Wave Propagation in Linear Media | re-examined

is most easily obtained from R 1
0
6Towards a quadrature routine
sin y
y dy with the substitution y = p x 2 , a 2 . For the purpose
of demonstration, we assume that we cannot calculate the zeros of the integrand directly.
Instead, we know that for x a, p x 2 , a 2 behaves like x, which is exactly the polynomial
part (6.11) of its expansion (6.10). So we compute the partition points as if the argument
of the sin-function was x, and we plot the sequence of partial sums for the parameter value
a =50 . Owing to the singularity of the integrand at x = a, we determine the value of the
rst subinterval analytically and include it explicitly in the sequence of the partial sums.
In[19]:= a = 50 Pi;
firstval = Integrate[Sin[Sqrt[x^2 - a^2]]/(x^2 - a^2) x,
fx,50 Pi,51 Pig]
Out[19]= SinIntegral[Sqrt[101] Pi]
In[20]:= sequ = Table[NIntegrate[Sin[Sqrt[x^2 - a^2]] x/(x^2 - a^2),
fx,i Pi,(i+1) Pig],
fi,51,300g];
partial = FoldList[Plus,N[firstval],sequ];
1.576
1.574
1.572
1.568
1.566
z1 = ListPlot[partial,PlotStyle->PointSize[0.006]];
50 100 150 200 250
In[21]:= SequenceLimit[partial] - N[Pi/2]
Out[21]= -14
1.59872 10
We emphasise that in this case the computation of the extrema instead of the zeros yields no
better result since the partial integrals in either case cover both negative and positive portions
146