For example when solving Romer’s model (1990) in continuous time, for the firm producing final goods, its production function is:

$ Y(t) = \int_{0}^{M(t)} (A(t) L_Y)^{1-\alpha} {x(i,t)}^{\alpha} di$,

where $A(t)$ is labor productivity, $L_Y$ the labor hired to produce the final good $Y$ and $x(i,t)$ is the quantity used of the $i$-th intermediate good, with $t$ being time.

$M(t)$ is the quantity of different (non-homogeneous) intermediate goods.

Suppose the final good $Y$ is the numeraire, i.e. its price is set to $p_Y = 1$.

With this, the profit function is given by

$\Pi^Y = \int_{0}^{M(t)} (A(t) L_Y)^{1-\alpha} {x(i,t)}^{\alpha} di - w(t) L_Y - \int_{0}^{M(t)} p(i,t) x(i,t) di$

where $w(t)$ is the wage and $p(i,t)$ is the price of $x_i$ (the $i$-th intermediate good).

The way I learned to get the f.o.c.’s for my class is

$\frac{\partial \Pi^Y}{\partial x_i} = \alpha (A(t) L_Y)^{1-\alpha} {x(i,t)}^{\alpha -1} - p(i,t) = 0$.

However, to do the actual differentiation, wouldn’t we have to use some sort of chain rule which would still leave us with an integral, instead of differentiating pretending there is no integral at all?

Similarly with the $\frac{\partial \Pi^Y}{\partial L_Y}$ f.o.c.

I asked my professor about it, and he said it’s just a hand wavy thing Economists do to solve this kind of models.

As a math guy, I would like to know the underlying reason why that works. I’d appreciate any insight about it.

This model appeared in my third course in Macroeconomics which covers many long-run growth models, such as Solow (the only one not involving summation/integration), Ramsey, Romer (1986), Lucas (1989), Aghion and Howitt (1992), as well as this one [Romer (1990)].

Whenever we work with continuous time versions of some of the above models, we take the f.o.c’s with the same method.

I remember I answered a question here months ago before getting to this topic, involving finding an f.o.c. for a function involving integrals; and the discrepancy between my answer and OP’s book’s answer was that I had extra integrals from the chain rule, probably due to the same method.


1 Answer 1


The justification for this rule of thumb is the calculus of variations, specifically with the functional derivative.

First, note that the problem is static, so for ease of notation I'll drop the dependence on $t$. Second, I'll rename the integration index, since I think it would otherwise just be a source of confusion. Hence we have,

$$ \Pi^Y(x) = \int_0^M(AL_Y)^{1- \alpha}x(j)^\alpha\mathrm{d}j - wL_Y - \int_0^Mp(j)x(j)\mathrm{d}j. $$

Note that $\Pi^Y$ is a functional, i.e. it takes the function $x$ and outputs a real number (for every $t$, but again, I'll drop that detail). Now we take an arbitrary function $\phi(j)$ and perturb this functional by a small amount $\varepsilon \phi(j)$. Then, we imagine taking the limit as $\varepsilon$ tends to zero. This defines the functional derivative of $\Pi^Y$

\begin{align} \frac{\delta \Pi^Y}{\delta x} &= \left[\frac{\mathrm{d}}{\mathrm{d}\varepsilon} \Pi^Y (x + \varepsilon \phi)\right]_{\varepsilon = 0}\\ &= \left[\frac{\mathrm{d}}{\mathrm{d}\varepsilon} \left(\int_0^M(AL_Y)^{1- \alpha}(x(j) + \varepsilon\phi(j))^\alpha \mathrm{d}j - wL_Y - \int_0^Mp(j)(x(j) + \varepsilon \phi(j))\mathrm{d}j\right)\right]_{\varepsilon = 0}\\ &= \left[\int_0^M\alpha(AL_Y)^{1- \alpha}(x(j) + \varepsilon\phi(j))^{\alpha - 1}\phi(j) \mathrm{d}j - \int_0^Mp(j) \phi(j)\mathrm{d}j\right]_{\varepsilon = 0} \\ &= \int_0^M\alpha(AL_Y)^{1- \alpha}x(j)^{\alpha - 1}\phi(j) \mathrm{d}j - \int_0^Mp(j) \phi(j)\mathrm{d}j \\ & = 0. \end{align} where I set this functional derivative to zero (by standard FOC logic). Hence,

$$ \int_0^M \left[\alpha (AL_Y)^{1 - \alpha}x(j)^{\alpha - 1} - p(j)\right]\phi(j)\mathrm{d}j = 0 $$ Since this must be true for all $\phi$, it is sufficient that the integrand is identically equal to zero (this means it is zero for all $j$, and in particular, for $j = i$). This yields the FOC.

I think another formal justification could work with approximating the integral by a sum, restricting the choice to step functions, maximizing as one would normally with a sum, and then taking the limit.

Hope this helps!

  • $\begingroup$ Thank you very much for your answer!!! I have taken a course in Calculus of Variations, and found your answer very clean and easy to follow! $\endgroup$ Jul 12, 2023 at 19:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.