Prove $\int_{0}^{\pi/2} \int_{0}^{\pi/2} \frac{\theta\cot\theta-\varphi\cot\varphi}{\cos\theta-\cos\varphi} \text{d}\varphi\text{d}\theta = \pi\ln2$

Question

Month ago I encounter a nice result numerically checked by Mathematica $$ \int_{0}^{\pi/2} \int_{0}^{\pi/2} \frac{\theta\cot\theta-\varphi\cot\varphi}{\cos\theta-\cos\varphi} \mathrm{d}\varphi\mathrm{d}\theta = \pi\ln2 $$ where the integrated function is actually well-defined even around its singularity $\theta=\varphi=0$.

At my first sight, I thought it might be a trivial conclusion derived from a kind of typical integral like $$ \int_{0}^{\pi} \frac{\cos n\theta}{\cos\theta-\cos\varphi} \mathrm{d}\theta = \pi\frac{\sin n\varphi}{\sin\varphi} $$ just using a proper series expansion. However, when I review it in detail, the result over $(0,\pi/2)$ will be awkwardly complicated. I realize this double integrals may not be done directly, or I may lack some essential insight to solve it.

So I question it here for some further discussion, and thanks in advance for any suggestion.

Answer 1

The following illustrates an approach that looks promising, but that I could not yet complete: wish you and the community here might help to finalize it.

Starting from the integrand $$ f(\theta ,\varphi ) = {{\theta \cot \theta - \varphi \cot \varphi } \over {\cos \theta - \cos \varphi }} = {{\theta {{\cos \theta } \over {\sin \theta }} - \varphi {{\cos \varphi } \over {\sin \varphi }}} \over {\cos \theta - \cos \varphi }} = f(\varphi ,\theta ) $$ it's structure suggests the idea that Green's Theorem might be applied.

Formulated as above, it is difficult to find the needed potential function.
But applying the transformation $$ \left\{ \matrix{ \theta = x + y \hfill \cr \varphi = y - x \hfill \cr} \right.\quad \Leftrightarrow \quad \left\{ \matrix{ x = {{\theta - \varphi } \over 2} \hfill \cr y = {{\theta + \varphi } \over 2} \hfill \cr} \right. $$ we get $$ \eqalign{ & g(x,y) = f(x + y,\;y - x) = \cr & = {{\left( {y - x} \right){{\cos \left( {y - x} \right)} \over {\sin \left( {y - x} \right)}} - \left( {y + x} \right){{\cos \left( {y + x} \right)} \over {\sin \left( {y + x} \right)}}} \over {\cos \left( {y - x} \right) - \cos \left( {y + x} \right)}} = \cr & = {{y\left( {{{\cos \left( {y - x} \right)} \over {\sin \left( {y - x} \right)}} - {{\cos \left( {y + x} \right)} \over {\sin \left( {y + x} \right)}}} \right) - x\left( {{{\cos \left( {y - x} \right)} \over {\sin \left( {y - x} \right)}} + {{\cos \left( {y + x} \right)} \over {\sin \left( {y + x} \right)}}} \right)} \over {2\sin x\sin y}} = \cr & = {{y\sin \left( {2x} \right) - x\sin \left( {2y} \right)} \over {2\sin x\sin y\sin \left( {y - x} \right)\sin \left( {y + x} \right)}} = \cr & = y{{\sin \left( {2x} \right)} \over {2\sin x\sin y\sin \left( {y - x} \right)\sin \left( {y + x} \right)}} - x{{\sin \left( {2y} \right)} \over {2\sin x\sin y\sin \left( {y - x} \right)\sin \left( {y + x} \right)}} = \cr & = y{{\cos x} \over {\sin y\left( {\sin ^2 y\cos ^2 x - \cos ^2 y\sin ^2 x} \right)}} - x{{\cos y} \over {\sin x\left( {\sin ^2 y\cos ^2 x - \cos ^2 y\sin ^2 x} \right)}} = \cr & = y{{\cos x} \over {\sin y\left( {\cos ^2 x - \cos ^2 y} \right)}} - x{{\cos y} \over {\sin x\left( {\cos ^2 x - \cos ^2 y} \right)}} = \cr & = g( - x,y) \cr} $$

