In footnote 3 of their paper "Patent Breadth, Patent Life, and the Pace of Technological Progress", O'Donoghue Scotchmer and Thisse take a discount rate of $r$ and write

If the flow of profit $\Delta$ lasts for length $t$, the discounted profit is $\Delta(1 − e^{−rt})/r$.

Can someone explain where the expression $\Delta(1 − e^{−rt})/r$ comes from?

Whilst I am familiar with continuous time discounting, I have never seen it formulated in this way before. I would usually expect the expression for discounted profit to be $\Delta e^{-rt}$.


1 Answer 1


From time $0$ to time $t$ the profit flow is constant $\Delta$. The discounted value of the profit flow of any instance $s$ is $\Delta \cdot e^{-r s}$. To get the total discounted profit we will need the integral of this from $0$ to $t$: $$ \int_0^t \Delta \cdot e^{-r s} \ ds = \left. \frac{\Delta \cdot e^{-r s}}{-r} \right]_0^t = \frac{\Delta \cdot e^{-r t}}{-r} - \frac{\Delta \cdot e^{-r \cdot 0}}{-r} $$ And as $e^{-r \cdot 0} = e^0 = 1$, this is $$ - \frac{\Delta \cdot e^{-r t}}{r} + \frac{\Delta \cdot 1}{r} = \Delta \cdot \frac{1 -e^{-r t}}{r}. $$

  • $\begingroup$ Yes, I had confused flows with stocks. Schoolboy error! thanks for for clearing it up. $\endgroup$
    – Ubiquitous
    Jul 30, 2015 at 15:28

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