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.

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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

lurflurf
Homework Helper
undefined83 said:
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.
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

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.

lurflurf
Homework Helper
undefined83 said:
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.
Don't over think it
$$|{\int_x^\infty \frac{\sin(\frac{1}{x})}{1+x^3}}|<\int_x^\infty \frac{1}{x^3}dx=\frac{1}{2x^2}$$

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

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.

saltydog
Homework Helper
undefined83 said:
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.
$$\int_{10}^{45} \frac{Sin(1/x)}{1+x^3}dx\approx 0.000329176$$

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

As per Mathematica's NIntegrate.

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lurflurf
Homework Helper
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)

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