How to evaluate $\int_0^\frac{\pi}{2}\int_0^\frac{\pi}{2}\frac{(\cos x\cos y)^5}{\left(\cos^2x+\cos^2y\right)^2}\mathrm{~d}x\mathrm{d}y$

Question

Question:

How to evaluate the following integral.

\begin{align} I=\int_0^\frac{\pi}{2}\int_0^\frac{\pi}{2}\frac{(\cos x\cos y)^5}{\left(\cos^2x+\cos^2y\right)^2}\mathrm{~d}x\mathrm{d}y. \end{align}

My attempt:

Firstly, we have \begin{align} I &= \int_0^\frac{\pi}{2}\int_0^\frac{\pi}{2}\frac{(\cos x\cos y)^5}{\left(\cos^2 x + \cos^2 y\right)^2} \, \mathrm{d}x \, \mathrm{d}y \>\>\>\>\> ({\cos^2 y \to y, \, \cos^2 x \to x} )\\ &{=}\ \frac{1}{4} \int_0^1 \int_0^1 \frac{x^2 y^2}{(x+y) \sqrt{1-x} \sqrt{1-y}} \, \mathrm{d}x \, \mathrm{d}y \>\>\>\>\> ({\sqrt{1-y} \to y})\\ &{=}\ \frac{1}{2} \int_0^1 \frac{x^2}{\sqrt{1-x}} \int_0^1 \frac{\left(1 - y^2\right)^2}{\left(1 + x - y^2\right)^2} \, \mathrm{d}y \, \mathrm{d}x, \end{align}

I'm currently stuck at this point. Can someone help me evaluate the last integral?

What should I add to the question to have it reopened?

Answer 1

Per symmetry, rewrite the integral as

$$I=\int_0^\frac{\pi}{2}\int_0^\frac{\pi}{2}\frac{\cos^5x\cos^5y}{\left(\cos^2x+\cos^2y\right)^2}dy\ dx\\ =\int_0^\frac{\pi}{2}\int_0^x\frac{2\cos^5x\cos^5y}{\left(\cos^2x+\cos^2y\right)^2}dy\ dx$$

Then, change variables for the inner integral from $y$ to $t$ according to $\sin y=\sin x\tan\frac t2$ \begin{align} I = &\int_0^\frac{\pi}{2}\int_0^{\frac\pi2} \frac{\sec^2\frac t2(\cos t+\sin^2\frac t2\cos^2x)^2\cos^4x\sin x}{\left(\cos t+\cos^2x\right)^2}dt\overset{z=\cos^2x}{dx}\\ = &\int_0^\frac{\pi}{2}\int_0^{1} \frac{\frac12\sec^2\frac t2(\cos t+z\sin^2\frac t2)^2 z^2}{\left(\cos t+z\right)^2}\overset{ibp}{dz}\ dt\\ = &\int_0^{\frac\pi2} \int_0^1 \bigg(\frac12 z^2 \sec^2\frac t2 -\frac14 +z^3 -\frac{z^4}{\cos t+z}\bigg)dz\ {dt}\\ =&\ \frac13 - \int_0^1\int_0^{\frac\pi2} \frac{z^4}{\cos t+z}dt\overset{z=\sin\theta}{dz}\\ =&\ \frac13 + \int_0^{\frac\pi2} \sin^4\theta\ln(\tan\frac {\theta}2 ) d\theta=\frac{19}{24}-\frac34G \end{align} where $$\int_0^{\frac\pi2} \sin^4\theta\ln\tan\frac {\theta}2 \ d\theta = \frac{11}{24}+\frac38 \int_0^{\frac\pi2} \ln\tan\frac {\theta}2 \ d\theta= \frac{11}{24}-\frac34G $$

