Differentiation under Integral Sign involving an Exponential

derivativesexponential functionintegration

An economics paper that I am reading has this expression:
$Q_{t} = \int_{t}^{\infty}\delta e^{-(\bar{r}+\delta)(s-t)-\int_{t}^{s}\pi_{u}du}ds$. After taking derivative with respect to time, the above relationship changes to
$Q_{t}(\bar{r} + \delta + \pi_{t}) = \delta + \dot{Q_{t}}$
where the dot over $Q_{t}$ refers to time derivative. I am trying to go from the first expression to the second one but I can't seem to get it right. Can you please help?

Here is what I did:

$\frac{d}{dt}Q_{t} = \dot{Q_{t}} = \int_{t}^{\infty}\frac{d}{dt}[\delta e^{-(\bar{r}+\delta)(s-t)-\int_{t}^{s}\pi_{u}du}]ds \Rightarrow
\dot{Q_{t}} = \int_{t}^{\infty}\delta e ^{-(\bar{r}+\delta)(s-t)-(\pi_{s} – \pi_{t})}ds \cdot (\bar{r}+\delta+\frac{d}{dt}(\pi_{t}))
= Q_{t}(\bar{r}+\delta+\frac{d}{dt}(\pi_{t}))$

This, of course, is not what the paper has. Will appreciate any help.

Best Answer

Thanks for showing your work—it makes it easy to help out!

There are two mistakes in your derivation. The first is the one I mentioned in my comment. If we have a function of the form $G(t) = \int_t^\infty f(s,t)\,ds$, then the formula for the derivative is that $G'(t) = -f(t,t) + \int_t^\infty \frac\partial{\partial t}f(s,t)$. (Mnemonic, which can be turned into a proof: if $t$ appears multiple times in the definition of a function, then the function's $t$-derivative is the sum of all the derivatives you get by considering each $t$ in turn, pretending all the other $t$s are actually constant, and taking the derivative of the function-with-just-that-one-variable-$t$. If you think about it, that's exactly what the product rule formula does—and even the formula for $t^n = t\times\cdots\times t$ works right under this mnemonic!) In this case, that tells us that \begin{align*} \dot Q_t &= -\delta e^{-(\bar r+\delta)(t-t)-\int_t^t \pi_u\,du} + \int_t^\infty \frac\partial{\partial t} \big[ \delta e^{-(\bar r+\delta)(s-t)-\int_t^s \pi_u\,du} \big] \,ds \\ &= -\delta + Q_t \frac\partial{\partial t} \bigg[ {-}(\bar r+\delta)(s-t)-\int_t^s \pi_u\,du \bigg]. \end{align*} The second mistake is that you wrote $\int_t^s \pi_u\,du = \pi_s-\pi_t$ (it would be the case that $\int_t^s \frac d{du}(\pi_u)\,du = \pi_s-\pi_t$, but that's not what we have), which caused an error in your evaluation of $\frac\partial{\partial t} \big[ {-}(\bar r+\delta)(s-t)-\int_t^s \pi_u\,du \big]$. We actually get $$ \frac\partial{\partial t} \bigg[ {-}(\bar r+\delta)(s-t)-\int_t^s \pi_u\,du \bigg] = \bar r+\delta + \pi_t, $$ which plugs into the above equation to yield $$ \dot Q_t = -\delta + Q_t(\bar r+\delta + \pi_t) $$ as desired.