The following passage is in my textbook:

$$A(S) = \int_0^{\infty} f(E) \max(S-E,0)dE$$

This simplifies to $$A(S) = \int_0^{S} f(E)(S-E) dE$$

Now this is from a finance textbook so it might not be fully rigorous, but I'm curious about what happens with the integral. I think this is an improper integral, and I haven't really studied those yet, but after reading up on them I'm still confused as to what is done here and why it's allowed.


This is because $\max(S - E, 0) = 0$ for $E \geq S$. Alternatively, it is equal to $S - E$ for $E \leq S$.

  • $\begingroup$ Yes, but how does this explain the changing integral bound from infinity to S? $\endgroup$ – YakSal Tafri May 29 '15 at 18:35
  • $\begingroup$ Because the function is zero there and $$\int_0^\infty f(x)dx = \int_0^S f(x)dx + \int_S^\infty f(x)dx$$ $\endgroup$ – muaddib May 29 '15 at 18:37
  • 1
    $\begingroup$ Oh of course, thanks. $\endgroup$ – YakSal Tafri May 29 '15 at 18:46

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