1. Aug 5, 2005

undefined83

Im supposed to solve
integral 10 to +infinity ((sin(1/x)/(1+x^3))dx with error precision of e=0.5*10^-4. Can someone please give me detailed explenation of solving this. (Supposedly by Simpson but i get lost in the way.

P.S. sorry for bad spelling and lack of proper formula notions.

2. Aug 5, 2005

undefined83

I process from integral i get
1/3[ln((x+1)/sqrt(x^2-x+1))+sqrt(3)*arctg(2*sqrt(3)*x/3-sqrt(3)/3)] from M to +infinity <=1/4*10^-2

than i get

-ln((M+1)/(sqrt(M^2-M+1))) +sqrt(3)*(pi/2-arctg(2*sqrt(3)*M/3-sqrt(3)/3))<=7.5*10^-3

and i cant find any exact solution to solve that

3. Aug 5, 2005

lurflurf

break the interval up
10=x0<x1<x2<...<xn-1<xn=infinity
for
int(10,x1)
int(x1,x2)
.
.
.
int(xn-2,xn-1)
use simpsons rule
for
int(xn-1,infinity)
chose xn-1 large
estimate the integral on each subinterval accurate enough so that total error is within limit

4. Aug 5, 2005

undefined83

Thanks, but i am totaly lost when i need to find int(xn-1,infinity). I try to make it lower than error estimate but i get either logaritmic inverse-trigonometric inequality as above or another insolvable integral.

5. Aug 5, 2005

lurflurf

Don't over think it
$$|{\int_x^\infty \frac{\sin(\frac{1}{x})}{1+x^3}}|<\int_x^\infty \frac{1}{x^3}dx=\frac{1}{2x^2}$$

Last edited: Aug 5, 2005
6. Aug 5, 2005

undefined83

Thank you very, very much. You are a life savior.

7. Aug 5, 2005

undefined83

Does someone who has Maple or Mathlab can give me aproximate solution of the integral, posiblly of int(10,45), and int(10,142). Error comarision goes to absurd. I get 3.3*10^-4 on pocket calculator for int(10,35). Simpson with above given boundries gives more than 10 times larger values.

8. Aug 5, 2005

saltydog

$$\int_{10}^{45} \frac{Sin(1/x)}{1+x^3}dx\approx 0.000329176$$

$$\int_{10}^{142} \frac{Sin(1/x)}{1+x^3}dx\approx 0.000332717$$

As per Mathematica's NIntegrate.

Last edited: Aug 5, 2005
9. Aug 6, 2005

lurflurf

I don't know if you want to stay with the original integral or not. In any case a substitution will make this much easier.
u=1/x
$$\int_{10}^\infty \frac{\sin(\frac{1}{x})}{1+x^2}dx=\int_0^\frac{1}{10}\frac{u\sin(u)}{1+u^3}du$$
then since x(sin(.1)/.1)<sin(x)<x for 0<x<.1
$$\frac{\sin(.1)}{.3}\log(1.001)=10\sin(1/10)\int_0^\frac{1}{10}\frac{u^2}{1+u^3}du<\int_0^\frac{1}{10}\frac{u\sin(u)}{1+u^3}du<\int_0^\frac{1}{10}\frac{u^2}{1+u^3}du=\frac{1}{3}\log(1.001)$$
The average of these yeild a good approximation.
simpsons rule will meet the error tolerance
I~(.1-0)/6(f(0)+4f(.05)+f(.1))
where f(x)=x sin(x)/(1+x^3)

Last edited: Aug 6, 2005
10. Aug 6, 2005

undefined83

Thank. This afternoon i finaly finished the monster with int(10,150) by Simpson. Got aprox 0.000324. I cant realy believe its finished.