Message Boards Message Boards

0
|
5878 Views
|
3 Replies
|
1 Total Likes
View groups...
Share
Share this post:

Another integration problem

Posted 10 years ago
Hi,

I wanted to integrate this:
Integrate[Simplify[
  Exp[-2*(m*c/hbar)*x*r] * (1 + 1/(2*x^2)) * (x^2 - 1)^(1/2) / x^2,
  {x, r, m, c, hbar} > 0 && {x, r, m, c, hbar} \[Element] Reals], {x,
   1, Infinity}]
where m,c,hbar are constants defined like this
m := 9.10938291*10^-28
hbar := 1.054571726*10^-27
c := 29979245800
and
r
is a parameter.

I tried to put in some simplifying assumtion that all values are Real and Positive in order to help solve the integral.
However the result of the integration is:
ConditionalExpression[
6.10161*10^10 r + 1.81857*10^31 r^3 +
  2.4674 MeijerG[{{}, {1/2, 3/2}}, {{0, 0}, {1/2, 1/2}},
    6.70605*10^20 r^2] +
  1.2337 MeijerG[{{}, {1/2, 5/2}}, {{0, 1}, {1/2, 1/2}},
    6.70605*10^20 r^2], Re[r] > 0]
So I was wondering if I miss something which causes the integration to result in this complicated conditional expression (for which I have no idea what it means).
In case some context might be of help: I'm trying to solve the integral part of Eq.1 of this paper: http://cds.cern.ch/record/966452/files/PhysRev.95.1048.pdf

Any ideas anyone ?
POSTED BY: xort dsc
3 Replies
Posted 10 years ago
Xort, the following form should work
m := 9.10938291*10^-28
hbar := 1.054571726*10^-27
c := 29979245800

function = Exp[(-2*(m*c/hbar)*x*r)]*(1 + 1/(2*x^2))*(x^2 - 1)^(1/2)/x^2

Integrate[function, {x, 1, Infinity}]
POSTED BY: Alejandro Braun
Posted 10 years ago
ah thank you very much. yes, i'm still struggling a lot with the syntax :/
POSTED BY: xort dsc
Correct syntax will help here.
Integrate[
Exp[-2*(m*c/hbar)*x*r]*(1 + 1/(2*x^2))*(x^2 - 1)^(1/2)/x^2, {x, 1,
  Infinity}, Assumptions -> Thread[{r, m, c, hbar} > 0]]
Alternatively, give those constants the intended values. The integral will evaluate much faster.
m = 9.10938291*10^-28;
hbar = 1.054571726*10^-27;
c = 29979245800;

Integrate[
Exp[-2*(m*c/hbar)*x*r]*(1 + 1/(2*x^2))*(x^2 - 1)^(1/2)/x^2, {x, 1,
  Infinity}, Assumptions -> r > 0]
Better might be to remove the constants, since they only are used with r, and put them back later.
Integrate[
Exp[-2*(m*c/hbar)*x*r]*(1 + 1/(2*x^2))*(x^2 - 1)^(1/2)/x^2, {x, 1,
  Infinity}, Assumptions -> Thread[{r, m, c, hbar} > 0]] /. r->(m*c/hbar)*r
POSTED BY: Daniel Lichtblau
Reply to this discussion
Community posts can be styled and formatted using the Markdown syntax.
Reply Preview
Attachments
Remove
or Discard

Group Abstract Group Abstract