Register to reply

Difficult integral

by Per Oni
Tags: difficult, integral
Share this thread:
Per Oni
#1
Nov2-09, 02:34 PM
P: 262
Please help required with this integral:
(x/(x-a))^0.5 where "a" is a start distance of 10^-3 and the final distance needs to be 10^-2

It looks simple but its not.
Wolfram integrator gave this answer:

Integrate[(x/(x - a))^0.5, x] ==
(0.*(x/(-a + x))^0.5*(-a + x)^0.5)/x^0.5 + (2.*(x/(-a + x))^0.5*(-a + x)^0.5*(-1.*a + x)^0.5* Hypergeometric2F1[0.5, -0.5, 1.5, 1. - (1.*x)/a])/ (0. + x/a)^0.5

Which is way way over my head. Is there a simpler solution?
Phys.Org News Partner Science news on Phys.org
Final pieces to the circadian clock puzzle found
A spray-on light show on four wheels: Darkside Scientific
How an ancient vertebrate uses familiar tools to build a strange-looking head
Gib Z
#2
Nov3-09, 02:09 AM
HW Helper
Gib Z's Avatar
P: 3,348
Assuming the integral you want to solve is [tex]\int \sqrt{ \frac{x}{x-a}} dx[/tex], make the substitution [itex] x= a \sec^2 \theta[/itex].
g_edgar
#3
Nov3-09, 06:47 AM
P: 607
Quote Quote by Per Oni View Post
Wolfram integrator gave this answer:

Integrate[(x/(x - a))^0.5, x] ==
(0.*(x/(-a + x))^0.5*(-a + x)^0.5)/x^0.5 + (2.*(x/(-a + x))^0.5*(-a + x)^0.5*(-1.*a + x)^0.5* Hypergeometric2F1[0.5, -0.5, 1.5, 1. - (1.*x)/a])/ (0. + x/a)^0.5

Which is way way over my head. Is there a simpler solution?
Tell Wolfram again, but this time use 1/2 and not 0.5 ... this tells Wolfram that the exponent is an exact number, and not just a decimal approximation to some number. If the exponent is very close to 1/2, but perhaps not equal to 1/2, then the answer will come out as an 2F1 as shown. But if the exponent is exactly 1/2, then you can get an answer in logarithms.

Per Oni
#4
Nov3-09, 07:20 AM
P: 262
Difficult integral

Quote Quote by g_edgar View Post
Tell Wolfram again, but this time use 1/2 and not 0.5 ... this tells Wolfram that the exponent is an exact number, and not just a decimal approximation to some number. If the exponent is very close to 1/2, but perhaps not equal to 1/2, then the answer will come out as an 2F1 as shown. But if the exponent is exactly 1/2, then you can get an answer in logarithms.
Thanks a lot. Using 1/2 gave me a sensible answer.


Register to reply

Related Discussions
A difficult integral! for help! Calculus & Beyond Homework 1
Difficult integral.... General Math 10
Difficult integral Calculus 0
A Difficult Integral Calculus & Beyond Homework 2
A difficult integral Calculus & Beyond Homework 2