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Then use the [http://en.wikipedia.org/wiki/Integration_by_parts integration by parts] formula <center><math>\int u\, \frac{dv}{dx}\; dx=uv-\int v\, \frac{du}{dx} \; dx\!</math></center>
 
 
Observe that as the expected purchase price <math>\left [ \underline{p} + \int_{\underline{p}}^{\overline{p}} (1-F(p))^n \; dp \right ]\,</math> is decreasing in <math>n\,</math> and that the cost of search is positive (<math>c > 0\,</math>) the optimum will be finite.
 
 
The expected benefit for a customer to increase their sample size from <math>n-1\,</math> to <math>n\,</math> is:
 
 
<center><math>\mathbb{E}(B^{(n)}=\left ( \mathbb{E} (p_{min}^{n-1}) - \mathbb{E} (p_{min}^{n}) \right ) \times K\,</math></center>
 
 
This is decreasing in <math>n\,</math> and increasing in <math>K\,</math>. Also as the cost of the <math>nth search is independent of <math>K\,</math>, the equilibrium search intensity is increasing in <math>K\,</math>. Note that <math>K\,</math> may refer to either purchases in greater quantities or more frequent purchases.
 
 
Although customers inelastically purchase <math>K\,</math> units, a version of the [http://en.wikipedia.org/wiki/Law_of_demand law of demand] holds: Each firm's expected demand is a non-increasing function of its price.
 
 
A firm charging price <math>p is visited by <math>\mu n^* customers\,</math>
 
 
A firm charging price <math>p offers the lowest price with probability <math>(1-F(p))^{n^*-1}\,</math>
 
 
<center><math>\therefore Q(p) - \mu n^* K (1-F(p))^{n^*-1}\,</math></center>
 
<math>
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