Putting: $$ \eqalign{ & M(x,y) = - {y \over {2\sin ^2 y}}\ln \left( {{{ - \sin x + \sin y} \over {\sin x + \sin y}}} \right) \cr & L(x,y) = {x \over {2\sin ^2 x}}\ln \left( {{{ - \sin x + \sin y} \over {\sin x + \sin y}}} \right) \cr} $$ then we have the expression required for applying the Green's theorem $$ g(x,y) = {\partial \over {\partial x}}M(x,y) - {\partial \over {\partial y}}L(x,y) $$

With the help of this sketch

we have $$ \eqalign{ & I = \int_0^{\pi /2} {\int_0^{\pi /2} {f(\theta ,\varphi )\,d\varphi \,d\theta } } = \cr & = \int\!\!\!\int\limits_{square} {f(\varphi ,\theta )\,d\varphi \,d\theta } = 2\int\!\!\!\int\limits_{triang} {f(\varphi ,\theta )\,d\varphi \,d\theta } = \cr & = 4\int\!\!\!\int\limits_{triang} {g(x,y)\,dx\,dy} = 4\int\!\!\!\int\limits_{triang} {\left( {{\partial \over {\partial x}}M - {\partial \over {\partial y}}L} \right)\,dx\,dy} = \cr & = 4\oint\limits_{triang} {Ldx + Mdy} \cr} $$ that is $$ \eqalign{ & {I \over 4} = \oint\limits_{triang} {Ldx + Mdy} = \cr & = \left( \matrix{ \int_{y = 0}^{\pi /4} {\left( {L\left( {y,y} \right) + M(y,y)} \right)dy} + \hfill \cr + \int_{y = \pi /4}^{\pi /2} {\left( { - L\left( {\pi /2 - y,y} \right) + M(\pi /2 - y,y)} \right)dy} + \hfill \cr - \int_{y = 0}^{\pi /2} {M(0,y)dy} \hfill \cr} \right) = \cr & = I_{\,1} + I_2 + I_3 \cr } $$

Now $$ I_3 = - \int_{y = 0}^{\pi /2} {M(0,y)dy} = \int_{y = 0}^{\pi /2} {{y \over {2\sin ^2 y}}\ln \left( {{{\sin y} \over {\sin y}}} \right)dy} = 0 $$

$I_{\,1}$ instead, is not null as it might appear.
If the path is taken to approach $y=x$ while maintaining positive the difference $y-x$ then a numerical computation gives a value of $\approx -1.233688 \ldots$, which together with the computation of $I_{\, 2}$ returns the expected value for $I$.

I am blocked at this point: analytically establish and justify the path for $I_{\, 1}$ and find a close expression for it and for $I_{\, 2}$.

Answer 2

Here is a solution based on @G Cab's remarkable observation.

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Reduction. Let $I$ denote the integral, and apply the substitution $(\theta,\varphi)\mapsto(2\theta,2\varphi)$ to write

$$ I = 4 \int_{0}^{\pi/4} \int_{0}^{\pi/4} \frac{2\theta\cot(2\theta) - 2\varphi \cot(2\varphi)}{\cos(2\theta)-\cos(2\varphi)} \, \mathrm{d}\theta\mathrm{d}\varphi. $$

If we define the $1$-form $\omega$ by

$$ \omega = \left( \log\frac{\cot\theta}{\cot\varphi} \right) \biggl( \frac{\mathrm{d}( (\theta-\varphi)^2 )}{2\sin^2(\theta-\varphi)} - \frac{\mathrm{d}( (\theta+\varphi)^2 )}{2\sin^2(\theta+\varphi)} \biggr), $$

then $\omega$ is smooth in the region $(0, \pi/2)^2$ via continuation along the diagonal $\varphi = \theta$. Moreover, @G Cab's computation for the differential reduces to¹⁾

$$ \mathrm{d}\omega = 4 \left( \frac{2\theta\cot(2\theta) - 2\varphi \cot(2\varphi)}{\cos(2\theta)-\cos(2\varphi)} \right) \, \mathrm{d}\theta \wedge \mathrm{d}\varphi. $$

