Please help me evaluate the following integral analytically

Question

$$\int_{0}^{5} \left(\arccos\left(\frac{2}{x+2}\right)\right)^2 \, dx $$ I tried evaluating this integral using traditional techniques such as integration by parts, u-substitution and trigonometric identities, however, I was unsuccessful. If this integral is analytically derivable, please specify the technique or analysis framework that could yield results, including multivariable calculus, complex analysis or tricks (such as Feynman's trick). I would prefer a closed-form solution to this integral if possible, for the positive real numbers.

Thanks in advance!

Answer 1

Starting with

$$I = \int \left[ \arccos{ \left( \frac{2}{x+2} \right)} \right]^2 \, dx,$$

change variables to

$$y = \arccos{ \left( \frac{2}{x+2} \right)}$$

to get

$$I = 2 \int y^2 \frac{\sin{y} \, dy}{\cos^2{y}},$$

which can be integrated by parts:

$$I = 2 \int y^2 \, d\left( \frac{1}{\cos{y}} \right) = \frac{2y^2}{\cos{y}} - 4J$$

where

$$J = \int \frac{y\,dy}{\cos{y}}.$$

To integrate $J$, express it as

$$J = 2 \int \frac{y\,dy}{e^{iy}+e^{-iy}}$$

and change variables to

$$z = e^{iy}$$

to get

$$J = -2 \int \log{z}\,\frac{dz}{z^2 + 1}.$$

This can then also be integrated by parts:

$$J = -2 \int (\log{z}) \, d(\arctan(z)) = -2 (\log{z}) \arctan{z} + 2K$$

where

$$K = \int \frac{\arctan{z}}{z}\,dz.$$

Now use the identity

$$\arctan{z} = -\frac{i}{2}\log\frac{1+iz}{1-iz}$$

to write

$$K = -\frac{i}{2} \left[ \int \frac{\log{(1+iz)}}{z} \, dz - \int \frac{\log{(1-iz)}}{z} \right]$$

and recognize these integrals as dilogarithms:

$$K = -\frac{i}{2} \left[ \text{Li}_2(iz) - \text{Li}_2(-iz) \right].$$

Assembling these results,

$$I = \frac{2y^2}{\cos{y}} + 4y \left[ \log{(1+ie^{iy})} - \log{(1-ie^{iy})} \right] + 4i \left[ \text{Li}_2(ie^{iy}) - \text{Li}_2(-ie^{iy}) \right]$$

where

$$y = \arccos{ \left( \frac{2}{x+2} \right)}.$$

Plotting $I$ between $x = 0$ and $x = 5$ reveals that it is in fact real and well-behaved. The value at $x = 0$ is $-7.32772$ and the value at $x = 5$ is $-1.77109$, giving $5.55664$ for the definite integral, in agreement with numeric evaluation of the original integral you asked about.

Answer 2

If you make the substitution $\cos\theta=\frac{2}{x+2}$ then you get

$$ dx=2\sec\theta\tan\theta\,d\theta $$

and the integral becomes

$$ \int_0^{\arccos(2/7)}2\theta^2\sec\theta\tan\theta\,d\theta$$

I suspect Feynman's trick will work on this version since Wolfram gives an antiderivative. But it is not pretty:

Answer 3

$$I=\int \left(\arccos\left(\frac{2}{x+2}\right)\right)^2 \, dx$$ Let $t=\frac{2}{x+2}$ $$I=2\int \frac{ \big[\cosh ^{-1}(t)\big]^2}{t^2}\,dt$$ $$I=-\frac 2t \Bigg[\cosh ^{-1}(t) \left(\cosh ^{-1}(t)+4 t \cot ^{-1}\left(e^{\cosh ^{-1}(t)}\right)\right)-2 i t \left(\text{Li}_2\left(i e^{-\cosh ^{-1}(t)}\right)-\text{Li}_2\left(-i e^{-\cosh ^{-1}(t)}\right)\right)\Bigg]$$ No way to avoid the polylogarithms.

If you want a series solution, write $$\big[\cosh ^{-1}(t)\big]^2=\sum_{n=0}^\infty a_n\, t^n$$ where $$a_n=\frac {(n-2)^2} {n(n-1)}a_{n-2} \quad\quad \text{with}\quad\quad a_0=-\frac{\pi ^2}{4}\quad \quad a_1=\pi\quad\quad a_2=1$$ Integrating termwise $$I=\sum_{n=0}^\infty \frac {a_n}{n-1}\, t^{n-1}$$

Using the given bounds $$J=\int_0^5 \left(\arccos\left(\frac{2}{x+2}\right)\right)^2 \, dx=-5 a_0-2 a_1 \log \left(\frac{7}{2}\right)-\frac{10 }{7}a_2+$$ $$ 2\sum_{n=3}^\infty \frac {a_n}{n-1}\,\left(1-\left(\frac{2}{7}\right)^{n-1}\right)$$ Computing the partial sum after the expansion to $O(t^{k+1})$

$$\left( \begin{array}{cc} k & \text{partial sum} \\ 10 & 5.56163 \\ 20 & 5.55774 \\ 30 & 5.55707 \\ 40 & 5.55685 \\ 50 & 5.55676 \\ \cdots & \cdots \\ \infty &5.55664 \end{array} \right)$$