How to evaluate $\int_{0}^{1} \int_{0}^{1} \frac{4y \, da \, dy}{(y^2 + ay + 1)(y^2 - ay + 1)}$?

Question

Question: How to evaluate $$\int_{0}^{\pi/2} \ln \left( \frac{2 + \sin x}{2 - \sin x} \right) \, dx$$

My attempt

The original integral is: $$ J = \int_{0}^{\pi/2} \ln \left( \frac{2 + \sin x}{2 - \sin x} \right) \, dx $$

Using the substitution $y = \tan\left(\frac{x}{2}\right)$, we have: $$ \sin x = \frac{2y}{1 + y^2}, \quad dx = \frac{2 \, dy}{1 + y^2} $$

The integral becomes: \begin{align*} J &= \int_{0}^{1} \ln \left( \frac{2 + \frac{2y}{1 + y^2}}{2 - \frac{2y}{1 + y^2}} \right) \cdot \frac{2 \, dy}{1 + y^2} \\ &= \int_{0}^{1} \ln \left( \frac{2 \left( 1 + \frac{y}{1 + y^2} \right)}{2 \left( 1 - \frac{y}{1 + y^2} \right)} \right) \cdot \frac{2 \, dy}{1 + y^2} \\ &= \int_{0}^{1} \ln \left( \frac{\frac{2y^2 + 2y + 2}{1 + y^2}}{\frac{2y^2 - 2y + 2}{1 + y^2}} \right) \cdot \frac{2 \, dy}{1 + y^2} \\ &= \int_{0}^{1} \ln \left( \frac{y^2 + y + 1}{y^2 - y + 1} \right) \cdot \frac{2 \, dy}{1 + y^2} \end{align*}

Rewriting as a sum of integrals: $$ \int_{0}^{1} \ln \left( \frac{y^2 + ay + 1}{y^2 - ay + 1} \right) \cdot \frac{2 \, dy}{1 + y^2} $$

Splitting into parts: $$ = \int_{0}^{1} \int_{0}^{1} \left( \frac{y}{y^2 + ay + 1} + \frac{y}{y^2 - ay + 1} \right) \cdot \frac{2 \, da \, dy}{1 + y^2} $$

Combining \begin{align*} = \int_{0}^{1} \int_{0}^{1} \frac{4y \, da \, dy}{(y^2 + ay + 1)(y^2 - ay + 1)} \\ \end{align*}

maybe we can simplify the terms and get two different arctan integrals, but that doesn't seem like a good way to evaluate the integral. I tried simplifying the integral, and it resulted in two arctan integrals in my mind, so it might be incorrect.

edit: Thanks to Princess Eev for the link, but I'm actually interested in how to pick up where I left off. This isn't a duplicate question at all.

edit 2: Why was my question closed? Can someone explain? My main question is how to evaluate the last double integral.

Answer 1

$$I=\int_{0}^{1} \int_{0}^{1} \frac{4y }{(y^2 + ay + 1)(y^2 - ay + 1)} \, dy \, da$$ $$J=\int_{0}^{1} \frac{4y }{(y^2 + ay + 1)(y^2 - ay + 1)} \, dy=\frac 2 a \int_0^1 \Bigg(\frac{1}{y^2-a y+1}-\frac{1}{y^2+a y+1} \Bigg)\,dy$$ $$J=\frac{4 \left(2 \tan ^{-1}\left(\frac{a}{\sqrt{4-a^2}}\right)-\tan ^{-1}\left(\frac{a+2}{\sqrt{4-a^2}}\right)+\csc ^{-1}\left(\frac{2}{\sqrt{2-a}}\right)\right)}{a \sqrt{4-a^2}}=\color{red}{\frac{4 \sin ^{-1}\left(\frac{a}{2}\right)}{a \sqrt{4-a^2}}}$$

$$I=\int_0^1 \frac{4 \sin ^{-1}\left(\frac{a}{2}\right)}{a \sqrt{4-a^2}}\,da=2\int_0^{\frac \pi 6} t \csc (t)\,dt=\color{red}{\frac{8 }{3}C-\frac{\pi}{3} \cosh ^{-1}(2)}$$

All the trick is the simplifcation at the third lien.