Now let $\epsilon > 0$ be small and consider the triangular region

$$ \mathcal{T}_{\epsilon} = \bigl\{ (\theta, \varphi) : \epsilon \leq \varphi \leq \theta \leq \tfrac{\pi}{4} \bigr\}. $$

Its boundary $\partial\mathcal{T}_{\epsilon}$ consists of three oriented line segments

Then by the symmetry and the Stokes' theorem,

\begin{align*} I &= 2 \lim_{\epsilon \to 0^+} \int_{\mathcal{T}_{\epsilon}} \mathrm{d}\omega = 2 \lim_{\epsilon \to 0^+} \int_{\partial \mathcal{T}_{\epsilon}} \omega = 2 (I_{\rightarrow} + I_{\uparrow} + I_{\swarrow}), \end{align*}

where $\rightarrow$, $\uparrow$, and $\swarrow$ denote the oriented bottom/height/hypothenus of $\partial\mathcal{T}_{\epsilon}$, respectively, and

$$ I_{\bullet} = \lim_{\epsilon \to 0^+} \int_{\bullet} \omega $$

for each $\bullet \in \{ \rightarrow, \uparrow, \swarrow \}$.

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Integral Computations. We will evaluate each $I_{\bullet}$:

1. It is not hard to check that $\omega = 0$ along $\varphi = \theta$, and so, we get $I_{\swarrow} = 0$.

2. For $I_{\uparrow}$, we note that

\begin{align*} I_{\uparrow} &= \int_{\varphi = 0}^{\varphi = \pi/4} \omega|_{\theta=\pi/4} \\ &= \int_{\varphi = 0}^{\varphi = \pi/4} \left( - \log\cot\varphi \right) \biggl( \frac{\mathrm{d}( (\pi/4-\varphi)^2 )}{2\sin^2(\pi/4-\varphi)} - \frac{\mathrm{d}( (\pi/4+\varphi)^2 )}{2\sin^2(\pi/4+\varphi)} \biggr) \\ &= 2 \int_{0}^{\pi/4} \left( \log\cot\varphi \right) \biggl( \frac{\pi/4-\varphi}{1-\sin(2\varphi)} + \frac{\pi/4+\varphi}{1+\sin(2\varphi)} \biggr) \, \mathrm{d}\varphi \\ &= 2 \int_{-\pi/4}^{\pi/4} \frac{\pi/4+\varphi}{1+\sin(2\varphi)} \log \left| \cot\varphi \right| \, \mathrm{d}\varphi \\ &= \frac{1}{2} \int_{0}^{\pi} \frac{x}{1-\cos x} \log \left| \cot \left(\frac{x}{2} - \frac{\pi}{4}\right) \right| \, \mathrm{d}x, \end{align*}

where the last step is the result of the substitution $\varphi=\tfrac{x}{2}-\tfrac{\pi}{4}$. Then by noting that

$$ \log \left| \cot \left(\frac{\pi}{4} - \frac{x}{2} \right) \right| = \operatorname{Re}\log\left(\frac{1-ie^{ix}}{1+ie^{ix}}\right) = i (\arctan(e^{-ix}) - \arctan(e^{ix})), $$

we get

\begin{align*} I_{\uparrow} &= \frac{1}{2}\int_{0}^{\pi} \frac{ix}{1-\cos x} (\arctan(e^{-ix}) - \arctan(e^{ix})) \, \mathrm{d}x \\ &= \frac{1}{2}\int_{-\pi}^{\pi} \frac{ix}{1-\cos x} \left(\frac{\pi}{4} - \arctan(e^{ix})\right) \, \mathrm{d}x \\ &= i \int_{\mathcal{C}} \frac{(\log z )(\frac{\pi}{4} - \arctan z)}{(z-1)^2} \, \mathrm{d}z, \end{align*}

where the substitution $z = e^{ix}$ is applied to the last step and $\mathcal{C}$ is the path parametrized by $e^{ix}$ for $-\pi < x < \pi$. Now deforming the contour $\mathcal{C}$ so as to enclose the principal branch cut for $\log$,

\begin{align*} I_{\uparrow} &= 2\pi \int_{0}^{1} \frac{\frac{\pi}{4} + \arctan t}{(t+1)^2} \, \mathrm{d}t = 2\pi \int_{0}^{1} \frac{1}{(t^2+1)(t+1)} \, \mathrm{d}t = \frac{\pi^2}{4} + \frac{\pi}{2} \log 2. \end{align*}

