1
$\begingroup$

I am trying to find an analytical solution of an integral of the form $$ \int_a^{\infty} x^p [e^{\epsilon(x^{-m} - x^{-n})}-1]dx $$ where $0<a<+\infty$ and $p\in[0, 1, 2, 3]$

I understand that there is no way to express this integral using standard mathematical functions and that is fine. That said...

Is it possible to use a series expansion for the exponential term?

The term in square brackets can be expressed as: $$ \sum_{k=1}^\infty \frac{[\epsilon(x^{-m} - x^{-n})]^k}{k!} $$

but when I do this, I get an integral that diverges... I am most likely doing something wrong because I have computed the numerical solution of this integral and it does not diverge.


NOTE:

In the case where $m=2n$, we can use the exponential generating function of Hermite polynomials $H_n(x)$ : $$ \exp (2xt-t^2) = \sum_{n=0}^\infty H_n(x) \frac {t^n}{n!} $$

$\endgroup$
1
  • $\begingroup$ What are m, n, and p that get you that graph, as it stands it incredibly general. Or did you sample n and m also to plot it $\endgroup$
    – Triatticus
    Commented Oct 27, 2016 at 23:53

1 Answer 1

1
$\begingroup$

If $\min(m,n) > p+1$,

$$\eqalign{\int_a^\infty x^p \dfrac{(x^{-m} - x^{-n})^k}{k!} \; dx &= \sum_{j=0}^k \int_a^\infty {k \choose j} (-1)^{j} x^{p - j n - (k-j) m}\; dx\cr &= \sum_{j=0}^k {k \choose j} (-1)^{j+1} \dfrac{a^{p-jn-(k-j)m+1}}{p-jn-(k-j)m+1} \cr }$$

It's not pretty, but I don't see how you're getting a divergent integral out of it.

In the case $a=1$, this actually simplifies to an expression using the Gamma function:

$$ \dfrac{\Gamma \left( {\frac { \left( -k+1 \right) m-n+p+1}{m-n}} \right) k!}{\Gamma \left( {\frac { \left( -k+1 \right) m-n+p+1}{m-n}} + k \right) \left( mk-p-1 \right) } $$

$\endgroup$
0

You must log in to answer this question.

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