Evaluate the integration $\int_0^\pi \cos(\frac{\pi}{2}\cos(x)) dx$

Question

I need to evaluate the value of $$\int_0^\pi \cos \left(\frac{\pi}{2}\cos{(x)}\right) \, \mathrm{d}x$$ and show this is less than $\pi/2$.

I know this is equal to $\pi J_0(\pi/2)\approx 1.48$ by computer calculator where $J_0$ is the Bessel function. But I do know little about the insight of Bessel function and I do not know how to calculate (or evaluate) $J_0(\pi/2)$. I would appreciate it a lot if any help were offered.

Answer 1

Hint: $$\int_0^\pi\cos\left(\frac{\pi}{2}\cos x\right)dx=2\int_0^1\frac{\cos\left(\frac{\pi}{2}u\right)}{\sqrt{1-u^2}}du$$ Think about how you can bound $$\frac{\cos\left(\frac{\pi}{2}u\right)}{\sqrt{1-u^2}}$$

Answer 2

If you use the series expansion of the most outside cosine, $$\cos(\frac{\pi}{2}\cos(x))=\sum_{n=0}^\infty (-1)^n \frac 1{(2n)!}\left(\frac{\pi }{2}\right)^{2 n} \cos ^{2 n}(x)$$

Using $$\int_0^\pi \cos ^{2 n}(x)\,dx=\sqrt{\pi }\, \frac{ \Gamma \left(n+\frac{1}{2}\right)}{\Gamma (n+1)}$$ the integral is $$\sqrt{\pi }\,\sum_{n=0}^\infty (-1)^n \frac 1{(2n)!}\left(\frac{\pi }{2}\right)^{2 n} \, \frac{ \Gamma \left(n+\frac{1}{2}\right)}{\Gamma (n+1)}$$

$$a_n=\frac 1{(2n)!}\left(\frac{\pi }{2}\right)^{2 n} \, \frac{ \Gamma \left(n+\frac{1}{2}\right)}{\Gamma (n+1)}\quad \implies \quad \frac{a_{n+1}}{a_n}=\frac{\pi ^2}{16 (n+1)^2}$$

For sure, the result is $\pi\, J_0\left(\frac{\pi }{2}\right)$.

But, if you consider the partial sum $$S_p=\sqrt{\pi }\,\sum_{n=0}^p (-1)^n \frac 1{(2n)!}\left(\frac{\pi }{2}\right)^{2 n} \, \frac{ \Gamma \left(n+\frac{1}{2}\right)}{\Gamma (n+1)}$$ you will notice that it converges very fast to the exact value.

Using $p=10$, the result is $\color{red}{1.4828355519429372}315$

You could even know in advance how many terms have to be added in order to have $$a_{p+1}\leq 10^{-k}$$ since, taking logarithms and using Stirling approximation you need to solve for $p$ the equation $$2 p \left(\log \left(\frac{4 p}{\pi }\right)-1\right)=k\,\log(10)$$ the solution of which being given in terms of Lambert function. $$p\sim \frac{k \log (10)}{2 W\left(\frac{2k \log (10)}{e \pi }\right)}$$

For $k=20$, this gives $p=12.8361$, that is to say $13$ terms to be added.

To give an idea of the approximation, the exact solution is $p=11.2564$