3. For the integral along the bottom,

\begin{align*} \int_{\rightarrow} \omega &= \int_{\theta = \epsilon}^{\theta = \pi/4} \omega|_{\varphi=\epsilon} \\ &= \int_{\theta = \epsilon}^{\theta = \pi/4} \log\left( \frac{\cot\theta}{\cot\epsilon} \right) \biggl( \frac{\mathrm{d}( (\theta-\epsilon)^2 )}{2\sin^2(\theta-\epsilon)} - \frac{\mathrm{d}( (\theta+\epsilon)^2 )}{2\sin^2(\theta+\epsilon)} \biggr) \\ &= \int_{\epsilon}^{\pi/4} \log\left( \frac{\cot\theta}{\cot\epsilon} \right) \biggl( \frac{\theta-\epsilon}{\sin^2(\theta-\epsilon)} - \frac{\theta+\epsilon}{2\sin^2(\theta+\epsilon)} \biggr) \, \mathrm{d}\theta. \end{align*}

Substituting $\theta = \epsilon x$,

\begin{align*} \int_{\rightarrow} \omega &= \int_{1}^{\pi/4\epsilon} \log\left( \frac{\cot(\epsilon x)}{\cot\epsilon} \right) \biggl( \frac{\epsilon^2 (x-1)}{\sin^2(\epsilon(x-1))} - \frac{\epsilon^2 (x+1)}{2\sin^2(\epsilon(x+1))} \biggr) \, \mathrm{d}x. \end{align*}

It is not hard (although not entirely trivial) to show that the above integral converges to

\begin{align*} I_{\rightarrow} = \lim_{\epsilon \to 0^+} \int_{\rightarrow} \omega = \int_{1}^{\infty} (-\log x) \biggl( \frac{1}{x-1} - \frac{1}{x+1} \biggr) \, \mathrm{d}x = -\frac{\pi^2}{4}. \end{align*}

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Conclusion. Combining all these observations altogether, we have

$$ \bbox[10px,#ffd]{ I = 2 (I_{\rightarrow} + I_{\uparrow} + I_{\swarrow}) = \pi \log 2 } $$

as desired.

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${}^{1)}$ We include the sketch of computation for $\mathrm{d}\omega$, which is essentially @G Cab's computation in backward direction. Substituting $u=\theta-\phi$ and $v=\theta+\phi$,

$$ \omega = \left( \log \left( \frac{\sin v - \sin u}{\sin v + \sin u} \right) \right)\biggl( \frac{u \, \mathrm{d}u}{\sin^2 u} - \frac{v \, \mathrm{d}v}{\sin^2 v} \biggr) . $$

Taking $\mathrm{d}$,

\begin{align*} \mathrm{d}\omega &= \mathrm{d}\left( \frac{u}{\sin^2 u} \log \left( \frac{\sin v - \sin u}{\sin u + \sin v} \right) \right) \wedge \mathrm{d}u \\ &\qquad - \mathrm{d}\left( \frac{v}{\sin^2 v} \log \left( \frac{\sin v - \sin u}{\sin u + \sin v} \right) \right) \wedge \mathrm{d}v \\ &= \frac{u}{\sin^2 u} \left( \frac{\partial}{\partial v} \log \left( \frac{\sin v - \sin u}{\sin u + \sin v} \right) \right) \mathrm{d}v \wedge \mathrm{d}u\\ &\qquad - \frac{v}{\sin^2 v} \left( \frac{\partial}{\partial u} \log \left( \frac{\sin v - \sin u}{\sin u + \sin v} \right) \right) \mathrm{d}u \wedge \mathrm{d}v \\ &= 4 \left( \frac{v \cos u \csc v - u \cos v \csc u}{\cos(2u) - \cos (2v)} \right) \, \mathrm{d}u \wedge \mathrm{d}v. \end{align*}

