What is the best way to approximate a function of the following form,

$$ \text{exp}\left(-\int_{y}^{+\infty} f(x)\ dx \right)$$

Any approximation to this, does taylor series work? The reason I am doing so is because $f(x)$ cannot be integrated in closed form.

If the form of $f(x)$ is needed then I will edit my question and write it down.

Appreciate any advice.

  • $\begingroup$ Can you write yor integral as $\exp(-N\int_y^{\infty}f(x)dx)$. ? $\endgroup$ – tired Oct 31 '14 at 17:45
  • $\begingroup$ what is N? @tired $\endgroup$ – Tyrone Oct 31 '14 at 17:46
  • $\begingroup$ some constant which we may send to zero or infinity... $\endgroup$ – tired Oct 31 '14 at 17:46
  • $\begingroup$ Nope I cantt @tired $\endgroup$ – Tyrone Oct 31 '14 at 17:53

Assuming the improper integral exists, your function can be written as

$$F(y) = C \exp\left(\int_0^y f(x)\; dx\right)$$ where $C = \exp(-\int_0^\infty f(x)\; dx)$.
Assuming the required derivatives exist, the Taylor series of $F(y)$ around $y=0$ starts

$$ F(y) = C + C f(0)\; y + \dfrac{C}{2} \left( f'(0) + f(0)^2\right) y^2 + \dfrac{C}{6} \left( f''(0) + 3 f(0) f'(0) + f(0)^3\right) y^3 + \ldots $$

Is that what you're looking for?

  • $\begingroup$ Thanks thats a great approximation however not even $C$ is integrable in my case. I was hoping to approximate the integral and not have to deal with it. $\endgroup$ – Tyrone Oct 31 '14 at 17:52
  • $\begingroup$ Not integrable? Do you mean the integral doesn't exist, or that you don't know the integral? $\endgroup$ – Robert Israel Oct 31 '14 at 21:00
  • $\begingroup$ If you don't know the integral and want to approximate it, then we'd really have to know more about $f$. $\endgroup$ – Robert Israel Oct 31 '14 at 21:03
  • $\begingroup$ $f(x) = (1-\frac{1}{1+sx^{-1}}) \text{exp}(-x^{0.5})$ for some positive s ! $\endgroup$ – Tyrone Oct 31 '14 at 21:09
  • 1
    $\begingroup$ I used Maple... $\endgroup$ – Robert Israel Oct 31 '14 at 22:30

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