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

How is this different from your [previous question](https://stackoverflow.com/questions/74925507/integrate-the-integral-is-probably-divergent), posted 8 hours ago, and to which you have already received an excellent answer? — Limey, Dec 27 '22 at 11:55

score 3 · Answer 1 · answered Dec 27 '22 at 11:44

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

how to do a quick manual binary search to get number 6,056,915,224 — EDITHA, Dec 28 '22 at 02:10

G. Grothendieck · Accepted Answer · 2022-12-28T13:13:20.740

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

thankyouu fo the answer. I try with number 1 and 2, but the answer number 2 it doesn't work for me, because when I running the code, it take a long time, and I dot get the result. but answer number 1 its work for me — EDITHA, Dec 28 '22 at 02:09
(2) works for me and gives the answer shown quickly. Maybe you are trying it on a different function? Yacas Integrate is somewhat limited in what it can integrate. Suggest you also try (3). — G. Grothendieck, Dec 28 '22 at 13:20

ERROR, The integral is probably divergent

2 Answers2