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'''p*Ei(x(log(r)-log(p)))'''
 
'''p*Ei(x(log(r)-log(p)))'''
  
Note:  <math> Ei(x) = \int_{-\infty}^{x} \frac{e^t}{t} dt </math>
+
Note:  <math> Ei(x) = \int_{x}^{\infty} \frac{e^{-t}}{t} dt </math>
  
 
I don't know how to use this integral, but I did some manipulation and got this:
 
I don't know how to use this integral, but I did some manipulation and got this:

Revision as of 10:55, 2 October 2008

First off, this is not part of homework. This equation (if I did it right) is the summation of an investment 'A' that gains interest over period 'p' and time 't' in years. The investment 'A' is added every period. I originally got $ A \sum_{t=0}^{n} \frac{r^t}{p^{t}} dt $, but I wanted to turn it into an integral and pulled out a $ \frac{t}{p} $ so I would have a dt. That led me to the integral below. Does it make sense and does anyone know how to integrate the problem?

Integrate this:

$ A \int_{0}^{n} \frac{r^t}{t*p^{t-1}} dt $


I searched how to do it on matlab, but could not find it. Then, I found this website on Wolfram. It integrates it using mathematica. Here is what it got:

p*Ei(x(log(r)-log(p)))

Note: $ Ei(x) = \int_{x}^{\infty} \frac{e^{-t}}{t} dt $

I don't know how to use this integral, but I did some manipulation and got this:

$ Total = \frac{A[(r+1)^{t+1}-(r+1)]}{r} $

---Gary Brizendine II

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Ruth Enoch, PhD Mathematics