1
$\begingroup$

I'm looking for a solution to the following integral.

$$\int_{\lambda}^{y}(x-a)^{-b}x^{-c}\exp\left( -d x^{-e} \right)dx,$$ where $b,c,d,e> 0$ and $0< a < \lambda < y$.

This equation appears in the context of Physical Layer Security, which is an area of study in digital communications (telecom).

$\endgroup$
7
  • $\begingroup$ without any further conditions (smallness of some parameters?) a closed-form solution will not be forthcoming; also note the pole at $x=a$, how is it avoided? $\endgroup$ Commented Dec 30, 2021 at 14:39
  • $\begingroup$ Dear @CarloBeenakker , do you think the additional information I've just added to the question can help solve it? $\endgroup$ Commented Dec 30, 2021 at 18:05
  • $\begingroup$ Are you looking for a closer form for the integral? for a single zero in closed form? for a an integral expression for $y$ in terms of the other variables which makes the integral zero? for something else? $\endgroup$
    – user44143
    Commented Dec 30, 2021 at 18:17
  • $\begingroup$ Dear @MattF., I'm looking for a closed form solution. $\endgroup$ Commented Dec 30, 2021 at 18:18
  • 1
    $\begingroup$ I think that Matt may be referring to is the following question: in what limits of your parameters are you interested? Even if there was a closed form for your integral, you are most likely interested in its behavior as certain parameters are small/large. It is not very interesting to just ask for a closed form without more mathematical context as right now your parameters are free to be located in essentially any part of a 7-dimensional hyperplane. $\endgroup$
    – Dispersion
    Commented Dec 30, 2021 at 22:11

1 Answer 1

3
$\begingroup$

There is no closed-form expression in terms of known functions, but if $a$ is small, you could use the power series in terms of the exponential integral, $$\int_{\lambda}^{y}(x-a)^{-b}x^{-c}\exp\left( -d x^{-e} \right)\,dx=\sum_{p=0}^\infty \frac{b\Gamma(p+b)}{p!\Gamma(1+b)}\frac{a^p}{e} \left[y^{-k}E_{1-k/e}\left(d y^{-e}\right)-\lambda^{-k}E_{1-k/e}\left(d \lambda^{-e}\right)\right]. $$ where $k=b+c+p-1$.

$\endgroup$
5
  • $\begingroup$ Dear. could you provide some further explanation on how you found that solution and what you mean by small "a" values, please? $\endgroup$ Commented Jan 3, 2022 at 19:09
  • 1
    $\begingroup$ I make a Taylor expansion of $(x-a)^{-b}$ in powers of $a$, then the integrand is of the form $x^{-b-c-p+1} e^{-dx^{-e}}$ and you have to sum over integers $p$; each term in the sum can be evaluated and gives the exponential integral; by "small a" I mean that series converges rapidly for $a\ll 1$, so then it is a helpful expansion (you would only need to retain a few terms) $\endgroup$ Commented Jan 3, 2022 at 20:08
  • $\begingroup$ Dear Carlo, one more question, in the exponential integral, is the subscript 1 - (k/e) or (1-k)/e? Thanks! $\endgroup$ Commented Jan 4, 2022 at 19:05
  • 1
    $\begingroup$ $1-k/e=1-(k/e)$ $\endgroup$ Commented Jan 4, 2022 at 19:28
  • $\begingroup$ thanks for the clarification. $\endgroup$ Commented Jan 4, 2022 at 19:29

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .