# Numerically integrating $\int_a^b \sin\left(100 \pi \sqrt{x^2 + 31364}\right) dx$, with $a=0$ and $b=10^{1000}$

I would like to know which numerical integral method to use to effectively calculate the definite integral of this trigonometric function from intervals a=$$0$$ to b=$$10^{1000}$$.

$$\int_a^b \sin\left(100\pi \sqrt{x^2 + 31364}\right) dx$$

Wolfram alpha calculates the integral from $$a=0$$ to $$b = \infty$$ to be: $$-0.4195181238484021201299757464$$

How did they arrive at this solution?

I've tried the Simpsons method between two consecutive root values so as to generalize the integral behavior across the function roots, but the function is not really periodic.

Rewrite the integrand as

$$\sin\left(100\pi\sqrt{x^2+31364}\right) = \sin\left(100\pi x\sqrt{1+\frac{31364}{x^2}}\right)$$

which for small and large $$x$$ we can approximate as

$$\begin{cases}\sin\left(100\pi\sqrt{x^2+31364}\right) \sim \sin(100\pi\sqrt{31364}) & x \ll \sqrt{31364} \\ \sin\left(100\pi x\sqrt{1+\frac{31364}{x^2}}\right) \sim \sin(100\pi x)+\frac{1578200\pi}{x}\cos(100\pi x) & x \gg \sqrt{31364}\end{cases}$$

We can see that the integral actually does not converge for $$b=\infty$$ as it infinitely oscillates, but you can choose indefinitely higher bounds that are integers and it will conditionally converge. That is the danger of using test bounds that may all end up secretly falling into a special case of the phenomenon you are investigating.

• Thanks this makes things clearer but then how do I compute an integral from both that converges at higher bounds though? Aug 3, 2022 at 8:22
• @SmithMayowa basically your normal methods will work for the first two regions - feel free to simpsons rule or your preferred method of choice. The last region once you reach a suitable region of tolerance where the approximation fits your needs better, integrate directly, it's just the cosine integral function. Aug 3, 2022 at 8:36
• Thanks much clearer now Aug 3, 2022 at 8:46