By noting that $\mathrm{d}u \wedge \mathrm{d}v = 2 \, \mathrm{d}\theta \wedge \mathrm{d} \varphi$, the question boils down to showing

$$ 2 \left( \frac{v \cos u \csc v - u \cos v \csc u}{\cos(2u) - \cos (2v)} \right) = \frac{(v+u)\cot(v+u) - (v-u)\cot(v-u)}{\cos(v+u) - \cos(v-u)}. $$

This can be verified by expanding the right-hand side.

Answer 3

Preparation Consider the bivariate function $$f(x, y) = \frac{x \cot x - y \cot y}{\cos x - \cos y}$$ Examine its symmetry and find

\begin{align*} f(x, y) &= \frac{x \cot x - y \cot y}{\cos x - \cos y} \\ &= \frac{y \cot y - x \cot x}{\cos y - \cos x} = f(y, x) \end{align*}

Therefore, the double integral's integration region can be divided into two equal parts, i.e.,

\begin{align*} D = D_1 \cup D_2 &: \left\{(x, y) \mid 0 \leq x \leq \frac{\pi}{2}, 0 \leq y \leq \frac{\pi}{2}\right\} \\ \Downarrow \,& \\ D_1 &: \left\{(x, y) \mid y \leq x \leq \frac{\pi}{2}, 0 \leq y \leq \frac{\pi}{2}\right\} \\ D_2 &: \left\{(x, y) \mid 0 \leq x \leq y, 0 \leq y \leq \frac{\pi}{2}\right\} \end{align*}

(Not obvious) Now, let's proceed with the change of variables. The transformation is given by

\begin{align*} \begin{cases} x &= u + v \\ y &= u - v \end{cases} \end{align*}

Then, the bivariate function $f(x, y)$ can be simplified to

\begin{align*} g(u, v) &= \frac{(u + v) \cot(u + v) - (u - v) \cot(u - v)}{\cos(u + v) - \cos(u - v)} \\ &= \frac{u[\cot(u + v) - \cot(u - v)] + v[\cot(u + v) + \cot(u - v)]}{\cos(u + v) - \cos(u - v)} \\ &= \frac{-2u\sin(v)\cos(v) + 2v\sin(u)\cos(u)}{-2\sin(u)\sin(v)\sin(u + v)\sin(u - v)} \\ &= \frac{\frac{v}{\sin v}\cos u - \frac{u}{\sin u}\cos v}{(\cos u - \cos v)(\cos u + \cos v)} \end{align*}

Next, let's find the Jacobian determinant of the transformation:

\begin{align*} \frac{\partial(x, y)}{\partial(u, v)} &= \begin{vmatrix} \frac{\partial x}{\partial u} & \frac{\partial y}{\partial v} \\ \frac{\partial x}{\partial u} & \frac{\partial y}{\partial v} \end{vmatrix} = \begin{vmatrix} 1 & 1 \\ 1 & -1 \end{vmatrix} = -2 \end{align*}

Hence, the absolute value of the Jacobian determinant is $$\left|\frac{\partial(x, y)}{\partial(u, v)}\right| = 2$$

Now, the transformed integration region for $D_2$ becomes

\begin{align*} D_2 &: \left\{(x, y) \mid 0 \leq x \leq y, 0 \leq y \leq \frac{\pi}{2}\right\} \\ \Downarrow \,& \\ D_3 &: \left\{(u, v) \mid v \leq u \leq \frac{\pi}{2} - v, 0 \leq v \leq \frac{\pi}{4}\right\} \end{align*}

