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12

As you notniced, the result is not correct. This is not uncommon with definite integrals, as the system needs to detect any special behaviour inbetween the integration bounds, which is hard. For this reason it's a good idea to verify such integrals numerically. We can get the correct result like this (please evaluate in a fresh kernel without ...


5

No, it is not doing a Gram-Schmidt procedure. One way to note that this is distinct from Gram-Schmidt is that Gram-Schmidt produces an orthonormal basis, whereas the example outputs in the Documentation Center page for LatticeReduce are neither normalized nor orthogonal. Instead, LatticeReduce returns a basis $B$ comprised of linear combinations of integer ...


3

The approach that kguler suggest in your precedent question is completely suitable for the current one: FindSequenceFunction[Table[With[{ d1 = HypoexponentialDistribution[Flatten[Table[{λ, μ}, {i - 1}]]], d2 = HypoexponentialDistribution[{λ, μ}], d3 = ExponentialDistribution[μ], d = ExponentialDistribution[λ]}, ...


2

Just a workaround. In[4]:= Clear["Global`*"] In[5]:= f[x_, t_] := Abs[Re[Exp[I*x]/(1 - t*Exp[I*x])]] In[6]:= Timing[ resAn[t_] = Integrate[f[x, EulerGamma], {x, 0, 2*Pi}] /. EulerGamma -> t] Out[6]= {23.104, -((2*(ArcCos[t] + 2*ArcTan[((1 + t)*Tan[ArcCos[t]/2])/(-1 + t)]))/t)} In[7]:= resNum[t_] := NIntegrate[f[x, t], {x, 0, 2*Pi}] ...


1

At least with Mathematica version 10.0.1, using option Assumptions does provide a correct answer: Integrate[Cos[m*x]*Cos[n*x], {x, 0, 2 Pi}, Assumptions -> Element[{m, n}, Integers]] (* (Sin[2*(m - n)*Pi]/(m - n) + Sin[2*(m + n)*Pi]/(m + n))/2 I don't know why this form of the input leads to a different result than the form using Assuming.



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