130 5 Numerical Integration
I(x) =P∫+ 1
− 1dtet
t. (5.22)
The integrand diverges atx=t= 0. We rewrite it using Eq. (5.20) as
P∫+ 1
− 1dt
et
t=
∫+ 1
− 1et− 1
t, (5.23)
sinceex=e^0 = 1. With Eq. (5.21) we have then
∫+ 1
− 1et− 1
t≈
N
∑
i= 1ωi
eti− 1
ti. (5.24)
The exact results is 2. 11450175075 .....With just two mesh points we recall from the previous
subsection thatω 1 =ω 2 = 1 and that the mesh points are the zeros ofL 2 (x), namelyx 1 =− 1 /
√
3
andx 2 = 1 /
√
3. SettingN= 2 and inserting these values in the last equation givesI 2 (x= 0 ) =√
3
(
e^1 /√ 3
−e−^1 /√ 3 )
= 2. 1129772845.
With six mesh points we get even the exact result to the tenth digit
I 6 (x= 0 ) = 2 .11450175075!We can repeat the above subtraction trick for more complicated integrands. First we mod-
ify the integration limits to±∞and use the fact that
∫∞
−∞
dk
k−k 0 =∫ 0
−∞dk
k−k 0 +∫∞
0dk
k−k 0 =^0.A change of variableu=−kin the integral with limits from−∞to 0 gives
∫∞
−∞
dk
k−k 0=
∫ 0
∞−du
−u−k 0+
∫∞
0dk
k−k 0=
∫∞
0dk
−k−k 0+
∫∞
0dk
k−k 0= 0.
It means that the curve 1 /(k−k 0 )has equal and opposite areas on both sides of the singular
pointk 0. If we break the integral into one over positivekand one over negativek, a change of
variablek→−kallows us to rewrite the last equation as
∫∞
0
dk
k^2 −k^20= 0.
We can use this to express a principal values integral as
P∫∞
0f(k)dk
k^2 −k 02=
∫∞
0(f(k)−f(k 0 ))dk
k^2 −k^20, (5.25)
where the right-hand side is no longer singular atk=k 0 , it is proportional to the derivative
d f/dk, and can be evaluated numerically as any other integral.
Such a trick is often used when evaluating integral equations, as discussed in the next
section.