Construct bivariate functions $P(u, v)$ and $Q(u, v)$ as follows:

\begin{align*} P &\approx P(u, v) = \frac{u}{2\sin^2u}\ln\left|\frac{\sin(u) - \sin(v)}{\sin(u) + \sin(v)}\right| \\ &= \frac{u}{2\sin^2u}\ln\left|\frac{\cot\left(\frac{u + v}{2}\right)}{\cot\left(\frac{u - v}{2}\right)}\right| \\ Q &\approx Q(u, v) = -\frac{v}{2\sin^2v}\ln\left|\frac{\sin(u) - \sin(v)}{\sin(u) + \sin(v)}\right| \\ &= -\frac{v}{2\sin^2v}\ln\left|\frac{\cot\left(\frac{u + v}{2}\right)}{\cot\left(\frac{u - v}{2}\right)}\right| \end{align*}

Therefore,

\begin{align*} \frac{\partial P}{\partial v} &= -\frac{u}{2\sin^2u}\left[\frac{\cos(v)}{\sin(u) - \sin(v)} + \frac{\cos(v)}{\sin(u) + \sin(v)}\right] \\ &= -\frac{u\cos v}{2\sin^2u}\left[\frac{2\sin(u)}{\sin^2(u) - \sin^2(v)}\right] \\ &= -\frac{u\cos v}{\sin u}\cdot\frac{1}{\sin^2(u) - \sin^2(v)} \\ &= \frac{u\cos v}{\sin u}\cdot\frac{1}{\cos^2(u) - \cos^2(v)} \\ \frac{\partial Q}{\partial u} &= -\frac{v}{2\sin^2v}\left[\frac{\cos(u)}{\sin(u) - \sin(v)} - \frac{\cos(u)}{\sin(u) + \sin(v)}\right] \\ &= -\frac{v\cos u}{2\sin^2v}\left[\frac{2\sin(v)}{\sin^2(u) - \sin^2(v)}\right] \\ &= -\frac{v\cos u}{\sin v}\cdot\frac{1}{\sin^2(u) - \sin^2(v)} \\ &= \frac{v\cos u}{\sin v}\cdot\frac{1}{\cos^2(u) - \cos^2(v)} \end{align*}

Thus,

$$\frac{\partial Q}{\partial u} - \frac{\partial P}{\partial v} = \frac{\frac{v}{\sin v}\cos u - \frac{u}{\sin u}\cos v}{(\cos u - \cos v)(\cos u + \cos v)} = g(u, v)$$

Now, Examine the continuity of the bivariate functions $P, Q$ and their partial derivatives $\frac{\partial P}{\partial v}, \frac{\partial Q}{\partial u}$ in the region $D_3$.

1. When $v = (2k - 1)\pi - u$ or $v = 2k\pi + u$ (where $k \in \mathbb{Z}$), $\frac{\cot\left(\frac{u + v}{2}\right)}{\cot\left(\frac{u - v}{2}\right)} = 0$.

   • For $v = (2k - 1)\pi + u$ (where $k \in \mathbb{Z}$), $\cot\left(\frac{u - v}{2}\right) = 0$.
   • Only the case $k = 0$ with $v = u$ falls into $D_3$.

2. When $u = k\pi$ (where $k \in \mathbb{Z}$), $\sin u = 0$.

   • Only the case $k = 0$ with $(u, v) = (0, 0)$ falls into $D_3$.

3. When $v = k\pi$ (here $k \in \mathbb{Z}$, $\sin v = 0$.

   • Only the case $k = 0$ with $v = 0$ falls into $D_3$.

In summary, define the set of points where $P, Q$ are discontinuous in $D_3$ as region $D_4$:

\begin{align*} D_4 = D_5 \cup D_6 & \\ \Downarrow \quad & \\ D_5 & : \left\{(u, v) \mid v = 0, 0 < u \leq \frac{\pi}{2}\right\} \\ D_6 & : \left\{(u, v) \mid u = v, 0 \leq u \leq \frac{\pi}{4}\right\} \end{align*}

It's worth noting that due to

\begin{align*} \lim_{v \to 0} P(u, v) &= \lim_{v \to 0} \frac{u}{2\sin^2u}\ln\left|\frac{\sin(u) - \sin(v)}{\sin(u) + \sin(v)}\right| = 0 \\ \lim_{v \to 0} Q(u, v) &= \lim_{v \to 0} -\frac{v}{2\sin^2v}\ln\left|\frac{\sin(u) - \sin(v)}{\sin(u) + \sin(v)}\right| = \frac{1}{\sin u} \end{align*}

