# How to approximate the following improper integral by using Simpson's rule

I am trying to approximate the following integral by using Simpson's Rule with $n=6$

$$I = \int_{1}^{\infty} \frac{\sin(x)}{x^4} dx$$

The textbook I am using for Numerical Analysis (Burden, Faires 9-th Edition) suggests the following substitution:

$$t = \frac{1}{x}, dx = {-t}^{-2}dt.$$

which yields the following $$I = \int_{0}^{1} t^{-2} \cdot \frac{\sin\left(\frac{1}{t}\right)}{\left(\frac{1}{t}\right)^4} dt$$

$$I = \int_{0}^{1} t^{2} \cdot \sin\left(\frac{1}{t}\right) dt$$

Now we have a left endpoint singularity, but how to proceed further

• The endpoint singularity is a good indication that using Simpson's rule is not a good idea in this case. – Somos May 6 '18 at 19:09
• But this is the statement of the problem I am trying to solve: Use Composite Simpson's Rule to approximate the integral – S.19LaBG May 6 '18 at 19:10
• I don't know what "Composite Simpson's Rule" is, but I suggest using a slightly different endpoint instead. That is, use $a>0$ but still small and try to find $\int_a^1$. Or else, just pretend that the integrand is zero at zero. – Somos May 6 '18 at 19:13
• There is a singularity at $t=0$, but the function (taking the value $0$ there) is still continuous there. – Robert Israel May 6 '18 at 19:57
• I hope your first integral is $\int_1^\infty \ldots$, not $\int_0^\infty \ldots$. – Robert Israel May 6 '18 at 20:12

## 1 Answer

$$f(t)=t^2 \sin\frac1t$$

To apply Simpson's rule for $n=6$, you need to calculate values $f_i$ for $t=i\Delta t$ where $i=0,1,2,...6$ and $\Delta t=1/6$

$f_0=f(0)$ is not defined but you can replace it with $\lim_{t\to0}f(t)=0.$

$f_1 = f(1/6)= -0.00776154$

$f_2 = f(2/6)= 0.01568$

$f_3 = f(3/6)= 0.227324$

$f_4= f(4/6)= 0.443331$

$f_5= f(5/6)= 0.647249$

$f_6= f(6/6)= 0.841471$

$I\approx\frac{\Delta t}{3}(f_0+4f_1+2f_2+4f_3+2f_4+4f_5+f_6)=0.290375$

Mathematica provides much better approximation (0.28653) but we are still within 2% margin.