用数值积分吧
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. }1 `4 h0 N4 Eclear all; x# {2 k" L8 b- [4 u
format long
' I$ s- P/ G" O+ ]! Ya=0;9 l4 [! v- E1 A2 z5 O! \' R3 e+ h
b=1;
% c. N4 ?) }% e/ z0 N& W: Sepsilon=10^(-6);& o6 o) a4 G: x' r% r. ]% m5 ~
syms x;
& T* A( k/ f( ^. dfun=log(x^2 + 1)/(x + 1);
: Y" \! ^+ e4 [4 y" nHfun=@ Remberg;
& M8 f. f0 X( T- S4 \2 N" FIvalue= feval(Hfun,fun,a,b,epsilon);
% q& | Y6 g! t6 b' }" E
( o' T% O+ M2 Y* E% ~, e: g+ ^2 f%Remberg.m x2 H- E8 y1 I- S
%a,b为积分限,epsilon为精度,s为返回积分值,fun为被积函数
. A B4 e5 v ~9 }- ~5 V%R(n,m)表示计算值,(n-1)为变步长指标,(m-1)为加速次数, e, j m( D# G% ~$ B
function s=Remberg(fun,a,b,epsilon)
# O6 N- v- s4 p+ j8 J8 j! asyms x ;
6 U. T. f( u% l; E$ @2 @fvalue=zeros(1,1000);
/ q& E% P; s2 C2 ^$ n; W2 w5 JR=zeros(100,100);! T% Q. Q' [( @* } m: Z( v/ Y$ C
fvaluea=double(subs(fun,x,a));
0 t- o1 Z1 [& _: P0 ]; vfvalueb=double(subs(fun,x,b));
. ?8 p: s0 S! U, ?: NR(1,1)=(b-a)/2*(fvaluea+fvalueb); %梯形公式9 k8 I- b; q8 N' N& P
km=1;2 {; u; K6 r3 o) J1 f- q9 _
for k1=1:100; %设置一个比较大的循环量6 K- i; ]# p5 E/ X/ |4 ]# D
h=(b-a)/(2^(k1));
$ F( F- j& `* I R(k1+1,1)=1/2*R(k1,1);
# b: b# a% e8 t3 c6 M for k2=1:2^(k1-1);2 r$ h2 C4 k: m. B& e2 J# A
fvalue(2*k2)=double(subs(fun,x,a+(2*k2-1)*h));1 J. `" [/ V. \/ {
R(k1+1,1)=h*fvalue(2*k2)+R(k1+1,1); %变步长值
4 I$ t( y2 U' W Z9 R9 [ end
, b, V* b5 o: b- A, V1 F, _ for k3=1:km; %加速计算- w! }9 m. o' s3 ?0 ^ q
R(k1+1,k3+1)=1/(4^(k3)-1)*(4^(k3)*R(k1+1,k3)-R(k1,k3));
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if abs(R(k1+1,km+1)-R(k1+1,km))<epsilon %控制精度9 l0 L, f/ B/ ], ^& c4 o
s=R(k1+1,km+1);
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else& y5 _5 U+ e& I
km=km+1;" p; P9 l( t! L7 J v3 R/ O
end9 Q# n( g( Z/ C$ _
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end* M& a e! e& I
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