Wave Propagation in Linear Media | re-examined

6.3 How to compute the rst integral
GaussKronrod DoubleExponential Trapezoidal
xm =40
R z510
0 f(x)dx 103.81 20.59 2968.07
P 509
k=0
R z6
0
xm =4
P 5
k=0
R zk+1
z k
f(x)dx 20.05 169.56 3119.32
f(x)dx 0.61 0.82 47.78
R zk+1
z k
f(x)dx 0.38 1.59 36.63
Table 6.1: Computing times in seconds for di erent quadrature algorithms.
Out[16]= f103.81 Second, 0.0491901g
Remark (Recursion depth) The function Timing returns the evaluation time as well
as the result. Note that because the integration interval spans many zeros of f(x), we
must increase the maximum number of recursive subdivisions Mathematica tries. The
according option is MaxRecursion. Otherwise the evaluation would terminate with an
error message and most likely a wrong result.
Another possibility istosubdivide the integration range at the zeros of the integrand | as
we would do for the interval to the right of the rst partition point | and to compute the
sum of these partial integrals. The computation is now much faster because the individual
integrals are smooth and we thus bene t from the simplicity of the Gauss-Kronrod formula.
In[17]:= Timing[
Sum[NIntegrate[f[x],fx,zero[i],zero[i+1]g,
Method->GaussKronrod],fi,0,509g]
]
Out[17]= f20.05 Second, 0.0491901g
We can carry out the same computations for the double exponential quadrature formula and
for a di erent value of xm = 4, which yields o = 5. The various results are summarised in
tab. 6.1 . We recognise rst of all that the trapezoidal rule indeed has a poor performance.
The double exponential algorithm seems to be a suitable choice, even if the transformation of
the integration variable takes some time, which slows down the computation in comparison
to the Gauss-Kronrod method if there are only few zeros in the integration interval. The
DE rule, on the other hand, needs no evaluation of any zeros, which can be a signi cant
advantage if the zeros cannot be found explicitly. In such a case, a subdivision strategy using
Gauss-Kronrod quadrature would be inferior as well. To demonstrate this, we assume that
we have no information about the zeros of the integrand in (6.8). Instead, we use equidistant
points to partition the interval and determine the total computing time depending on the
number of subintervals (again with xm = 40).
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