ERROR, The integral is probably divergent
Question:
Haii guys, i have problem with code in R
i wanna running the integral code, but the code its doesnt work. is there anyone can help me?
retensi = 1136074
b = 1/1.230219e-07
sx = function(x)
{exp(-x/b)}
integrate(sx, retensi, Inf)
when i run this code, the ruslt is error Error in integrate(sx, retensi, Inf) : the integral is probably divergent i dont get it, where the mistake that i made can anyone help me? Please tell me detail of the correct code guys
the pov is how to solve this integral in R studio
Answers:
I think the problem is that the expression exp(-x/b) has such a small value at higher values of x that you are running into the limits of floating point arithmetic. For example:
retensi = 1136074
b = 1/1.230219e-07
sx <- function(x) exp(-x/b)
sx(1e9)
#> [1] 3.734803e-54
sx(1e10)
#> [1] 0
In fact, a quick manual binary search shows that numbers above 6,056,915,224 will return 0
sx(6056915224)
#> [1] 4.940656e-324
sx(6056915225)
#> [1] 0
This means that you will get the best approximation of you integral if the upper limit of your integral is set to 6,056,915,224:
integrate(sx, retensi, 6056915224)
#> 7068377 with absolute error < 19
We can confirm this is right by simply finding the indefinite integral of your expression, which is: -b e^(-x/b) + c and noting that this is 0 when x is infinite, so the manual calculation is:
0 - (-b * exp(-retensi/b))
#> [1] 7068377
And if we are still unsure, we can confirm this in Wolfram Alpha
1) Separate into two parts Sum the integral from retensi to a plus the integral from a to Inf for an a that works. We can find such as a by trying 10^i for i = 7, 8, … The code stops at the first a that works.
retensi <- 1136074
b <- 1/1.230219e-07
sx <- function(x) exp(-x/b)
for(a in 10^(7:12)) {
res <- integrate(sx, a, Inf, stop.on.error = FALSE)
if (res$message == "OK") break
}
a
## [1] 1e+09
res
## 5.21431e-51 with absolute error < 9.7e-51
so at a value of a = 10^9 the integral from a to Inf is essentially zero so we can just calculate the integral from retensi to a = 10^9
res <- integrate(sx, retensi, a); res
## 7068377 with absolute error < 0.0044
2) Symbolic integration We can calculate the integral symbolically
library(Ryacas0)
x <- Sym("x")
b <- Sym("b")
Integrate(exp(-x/b), x)
## yacas_expression(-(exp(-x/b) * b))
and using that symbolic result
b <- 1/1.230219e-07
retensi <- 1136074
(-exp(-Inf/b) * b) - (-exp(-retensi/b) * b)
## [1] 7068377
3) Change of variables Another method is to use a change of variables replacing x with a function that goes to infinity when the input to that function goes to some finite value. Trying x = tan(y) we have dx = dy/cos(y)^2:
D(quote(tan(y)), "y")
## 1/cos(y)^2
The answer below checks with the above two.
b <- 1/1.230219e-07
retensi <- 1136074
sx <- function(x) exp(-x/b)
sy <- function(y) sx(tan(y)) / cos(y)^2
integrate(sy, atan(retensi), pi/2)
## 7068377 with absolute error < 15
Haii guys, i have problem with code in R
i wanna running the integral code, but the code its doesnt work. is there anyone can help me?
retensi = 1136074
b = 1/1.230219e-07
sx = function(x)
{exp(-x/b)}
integrate(sx, retensi, Inf)
when i run this code, the ruslt is error Error in integrate(sx, retensi, Inf) : the integral is probably divergent i dont get it, where the mistake that i made can anyone help me? Please tell me detail of the correct code guys
the pov is how to solve this integral in R studio
I think the problem is that the expression exp(-x/b) has such a small value at higher values of x that you are running into the limits of floating point arithmetic. For example:
retensi = 1136074
b = 1/1.230219e-07
sx <- function(x) exp(-x/b)
sx(1e9)
#> [1] 3.734803e-54
sx(1e10)
#> [1] 0
In fact, a quick manual binary search shows that numbers above 6,056,915,224 will return 0
sx(6056915224)
#> [1] 4.940656e-324
sx(6056915225)
#> [1] 0
This means that you will get the best approximation of you integral if the upper limit of your integral is set to 6,056,915,224:
integrate(sx, retensi, 6056915224)
#> 7068377 with absolute error < 19
We can confirm this is right by simply finding the indefinite integral of your expression, which is: -b e^(-x/b) + c and noting that this is 0 when x is infinite, so the manual calculation is:
0 - (-b * exp(-retensi/b))
#> [1] 7068377
And if we are still unsure, we can confirm this in Wolfram Alpha
1) Separate into two parts Sum the integral from retensi to a plus the integral from a to Inf for an a that works. We can find such as a by trying 10^i for i = 7, 8, … The code stops at the first a that works.
retensi <- 1136074
b <- 1/1.230219e-07
sx <- function(x) exp(-x/b)
for(a in 10^(7:12)) {
res <- integrate(sx, a, Inf, stop.on.error = FALSE)
if (res$message == "OK") break
}
a
## [1] 1e+09
res
## 5.21431e-51 with absolute error < 9.7e-51
so at a value of a = 10^9 the integral from a to Inf is essentially zero so we can just calculate the integral from retensi to a = 10^9
res <- integrate(sx, retensi, a); res
## 7068377 with absolute error < 0.0044
2) Symbolic integration We can calculate the integral symbolically
library(Ryacas0)
x <- Sym("x")
b <- Sym("b")
Integrate(exp(-x/b), x)
## yacas_expression(-(exp(-x/b) * b))
and using that symbolic result
b <- 1/1.230219e-07
retensi <- 1136074
(-exp(-Inf/b) * b) - (-exp(-retensi/b) * b)
## [1] 7068377
3) Change of variables Another method is to use a change of variables replacing x with a function that goes to infinity when the input to that function goes to some finite value. Trying x = tan(y) we have dx = dy/cos(y)^2:
D(quote(tan(y)), "y")
## 1/cos(y)^2
The answer below checks with the above two.
b <- 1/1.230219e-07
retensi <- 1136074
sx <- function(x) exp(-x/b)
sy <- function(y) sx(tan(y)) / cos(y)^2
integrate(sy, atan(retensi), pi/2)
## 7068377 with absolute error < 15