---

The procedure above can be generalized to evaluate $$\int_0^\frac{\pi}{2}\int_0^\frac{\pi}{2}\frac{(\cos x\cos y)^{2n+1}}{\left(\cos^2x+\cos^2y\right)^2}dy\ dx$$ In particular \begin{align} &\int_0^\frac{\pi}{2}\int_0^\frac{\pi}{2}\frac{(\cos x\cos y)^{3}}{\left(\cos^2x+\cos^2y\right)^2}dy\ dx=\frac12G-\frac14 \\ &\int_0^\frac{\pi}{2}\int_0^\frac{\pi}{2}\frac{(\cos x\cos y)^{5}}{\left(\cos^2x+\cos^2y\right)^2}dy\ dx =-\frac34G+ \frac{19}{24}\\ &\int_0^\frac{\pi}{2}\int_0^\frac{\pi}{2}\frac{(\cos x\cos y)^{7}}{\left(\cos^2x+\cos^2y\right)^2}dy\ dx =\frac{15}{16}G- \frac{379}{480}\\ &\int_0^\frac{\pi}{2}\int_0^\frac{\pi}{2}\frac{(\cos x\cos y)^{9}}{\left(\cos^2x+\cos^2y\right)^2}dy\ dx =-\frac{35}{32}G+\frac{35389}{33600}\\ &\int_0^\frac{\pi}{2}\int_0^\frac{\pi}{2}\frac{(\cos x\cos y)^{11}}{\left(\cos^2x+\cos^2y\right)^2}dy\ dx =\frac{315}{256}G\ -\cdots \\ \end{align}

Answer 2

Expanded comment: By rotating the region counterclockwise by $\dfrac\pi4$ rad, i.e. substituting $$(x,y)=\left(\frac{u-v}{\sqrt2},\frac{u+v}{\sqrt2}\right)$$ then rescaling the variables by $\sqrt2$, we have

$$\begin{align*} I &= \int_{x=0}^\tfrac\pi2 \int_{y=0}^\tfrac\pi2 \frac{\cos^5x \cos^5y}{\left(\cos^2x + \cos^2y\right)^2} \, dy\,dx \\ &= \frac1{32} \int_{u=0}^\tfrac\pi{\sqrt2} \int_{v=0}^{\tfrac\pi{\sqrt2}-u} \frac{\left(\cos\left(\sqrt2\,u\right)+\cos\left(\sqrt2\,v\right)\right)^5}{\left(1+\cos\left(\sqrt2\,u\right)\cos\left(\sqrt2\,v\right)\right)^2} \, dv\,du \\ &= \frac1{64} \int_{u=0}^\pi \int_{v=0}^{\pi-u} \frac{(\cos u+\cos v)^5}{\left(1+\cos u\cos v\right)^2} \, dv \, du \end{align*}$$

Per Mathematica, the indefinite integral wrt $v$ is elementary; plugging in the limits, observing that the argument of the $\arctan$ component approaches $+\infty$ from below the line $v=\pi-u$, yields

$$I = \frac1{24\,576} \int_0^\pi \frac{f(u)}{\cos^5u} \, du$$

where

$$\begin{align*} f(u)&=288\sin(2u)-66\sin(4u)+112\sin(6u)-3\sin(8u) \\ &\qquad + \pi\left[840\sin u-504\sin(3u)+168\sin(5u)-24\sin(7u)-768\sin^7u\right]\\ &\qquad +(\pi-u)\left[120\cos(6u)-120\cos(4u)+1128\cos(2u)-168\right] \end{align*}$$

This can be further folded up and slightly simplified by substituting $u\to\dfrac\pi2-u$ on $\left[0,\dfrac\pi2\right]$, and $u\to\pi-u$ on $\left[\dfrac\pi2,\pi\right]$:

$$I = \frac1{12\,288} \int_0^\tfrac\pi2 \frac{g(u)}{\sin^5u} \, du$$

where

$$\begin{align*} g(u) &= 3\sin(8u)+112\sin(6u)+66\sin(4u)+288\sin(2u)\\ &\qquad -u\left[120\cos(6u)+120\cos(4u)+1128\cos(2u)+168\right] \end{align*}$$

It turns out that the integral does appear to have an exact form,

$$I = \frac{19}{24} - \frac34 G$$

where $G$ is Catalan's constant, whose presence maybe shouldn't come as a surprise considering one of its integral representations. Compare WA's output for the integral and the alleged closed form.