用數值積分吧
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clear all+ R- Y" e2 P" Q3 w! y. q
format long
d0 s* ]( W" i6 R! Ta=0;1 y( ^% E% y3 r! h& L8 s! q9 f& L2 M. N
b=1;* P) U- Q1 U( |0 _
epsilon=10^(-6);
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fun=log(x^2 + 1)/(x + 1);
6 k3 w# {) A# W( m9 k9 jHfun=@ Remberg;
% t2 {' a1 L7 N5 j+ N! x. \% gIvalue= feval(Hfun,fun,a,b,epsilon); ~: M$ b: P/ y4 F
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%Remberg.m
& y4 v! J4 G' D) b) N. c4 \1 O6 d0 M%a,b為積分限,epsilon為精度,s為返回積分值,fun為被積函數6 p8 x& l1 Q, _& [
%R(n,m)表示計算值,(n-1)為變步長指標,(m-1)為加速次數
. w: F$ U" ^ Afunction s=Remberg(fun,a,b,epsilon) x1 w* _: k+ U7 f. A! G+ [8 b
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fvalue=zeros(1,1000);
% H) f7 S$ I1 L7 l. w# x; x% yR=zeros(100,100);' v) V& |* I. o0 \
fvaluea=double(subs(fun,x,a));
& y: W$ z4 P. c4 e& J- Z( A6 lfvalueb=double(subs(fun,x,b));( v6 C4 C7 g# w9 V. Y
R(1,1)=(b-a)/2*(fvaluea+fvalueb); %梯形公式
1 M0 H3 A+ v$ P' r3 nkm=1;
b" V. S1 }, Efor k1=1:100; %設置一個比較大的循環量2 T. |$ {, j- h. X6 d3 m
h=(b-a)/(2^(k1));( f$ g5 c& o0 v
R(k1+1,1)=1/2*R(k1,1);
1 F( \& Q8 E) \' a0 B: h for k2=1:2^(k1-1);) n r" A3 f8 S2 [3 h
fvalue(2*k2)=double(subs(fun,x,a+(2*k2-1)*h));
+ M0 K! k$ {0 C' Q R(k1+1,1)=h*fvalue(2*k2)+R(k1+1,1); %變步長值
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for k3=1:km; %加速計算7 i& ~. {4 g1 E# 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' D% w! X" y# X; Z! y& Y
s=R(k1+1,km+1);
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8 j" }. s: e8 k1 }' A7 q b km=km+1;
' l# O, ~' w# R end
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) i/ ^4 u9 K) A5 yend
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