I asked this question on math.stackexchange but deleted it from there and brought it here.

I had a question about Constant Elasticity of substitution type preferences of the form: $$U=\int_{0}^{1}(c(\omega)^{\rho}d\omega)^{\frac{1}{\rho}}$$ where the paramter $\rho$ governs the degree of substitutability between goods. Here, $c(\omega)$ represents consumption of good $\omega$ which exists on the unit interval. As such, this type of preference specification aggregates over consumption of different goods.

My question is as follows. I have always thought of integrals of the form: $$I=\int f(x)dx$$ as approximating sums of areas of infinitesimally small rectangles (in terms of their base) and heights being determined by $f(x)$ . In the case of the example above, what really is $c(\omega)$ ? Is it a function?


2 Answers 2


It is my impression that the correct expression for "Dixit-Stiglitz" preferences is


which then can be seen as a continuous incarnation (in [0,1]) of, say

$$\left (\sum_{i=1}^na_i\omega_i^{\rho}\right)^{\frac{1}{\rho}}$$

with $c(\omega_i) = a_i^{1/\rho}\omega_i$.

In other words, a definite (Riemann) integral is indeed conceived as a sum of infinitesimally small rectangles, but it can also be seen as the continuous incarnation of a sum.

A formal link between an integral and a sum is provided by the Euler-MacLaurin formula.

  • $\begingroup$ Could you elaborate a bit more on this? I'm having a similar conceptual problem. It's not clear to me that this is a Riemann integral, given that $c(\omega)$ is not necessarily continuous and hence the Riemann integral not necessarily exists.. $\endgroup$
    – FooBar
    Jun 10, 2018 at 12:09

Yes it is a function, just like you have $f(x)$, read as $f$ of $x$, $c(w)$ is $c$ of $w$ (in layman's words). Simply $c$ is the function of $w$ goods. Depending on the shape of the consumption function, the result is the area under the $c$ curve evaluated at each infinitesimally small intervals on $w$.


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