Rewrite $\int\limits_{x=0}^\infty{2\sqrt{a -\frac{b}{x}}{K_1}({2\sqrt{a-\frac{b}{x}}}){x^{M-1}}\exp({-\frac{x}{c}})dx}$ as non-elementary function?

Question

Is it possible to rewrite this integral $I=\int\limits_{x = 0}^\infty {2\sqrt {a - \frac{b}{x}} {K_1}\left( {2\sqrt {a - \frac{b}{x}} } \right){x^{M - 1}}\exp \left( { - \frac{x}{c}} \right)dx}$ as non elementary function (For exaple $Ei(x)$, $Li(x)$) ?

${K_1}\left( x \right)$ is the modified Bessel function of the second kind.

$a,b,c$ are positive real number and $M$ is a positive integer

It is ok if someone can help me express the integral $I$ as an infinite series. I have also think of using the expansion ${e^x} = \sum\limits_{k = 0}^\infty {\frac{{{x^k}}}{{k!}}}$ to deal with the exponential term but I cannot proceed.

For clarification, I am an engineer working in Telecommunications. This integral is something come out of the Probability the signal fall below some decoding threshold.

Thank you for your enthusiasm !

If I may ask, where did you catch this monster ? Back to serious, could you precise the context ? — Claude Leibovici, Nov 13 '20 at 13:41
You mean find an analytic closed-form expression for the integral? It is very probable to be impossible. — Somos, Nov 13 '20 at 14:28
@Somos yes Claude I have added the origin of this integral in the original post for clarification — Tuong Nguyen Minh, Nov 13 '20 at 14:34

Raffaele · Accepted Answer · 2020-12-22T10:14:32.343

1

No way to find a closed form.

I found an approximation which uses $K_2$, too $$\int_0^{\infty } 2 x^{M-1} \sqrt{a-\frac{b}{x}} K_1\left(2 \sqrt{a-\frac{b}{x}}\right)e^{-x/c} dx\approx \\2 c^{M-1} \Gamma (M-1) \left(b K_2\left(2 \sqrt{a}\right)-\frac{K_1\left(2 \sqrt{a}\right) (b-a c (M-1))}{\sqrt{a}}\right)$$

Mathematica code

Integrate[
  Series[2 E^(-(x/c)) Sqrt[a - b/x] x^(-1 + M)
      BesselK[1, 2 Sqrt[a - b/x]], {x, Infinity, 1}] // Normal, {x, 0,
    Infinity}] // FullSimplify

At $x\to\infty$

$$2 x^{M-1} \sqrt{a-\frac{b}{x}} K_1\left(2 \sqrt{a-\frac{b}{x}}\right)e^{-x/c}\sim \\e^{-\frac{x}{c}} x^{M-1} \left(\frac{2 b \left(\sqrt{a} K_2\left(2 \sqrt{a}\right)-K_1\left(2 \sqrt{a}\right)\right)}{\sqrt{a} x}+2 \sqrt{a} K_1\left(2 \sqrt{a}\right)\right)$$

edited Dec 22 '20 at 10:14

answered Nov 13 '20 at 16:01

Raffaele

26,371

Wow can you provide some detail about this approximation ? I did not quite catch up with your idea :) – Tuong Nguyen Minh Nov 14 '20 at 00:02
@TuongNguyenMinh I used Mathematica to get a Taylor approximation as $x\to +\infty$ – Raffaele Nov 14 '20 at 00:08
Could you kindly provide me with the code please ? – Tuong Nguyen Minh Nov 14 '20 at 04:55
1

@TuongNguyenMinh I have edited my answer to include the code – Raffaele Nov 14 '20 at 07:34
May I ask a stupid question about obtaining this answer. First you do the taylor series expansion at infinity for the integrand. Then you collect only one term and then integrate that term right ? – Tuong Nguyen Minh Dec 22 '20 at 02:35
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@TuongNguyenMinh Right! I added the approximation – Raffaele Dec 22 '20 at 10:15

Rewrite $\int\limits_{x=0}^\infty{2\sqrt{a -\frac{b}{x}}{K_1}({2\sqrt{a-\frac{b}{x}}}){x^{M-1}}\exp({-\frac{x}{c}})dx}$ as non-elementary function?

1 Answers1