Hedging with actual volatility: problem understanding the math behind the result

From this paper. page 3

We get that the total profit at expiration is the difference in value between the price of the option with actual volatility and the one with implied volatility.

I have tried solving the integral but I don't get the same result. I especially don't understand why the incremental value in both options cancel each other out.

$$e^{r\cdot t_0} \int_{t_0}^T d\left(e^{-r\cdot t}(V^i - V^a)\right) = V^a - V^i$$

I used the expression of the differential to solve the integral in the hope of getting the same result, but I get stuck with terms in $dV^i$ and $dV^a$

I get

$$\left(e^{-r\cdot T} - e^{-r.\cdot t_0}\right)\left[ (dV^a - dV^i)/r - (V^a - V^i) \right]$$

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Could you please include the integral in the question and show us exactly where you're stuck in your equation solving? –  SRKX Mar 25 '13 at 14:02

The integration is over a full differential, meaning we can write:

$$\int_{t_i}^T df(t) = f(T) - f(t_i)$$

Now, $V^i$ and $V^a$ represent the 'implied' and 'actual' value of the option, meaning they are time-dependent. This gives:

$$e^{r\cdot t_0} \int_{t_0}^T d\left(e^{-r\cdot t}(V^i - V^a)\right) = e^{-r(T-t_0)} (V^i(T) - V^a(T)) - (V^i(t_i) - V^a(t_i))$$

Next, we use the fact that at the time of expiration the value of the options is completely determined by the stock price, and independent of the volatility. This means: $V^i(T) = V^a(T)$. What remains is:

$$e^{r\cdot t_0} \int_{t_0}^T d\left(e^{-r\cdot t}(V^i - V^a)\right) = V^a(t_i) - V^i(t_i) = V^a - V^i$$

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Thank you, i feel so stupid now. I think it's e^{-r(T-t_0)} –  user1627466 Mar 25 '13 at 16:11
Yea, that was a typo. –  Olaf Mar 25 '13 at 16:25