->0 will not give "nice" results. > > > Well. But how I can get the partial answer for the case a=1 with > Limit? > I have tried the foolowing code: > In[1]:= Limit[Integrate[Cos[a*z]/Sin[z],z],a->1] > Out[1]= \[Infinity] > In[2]:= Limit[Integrate[Exp[a*z]*Sinh[z],z],a->1] > Out[2]= -\[Infinity] > In[3]:= Limit[Integrate[Exp[z]*Sinh[a z],z],a->1] > Out[3]= E^z DirectedInfinity[E^(-I*Im[z])] > In[4]:= Limit[Integrate[Exp[(a-1)*x],x],a->1] > Out[4]= \[Infinity] > All results are incorrect! > As I understand, Integrate does not give a generic result that behaves > sensibly in the limit as a->1! Sorry, I should have stated the situation more clearly. Yes, the antiderivatives you see have singularities at the parameter value in question (in all cases, when a is 1). Hence the results you show above. What I should have said is that one can get these singularities to cancel, if using the parametrized antiderivatives to obtain definite integrals a la Newton-Leibniz. Here is a standard simple example (modified slightly so as to have a->1 as the bad value). In[22]:= InputForm[i1 = Integrate[x^(-a),x]] Out[22]//InputForm= x^(1 - a)/(1 - a) Certainly it blows up as a->1. In[23]:= Limit[i1, a->1] Out[23]= -Infinity In[24]:= Limit[i1, a->1, Direction->1] Out[24]= Infinity But if we take a pair of values for x, say 2 and 3, to evaluate the definite integral, we in fact get the expected/desired logarithm result in the limit as a->1 (and independent of direction). In[27]:= InputForm[Limit[(i1/.x->3)-(i1/.x->2), a->1]] Out[27]//InputForm= Log[3/2] In[28]:= InputForm[Limit[(i1/.x->3)-(i1/.x->2), a->1, Direction->1]] Out[28]//InputForm= Log[3/2] To see that this really is reproducing the Fundamental Theorem of Calculus result (or at least doing a heckuva job to fool me), I'll show this instead as a definite integral from x0 to x. In[29]:= InputForm[Limit[i1 - (i1/.x->x0), a->1]] Out[29]//InputForm= Log[x] - Log[x0] I believe most of your examples will also behave fine when processed in this way. The Cos[a*z]/Sin[z] integrand might be an exception, but that indicates a limitation of Limit rather than a bug in Integrate. Depending on whether you give symbolic (z,z0) or exact numeric (3,2) input to Limit it either returns a complicated, but seemingly correct result, or else unevaluated. Daniel Lichtblau Wolfram Research

**References**:**Re: A Problem with Simplify***From:*Alexey Popkov <popkov@gmail.com>

**Re: A Problem with Simplify***From:*Alexey Popkov <popkov@gmail.com>