Present value of a perpetuity with continuous stream of cash flow The present value of a perpetuity (cash flows paid at the end of each year) is $PV = CF / r$ where $r$ is the interest rate. This formula is proved in the book that I'm studying, Principles of Corporate Finance.
However, then it is stated that if instead the cash flows are spread evenly throughout each year like a continuous stream of payments, we can use the same formula but replace $r$ with the continuously compounded rate $r_c$. This is not proved in the book and I cannot see how this would automatically hold. Can someone please show how this holds?
 A: If we have continuous compounding the future value after one year is $C\cdot e^r$
For $n$ yearly payments we have the series
$FV=C+Ce^r+Ce^{2r}+Ce^{3r}+\ldots+Ce^{(n-1)r}$
Using the closed form of a geometric series we get
$FV=C\cdot \frac{e^{rn}-1}{e^r-1}$
To get the present value the FV has to be discounted $n$ times
$PV=C\cdot\frac{1}{e^{rn}}\cdot \frac{e^{rn}-1}{e^r-1}=C\cdot \left( 1-\frac{1}{e^{rn}}\right)\cdot \frac{1}{e^r-1}$
Now let $n$ go to infinity
$$PV=\lim_{n \to \infty} C\cdot\left( 1-\frac{1}{e^{rn}}\right)\cdot \frac{1}{e^r-1}$$
$=C\cdot(1-0)\cdot \frac{1}{e^r-1}=\boxed{C\cdot\frac{1}{e^r-1}}$
A: Let the cashflow be evenly spread across time to perpetuity.  Then the present value of such a stream would be ( here the discount rate is continuously compounded rate $r_c$)
$$ PV = \int_{0}^{\infty} CF.e^{-r_ct}dt$$
$$ PV = -\frac{CF}{r_c}e^{-r_ct}\|_0^\infty$$
$$ PV = \frac{CF}{r_c}e^{-r_ct}\|_\infty^0$$
$$PV = \frac{CF}{r_c}\left(e^0 - e^{-\infty}\right)$$
$$PV = \frac{CF}{r_c}(1-0) = \frac{CF}{r_c}$$
Edit:
The discount factor for discrete compouding is $\frac{1}{(1+r)}$.  The discount factor for continuous compounding is $e^{-r_c}$.  Equating these you have 
$\frac{1}{(1+r)}. = e^{-r_c}$ 
=> $e^{r_c} = (1+r)$
=>$r_c = ln(1+r)$
Goodluck
A: The integration answer is correct. An integral is a sum. The Integrand is discounted correctly. The answer that sums the cash flow adds dollars in different units, Ce^r + Ce^2r for example adds dollars of two different years which have different values. If r is a continuously compounded yield like from zero yield Treasury curve just plug that rate into C/r. If r is an annual rate compounded once, use r’= ln(1+r) with r’~= ln (r) is f r very small.
A: The textbook example (from Brealey, et. al. I presume) involves a perpetituity with annual compounding and compares two scenarios: (1) one payment, $C$, at the end of each year and (2) "continuous cash flows" over the year (summing to C): a good approximation to daily payments of C/365.
I derived the following equation from first principles:
$$
\frac{C/n}{\left(1+\frac{r}{m}\right)^{m/n}-1}\left( 1-  \left(1+\frac{r}{m}\right)^{-mT} \right)
$$
This calculates the value of a $T$-year annuity, compounded $m$ times a year, with $n$ payments during the year adding up to an annual payment of $C$.
If we take the limit as $ n \rightarrow \infty$, we get the special case with continuous cash flows  (after applying La Hopital's Rule):
$$
\frac{C}{\ln\left(1+\frac{r}{m}\right)^{m}}\left( 1-  \left(1+\frac{r}{m}\right)^{-mT} \right)
$$
In the textbook, $m=1$ (annual compounding) and $T \rightarrow \infty$ (perpetuity). I suspect this gives the equation you are looking for:
$$
\frac{C}{\ln\left(1+r\right)}
$$
Notice that $\ln\left(1+r\right)$ is the continuous compounded rate with an effective annual interest rate of $r$. Since the continuously compounded interest rate is less than its corresponding effective annual rate then $\ln\left(1+r\right) < r$, so continuous payments are worth more than a single end-of-year payment.
This is consistent with the textbook Example 2 since $r =18.5\%$ and so $\ln\left(1+r\right)=17\%$. For $C=\$100$, the perpetuity with continuous cash flow is valued at $100/.17$, which is compared with the standard formula for a pertituity namely $C/r = 100/.185$.
The more general formulation can be used with Example 3, where $T=20$ and  $C=\$200,000$.
Finally, take the limit as $m \rightarrow \infty$, and we get the formula for continuous compounding:
$$
\frac{C/n} {e^{r/n}-1}\left(1-  e^{-rT}\right)
$$
And as $n \rightarrow \infty$, again using La Hopital's Rule,
$$
\frac{ A}{r}\left(1-  e^{-rT}\right)
$$
