Mystery integral in the first issue of Annals of Mathematics

Question

Vol 1. No. 1 of Annals of Mathematics has an "Exercises" section with this unusual integral sent in by a Professor Lewis Green Barbour:

$$\int_{\frac \pi 2}^\pi \sqrt{1-\frac 1 2 \cos^2\vartheta + \sin\vartheta\sin2\vartheta}\,{\rm d}\vartheta$$

The other exercises are answered in later issues but this one seems to have been quietly forgotten.

I've tried modern software and it's stumped on solving the indefinite integral. Numerical methods yield a value:

$$\approx 0.8277600029391442$$

However, this doesn't seem to be a closed-form number in any obvious way. Searches of various number databases turn up nothing.

Does this integral have a closed-form solution? If not, what methods would a reader in 1884 have used to solve it? Given how famous the Annals became, does this unsolved exercise from the very first issue have any lore attached to it?

Answer 1

$$I=\int\sqrt{1-\frac{1}{2} \cos ^2(x)+\sin (x) \sin (2 x)}\,dx$$ Using multiple angle formulae $$I=\frac 12 \int \sqrt{3+2 \cos (x)-\cos (2 x)-2 \cos (3 x)}\,dx$$ $$x=\cos^{-1}(t) \quad \implies \quad I=-\frac 12 \int \sqrt{\frac{4+8 t-2 t^2-8 t^3 }{1-t^2 }}\,dt$$

The integrand is minimum at a point where $$2t^4-4 t^2+t+2=0$$ and we know the exact value of the root $a$ of interest.

$$J=\int_{\frac \pi 2}^\pi\sqrt{1-\frac{1}{2} \cos ^2(x)+\sin (x) \sin (2 x)}\,dx$$ $$J=\frac 12 \int_{-1}^a \sqrt{\frac{4+8 t-2 t^2-8 t^3 }{1-t^2 }}\,dt+\frac 12 \int_{a}^0 \sqrt{\frac{4+8 t-2 t^2-8 t^3 }{1-t^2 }}\,dt$$ For each integral, using series expansion around each bound of $t$. Integrating termwise and converting to decimals

$$\frac 12 \int_{-1}^a \cdots \,dt =0.418355365493$$ $$\frac 12 \int_{a}^0 \cdots \,dt =0.409404637296$$ make a total of $$\frac 12 \int_{-1}^0 \cdots \,dt= \color{red}{0.827760002}79$$ to be compared to $\color{red}{0.82776000294}$

Answer 2

The problem being to compute $$I=\frac 12 \int_{\frac \pi 2}^\pi \sqrt{3+2 \cos (x)-\cos (2 x)-2 \cos (3 x)}\,dx$$ the integrand is minimum at $$x_*=\pi -\tan ^{-1}\left(\frac{\sqrt{5}}{2}\right)$$

Using series expansion around $x=\frac \pi 2$ $$\frac 12 \sqrt{3+2 \cos (x)-\cos (2 x)-2 \cos (3 x)}=\sum_{n=0}^\infty a_n\left(x-\frac{\pi }{2}\right)^n$$ $$I_1=\frac 12 \int_{\frac \pi 2}^{x_*} \sqrt{3+2 \cos (x)-\cos (2 x)-2 \cos (3 x)}\,dx$$ $$\color{blue}{I_1=\sum_{n=0}^\infty \frac{a_n}{(n+1)}\left(\frac{\pi }{2}-\tan ^{-1}\left(\frac{\sqrt{5}}{2}\right)\right)^{n+1}}$$ Around $x=\pi$ $$\frac 12 \sqrt{3+2 \cos (x)-\cos (2 x)-2 \cos (3 x)}=\sum_{n=0}^\infty b_n\left(\pi-x\right)^{2n}$$ $$I_2=\frac 12 \int_{x_*}^{\pi} \sqrt{3+2 \cos (x)-\cos (2 x)-2 \cos (3 x)}\,dx$$ $$\color{blue}{I_2=\sum_{n=0}^\infty \frac{b_n}{(2n+1)}\left(\tan ^{-1}\left(\frac{\sqrt{5}}{2}\right)\right)^{2 n+1}}$$

The required coefficients are explicitly known from those of the square of the integrand (they are simple).

Then, the result.

Edit

Defining $$S_p=\sum_{n=0}^{2p} \frac{a_n}{(n+1)}\left(\frac{\pi }{2}-\tan ^{-1}\left(\frac{\sqrt{5}}{2}\right)\right)^{n+1}+\sum_{n=0}^p \frac{b_n}{(2n+1)}\left(\tan ^{-1}\left(\frac{\sqrt{5}}{2}\right)\right)^{2 n+1}$$ we have a very slow converging oscillating series (as @user2249675 mentioned here) and the first time the string $0.8277600$ appears is for $p=90$.

I used the same procedure for the integral between $0$ and $\pi$, splitting the interval in three regions $$0 \leq x \leq \frac \pi 3 \qquad \frac \pi 3 \leq x \leq \pi -\tan ^{-1}\left(\frac{\sqrt{5}}{2}\right)\qquad \pi -\tan ^{-1}\left(\frac{\sqrt{5}}{2}\right) \leq x \leq \pi$$

Answer 3

By using the Maclaurin series $\sqrt{1+x}=\sum_{n=0}^{\infty}\frac{(-1)^{n+1}{2n\choose n}}{4^n(2n-1)}x^n$, the given integral $\int_0^1\frac{\sqrt{1+(1-t^2)(1-4t)}}{\sqrt2\,\sqrt{1-t^2}}dt$ can be written as $$\sum_{n=0}^{\infty}\frac{(-1)^{n+1}{2n\choose n}}{4^n\,\sqrt2\,(2n-1)}\int_0^1(1-4t)^n(1-t^2)^{n-\frac12}dt$$ and $$\sum_{n=0}^{\infty}\frac{(-1)^{n+1}{2n\choose n}}{4^n\,\sqrt2\,(2n-1)}\int_0^{\pi/2}(1-4\sin\theta)^n(\cos\theta)^{2n}\,d\theta$$ where the integral $\int_0^{\pi/2}(1-4\sin\theta)^n(\cos\theta)^{2n}\,d\theta=\sum_{k=0}^n(-1)^k2^{2k-1}{n\choose k}B(k+1,2n+1)$ can be expressed in terms of beta functions.