Answer 2

Ramanujan found that

$$\int_{0}^{\pi/2} \ln \left( \frac{2 + \sin x}{2 - \sin x} \right) \, dx \, = \, \color{blue}{\displaystyle\sum_{n=0}^\infty \displaystyle\frac{1}{\displaystyle\binom{2n}{n}\left(2n+1\right)^2}} \, = \,\boxed{\displaystyle\frac{8C-\pi\ln\left(2+\sqrt{3}\right)}{3}}$$

where $C$ is Catalan's constant.

---

The first few terms are:

$$\int_{0}^{\pi/2} \ln \left( \frac{2 + \sin x}{2 - \sin x} \right) \, dx \, = \, 1 + \frac{1}{18}+\frac{1}{150}+\frac{1}{980} + \cdots$$

The denominators are listed in OEIS sequence A294486.

This series can also be expressed as the value of a hypergeometric function: $$_3F_2\left(1,1, \frac{1}{2}\, ;\,\,\frac{3}{2}, \frac{3}{2}\, ; \,\,\frac{1}{4}\right)$$

Answer 3

Too long for comment:

---

This is more of an aside rather than a full solution to the problem. It turns out your integral has a natural expression as a special value of the inverse tangent integral function, and I thought you might find that interesting:

$$\begin{align} \mathcal{I} &=\int_{0}^{1}\mathrm{d}y\int_{0}^{1}\mathrm{d}x\,\frac{4y}{\left(y^{2}-xy+1\right)\left(y^{2}+xy+1\right)}\\ &=\int_{0}^{1}\mathrm{d}x\int_{0}^{1}\mathrm{d}y\,\frac{4y}{\left(y^{2}-xy+1\right)\left(y^{2}+xy+1\right)}\\ &=\int_{0}^{1}\mathrm{d}x\int_{1}^{0}\mathrm{d}t\,\frac{(-2)}{\left(1+t\right)^{2}}\cdot\frac{4\left(\frac{1-t}{1+t}\right)}{\left[\left(\frac{1-t}{1+t}\right)^{2}-x\left(\frac{1-t}{1+t}\right)+1\right]\left[\left(\frac{1-t}{1+t}\right)^{2}+x\left(\frac{1-t}{1+t}\right)+1\right]};~~~\small{\left[y=\frac{1-t}{1+t}\right]}\\ &=\int_{0}^{1}\mathrm{d}x\int_{0}^{1}\mathrm{d}t\,\frac{8\left(1-t^{2}\right)}{\left[\left(1-t\right)^{2}-x\left(1-t^{2}\right)+\left(1+t\right)^{2}\right]\left[\left(1-t\right)^{2}+x\left(1-t^{2}\right)+\left(1+t\right)^{2}\right]}\\ &=\int_{0}^{1}\mathrm{d}x\int_{0}^{1}\mathrm{d}t\,\frac{8\left(1-t^{2}\right)}{\left[2\left(1+t^{2}\right)-x\left(1-t^{2}\right)\right]\left[2\left(1+t^{2}\right)+x\left(1-t^{2}\right)\right]}\\ &=\int_{0}^{1}\mathrm{d}x\int_{0}^{1}\mathrm{d}t\,\frac{8\left(1-t^{2}\right)}{\left[\left(2-x\right)+\left(2+x\right)t^{2}\right]\left[\left(2+x\right)+\left(2-x\right)t^{2}\right]}\\ &=\int_{0}^{1}\mathrm{d}t\int_{0}^{1}\mathrm{d}x\,\frac{8\left(1-t^{2}\right)}{\left[\left(2-x\right)+\left(2+x\right)t^{2}\right]\left[\left(2+x\right)+\left(2-x\right)t^{2}\right]}\\ &=\int_{0}^{1}\mathrm{d}t\int_{0}^{\frac12}\mathrm{d}u\,\frac{4\left(1-t^{2}\right)}{\left[\left(1-u\right)+\left(1+u\right)t^{2}\right]\left[\left(1+u\right)+\left(1-u\right)t^{2}\right]};~~~\small{\left[x=2u\right]}\\ &=\int_{0}^{1}\mathrm{d}t\int_{\frac13}^{1}\mathrm{d}v\,\frac{2\left(1-t^{2}\right)}{\left(v+t^{2}\right)\left(1+vt^{2}\right)};~~~\small{\left[u=\frac{1-v}{1+v}\right]}.\\ \end{align}$$

