[R] Simplifying an expression with an integral

[David Winsemius]

On Dec 24, 2013, at 9:39 AM, Aurélien Philippot wrote:

> Dear R experts,
> I am computing the following integral.
> 
> [image: \int_{1,100} \frac{1}{x+Max(x-50,0)} g(x)dx], where g is the
> density of the standard normal, and [1,100] is the domain.
> 
> 1) I use the following code which works fine:
> integrand1<- function(x){1/x*dnorm(x)}
> integrand2<- function(x){1/(2*x-50)*dnorm(x)}
> 
> res1<- integrate(integrand1, lower=1, upper=50)$value+
> integrate(integrand2, lower=50, upper=100)$value
> res1
> 0.1116
> 
> In other words, I split the max function depending on the value of x in the
> domain.
> 
> 2) Alternatively, I can also compute it by vectorizing the max function
> integrand<- function (x) ifelse(x<50, 1/x*dnorm(x) , 1/(2*x-50)*dnorm(x))
> res4<- integrate(integrand, lower=1, upper=100)$value
> res4
> 0.1116
> 
> 3) However, in both cases, the syntax is a little bit heavy and not very
> convenient if I want to add more integrals, all of them with a max in the
> integrand. Is there a way to have a more concise syntax?
> Using max or pmax directly in the definition of the integrand does not work.
> For example:
> integrand<- function(x) {1/(x+max(x-50,0)*dnorm(x))}

Is boolean math more to your liking?

integr_both<- function(x){ 
                   (x < 50)*(1/x*dnorm(x))+
                   (x>= 50 & x<100)*(  1/(2*x-50)*dnorm(x) ) }
res_b<- integrate(integr_both, lower=1, upper=100)$value

> res_b
[1] 0.1116587

-- 
David

> 
> res<- integrate(integrand, lower=1, upper=100)$value
> res
> 4.60517
> 
> Thank you for any suggestion, and merry Christmas!
> 
> 	[[alternative HTML version deleted]]
> 
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David Winsemius
Alameda, CA, USA