# Why is the estimate of the order of error in Trapezoid converging to $2.5$?

The integral in question is: $\int_{0}^{\infty} \frac{x^{3/2}}{\cosh{(x)}}dx$

I coded a program to compute $p$, an estimate of the order of the error for the Trapezoid method of numerical integration. From class, I know that the order of the error for this method is $2$ as it goes like $h^2$ where $h$ is the step size (this is for the composite Trapeziod rule).

However when I run my program the the values for the estimate of $p$ seem to converge to $2.5$. I asked my lecturer and she said this is write, but I do not understand why?

I know that the error depends on the 2nd derivative of the integrand,and the function I am dealing with is not bounded on the interval of integration, but I do not get why that results in the convergence of the estimate to 2.5?

$p = \frac{\ln{r}}{\ln{2}}, r = \frac{I_n - I_{2n}}{I_{2n}-I_{4n}}$

Where $I_n$ represents the approximate value of the integral for $n$ sub-intervals.

• Hard to guess what $p$ is without an explicit definition. Morover, how the trapezoid method is carried out, since we have an integral over an unbounded domain? – Jack D'Aurizio Aug 27 '15 at 14:07
• I edited the question, also since the integrand converges very quickly, I just used an upper-limit of 100. – dable Aug 27 '15 at 14:21
• in your question, now $p$ is defined in terms of $I_n$, but $I_n$ is still undefined (however, this time is not difficult to guess what it is). – Jack D'Aurizio Aug 27 '15 at 14:49
• Fixed again! Thanks for pointing it out :) – dable Aug 27 '15 at 15:07
• What are your values for $I_n$? Also, have you tried computing to different precisions (you should be using at least double, if not more)? – marty cohen Aug 27 '15 at 15:34

The error of the trapezoid method is given in terms of the second derivative of a function. Our function is smooth and fast-decaying outside any right neighbourhood of the origin, but in a right neighbourhood of the origin the second derivative of $f$ is unbounded and behaves like $\frac{1}{x^{1/2}}$.

That is the reason for which $p\to 2+\frac{1}{2}$. You may also notice that you may "smoothen" the function by computing the equivalent integral: $$I = \int_{0}^{+\infty}\frac{x^4}{\cosh(x^2)}\,dx.$$ Try to see what happens by applying the trapezoid method to that function, instead of the original one. That gives a function with a bounded second derivative.

Some random babblings:

Let $f(x) =\frac{x^{3/2}}{\cosh(x)}$.

For large $x$, $f(x)$ is essentially zero, so you might be losing many significant digits in subtraction, especially if $I_{2n}$ is quite close to $I_n$.

Also, at the origin, $f(x) \approx x^{3/2}$, so there is no problem there.

$f'(x) =\frac{(3/2)x^{1/2}\cosh(x)-x^{3/2}\sinh(h)}{\cosh^2(x)}$, so $f'(x) = 0$ when $0 =(3/2)x^{1/2}\cosh(x)-x^{3/2}\sinh(x) =x^{1/2}\cosh(x)\left((3/2)-x\tanh(x)\right)$ or, according to Wolfy, $x\approx 1.62182$.

• What the location of the stationary points has to do with the order of convergence of the trapezoid method? – Jack D'Aurizio Aug 27 '15 at 15:46
• I don't know. As I said, those are random babblings. – marty cohen Aug 28 '15 at 3:29