Then,

$$\begin{align} \mathcal{I} &=\int_{\frac13}^{1}\mathrm{d}v\int_{0}^{1}\mathrm{d}t\,\frac{2\left(1-t^{2}\right)}{\left(v+t^{2}\right)\left(1+vt^{2}\right)}\\ &=\int_{\frac13}^{1}\mathrm{d}v\int_{0}^{1}\mathrm{d}t\,\frac{2}{\left(1-v\right)}\left[\frac{1}{v+t^{2}}-\frac{1}{1+vt^{2}}\right]\\ &=\int_{\frac13}^{1}\mathrm{d}v\,\frac{2}{\left(1-v\right)}\left[\int_{0}^{1}\mathrm{d}t\,\frac{1}{v+t^{2}}-\int_{0}^{1}\mathrm{d}t\,\frac{1}{1+vt^{2}}\right]\\ &=\int_{\frac13}^{1}\mathrm{d}v\,\frac{2}{\left(1-v\right)}\left[\int_{0}^{\frac{1}{\sqrt{v}}}\mathrm{d}x\,\frac{\sqrt{v}}{v+vx^{2}}-\frac{1}{\sqrt{v}}\int_{0}^{\sqrt{v}}\mathrm{d}x\,\frac{1}{1+x^{2}}\right]\\ &=\int_{\frac13}^{1}\mathrm{d}v\,\frac{2}{\left(1-v\right)\sqrt{v}}\left[\int_{0}^{\frac{1}{\sqrt{v}}}\mathrm{d}x\,\frac{1}{1+x^{2}}-\int_{0}^{\sqrt{v}}\mathrm{d}x\,\frac{1}{1+x^{2}}\right]\\ &=\int_{\frac13}^{1}\mathrm{d}v\,\frac{2}{\left(1-v\right)\sqrt{v}}\left[\arctan{\left(\frac{1}{\sqrt{v}}\right)}-\arctan{\left(\sqrt{v}\right)}\right]\\ &=\int_{\frac{1}{\sqrt{3}}}^{1}\mathrm{d}w\,\frac{4}{1-w^{2}}\left[\arctan{\left(\frac{1}{w}\right)}-\arctan{\left(w\right)}\right];~~~\small{\left[\sqrt{v}=w\right]}\\ &=\int_{\frac{1}{\sqrt{3}}}^{1}\mathrm{d}w\,\frac{4}{1-w^{2}}\left[\frac{\pi}{2}-2\arctan{\left(w\right)}\right]\\ &=\int_{\frac{1}{\sqrt{3}}}^{1}\mathrm{d}w\,\frac{8}{1-w^{2}}\left[\frac{\pi}{4}-\arctan{\left(w\right)}\right]\\ &=\int_{\frac{1}{\sqrt{3}}}^{1}\mathrm{d}w\,\frac{8}{1-w^{2}}\arctan{\left(\frac{1-w}{1+w}\right)}\\ &=4\int_{0}^{\frac{\sqrt{3}-1}{\sqrt{3}+1}}\mathrm{d}x\,\frac{\arctan{\left(x\right)}}{x};~~~\small{\left[w=\frac{1-x}{1+x}\right]}\\ &=4\int_{0}^{\tan{\left(\frac{\pi}{12}\right)}}\mathrm{d}x\,\frac{\arctan{\left(x\right)}}{x}\\ &=4\operatorname{Ti}_{2}{\left(\tan{\left(\frac{\pi}{12}\right)}\right)}.\\ \end{align}$$

---