<p>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
$$
- \frac{\Delta \cdot e^{-r t}}{r} + \frac{\Delta \cdot 1}{r} = \Delta \cdot \frac{1 -e^{-r t}}{r}.
$$</p>

CC BY-SA 3.0
2015-07-30T15:28:00.410Z
Yes, I had confused flows with stocks. Schoolboy error! thanks for for clearing it up.
Ubiquitous
CC BY-SA 3.0
2015-07-30T13:38:05.353Z
2015-07-30T15:19:37.350Z
2015-07-30T15:19:37.350Z
Giskard
<p>In footnote 3 of their paper "<a href="http://onlinelibrary.wiley.com/doi/10.1111/j.1430-9134.1998.00001.x/abstract" rel="nofollow">Patent Breadth, Patent Life, and the Pace of Technological Progress</a>", O'Donoghue Scotchmer and Thisse take a discount rate of $r$ and write</p>

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

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

<p>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 $\Delta e^{-rt}$.</p>

<mathematical-economics><continuous-time>
Help understanding expression for continuous discounting
Ubiquitous
# Help understanding expression for continuous discounting

(Asked by: username_0 on 2015-07-30)

In footnote 3 of their paper "[Patent Breadth, Patent Life, and the Pace of Technological Progress](http://onlinelibrary.wiley.com/doi/10.1111/j.1430-9134.1998.00001.x/abstract)", 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 $\Delta e^{-rt}$.

**Tags:** mathematical-economics, continuous-time

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## 1 answers

### Answer by username_1 on 2015-07-30. Score: 2 | Accepted answer

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
$$
- \frac{\Delta \cdot e^{-r t}}{r} + \frac{\Delta \cdot 1}{r} = \Delta \cdot \frac{1 -e^{-r t}}{r}.
$$

> **Comments:**
> **username_0**: Yes, I had confused flows with stocks. Schoolboy error! thanks for for clearing it up.  

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