---------- ---------- ---------- ---------- ---------- ---------- ---------- ---------- Integrals of F from A to B when A or B are infinite or F is undefined at A or B [ continuation from here ] # I have to calculate ∫[0,1] (x^3-x^5)/log(x) dx. f = function(x) (x^3-x^5)/log(x) # In 0 and 1 f is not defined. But if I make the graph: BF=3; HF=2.8 graphF( f,0,1, "brown") # I obtain the graph below on the left:# After all: f(0); f(1) # 0 NaN as x -> 1 f(x) -> -2, indeed: f(1-1e-10) # -2 # I could extend f like this: f(0)=1 f(1)=-2 # Therefore ∫[0,1] (x^3-x^5)/log(x) dx must be a negative number. # From the graph I understand it is near -1/2. We perform the calculation: integral(f, 0,1) # -0.4054651 # Is it the logarithm of something? exp(integral(f, 0,1)) # 0.6666667 fraction(exp(integral(f, 0,1))) # 2/3 log(2/3) # -0.4054651 # I can draw the graph of the "integral function" x -> ∫[0,x] f (the graph IF above): Plane(0,1, 0,-0.5) Gintegra(f, 0,1, "seagreen") POINT(1,log(2/3),"red") # # # I have to calculate ∫(-∞,∞) sin(x)/x dx
g = function(x) sin(x)/x # We know that g(x) -> 1 as x -> 0 g(10^-(1:5)) # g(0.1), ..., g(0.00001) # 0.9983342 0.9999833 0.9999998 1.0000000 1.0000000 BF=4.5; HF=3 RANGE0(g, -10*pi,10*pi) # "~ min F(min) Max F(Max) mFmF[1:4] = " # -4.49697447 -0.21723225 -0.03144737 0.99983519 g(x) sta tra -0.22 e 1 TICKx=pi; TICKy=0.1; Plane2(-10*pi,10*pi,-0.22,1) underX(c("-10pi","-6pi","-2pi","0","2pi","6pi","10pi"),2*pi*c(-5,-3,-1,0,1,3,5)) underY(c(-2,0,5,10)/10,c(-2,0,5,10)/10) graph2(g,-10*pi,10*pi,"brown") Z = function(x) 0 Diseq(Z,g, -10*pi,10*pi,"red") Diseq(g,Z, -10*pi,10*pi,"yellow") graph2(g,-10*pi,10*pi,"brown") # ∫(-∞,∞) sin(x)/x dx = 2 * ∫[0,∞) sin(x)/x dx (simmetry of the graphic) integral(g,0,1e2) # 1.562225 # integral(g,0,1e3) # Error: maximum number of subdivisions reached # It is advisable to proceed with successive sums (on [0,10] [10,20] ... [100,110], # then [0,10] ... [1000,1010], then [0,10] ... [10000,10010], then ...) s=0; b=10 n=1e1; s0=s; s=0; for(i in 0:n) s=s+integral(g,b*i,b*(i+1)); more( c(s,s-s0) ) # 1.57988048912506 1.57988048912506 n=1e2; s0=s; s=0; for(i in 0:n) s=s+integral(g,b*i,b*(i+1)); more( c(s,s-s0) ) # 1.57081912496797615 -0.00906136415708381 n=1e3; s0=s; s=0; for(i in 0:n) s=s+integral(g,b*i,b*(i+1)); more( c(s,s-s0) ) # 1.57073311569261e+00 -8.60092753700581e-05 n=1e4; s0=s; s=0; for(i in 0:n) s=s+integral(g,b*i,b*(i+1)); more( c(s,s-s0) ) # 1.57078774776842e+00 5.46320758134655e-05 n=1e5; s0=s; s=0; for(i in 0:n) s=s+integral(g,b*i,b*(i+1)); more( c(s,s-s0) ) # 1.57079730319125e+00 9.55542282965460e-06 # it is sufficient to convince oneself that is π/2 fraction(s/pi) # 1/2 # If you want you can go on (it takes a few seconds): n=1e6; s0=s; s=0; for(i in 0:n) s=s+integral(g,b*i,b*(i+1)); more( c(s,s-s0) ) # 1.57079622778987e+00 -1.07540137972428e-06 # # Therefore: ∫[0,∞) sin(x)/x dx = π/2, ∫(-∞,∞) sin(x)/x dx = π # # I have to calculate ∫(-∞,∞) 1/(pi*(1+x^2)) dx
g = function(x) 1/(pi*(1+x*x)); graphF(g,-5,5, "brown") integral(g, -Inf,Inf) # 1 g is a probability density function