On region $D_5$, both $P$ and $Q$ are bounded and belong to removable discontinuity regions. So, we can supplement the definition to make them continuous on $D_5$. Let's denote the supplemented bivariate functions as $P_0$ and $Q_0$:

\begin{align*} P_0(u, 0) &= 0 \\ Q_0(u, 0) &= \frac{1}{\sin u} \end{align*}

To ensure the continuity for the future application of Green's theorem for line integrals, we also need to construct an integral region $D_{\varepsilon}$ with a parameter $\varepsilon$ as follows:

$$D_{\varepsilon} : \left\{(u, v) \mid v + \frac{\varepsilon}{2} \leq u \leq \frac{\pi}{2} - v, 0 \leq v \leq \frac{\pi - \varepsilon}{4}\right\}$$

Certainly, it can be verified that $\frac{\partial Q_0}{\partial u}$ and $\frac{\partial P_0}{\partial v}$ are both continuous on the integral region $D_{\varepsilon}$, making the future application of Green's theorem for line integrals reasonable.

(Green's theorem for line integrals) If the bivariate functions $P_0, Q_0$ and their partial derivatives $\frac{\partial Q_0}{\partial u}$, $\frac{\partial P_0}{\partial v}$are continuous on the bounded closed region $D_{\varepsilon}$, then

\begin{align*} \iint_{D_{\epsilon}}\left(\frac{\partial Q_0}{\partial u} - \frac{\partial P_0}{\partial v}\right)dudv &= \oint_{\Gamma} P_0(u, v)dudv + Q_0(u, v)dv \end{align*}

where $\Gamma$ is the boundary closed curve enclosing the region $D_{\varepsilon}$ and counterclockwise direction is taken as the positive direction.

Now, a few indefinite and definite integrals are presented for reference (proofs are left to the reader):

\begin{align*} \int \frac{\ln x}{(1 + x)^2}dx &= \ln(x) - \frac{\ln(x)}{1 + x} - \ln(1 + x) + C \\ \int \frac{x\ln x}{(1 - x^2)^2}dx &= \frac{x^2\ln x}{2(1 - x^2)} + \frac{\ln|1 - x^2|}{4} + C \\ \int_0^1 \frac{\ln x}{1 - x^2}dx &= -\frac{\pi^2}{8} \\ \int_1^{\infty} \frac{\ln x}{x^2 - 1}dx &= \frac{1}{2}\int_1^{\infty}\left(\frac{1}{x - 1} - \frac{1}{x + 1}\right)\ln x dx = \frac{\pi^2}{8} \end{align*}

With all the preparation done, let's move on to the main topic.

[![image 1][1]][1] \begin{align*} \bbox[#EFF,15px,border:2px solid blue]{ \begin{split} I&=\int_0^{\frac{\pi}{2}}\!\!\int_0^{\frac{\pi}{2}}f(x,y)\mathrm{d}x\mathrm{d}y=2\int_0^{\frac{\pi}{2}}\!\!\int_y^{\frac{\pi}{2}}f(x,y)\mathrm{d}x\mathrm{d}y\\ &=2\int_0^{\frac{\pi}{4}}\!\!\int_v^{\frac{\pi}{2}-v}g(u,v)\left|\dfrac{\partial(x,y)}{\partial(u,v)}\right|\mathrm{d}u\mathrm{d}v\quad\,\left( \begin{split} x=u+v\\ y=u-v \end{split}\right)\\ &=4\int_0^{\frac{\pi}{4}}\!\!\int_{v}^{\frac{\pi}{2}-v}g(u,v)\mathrm{d}u\mathrm{d}v\\ &=4\lim\limits_{\varepsilon\to0}\int_0^{\frac{\pi-\varepsilon}{4}}\!\!\int_{v+\frac{\varepsilon}{2}}^{\frac{\pi}{2}-v}g(u,v)\mathrm{d}u\mathrm{d}v\\ &=4\lim\limits_{\varepsilon\to0}\kern{23pt}\bigcirc\kern{-29.5pt}\int\limits_{\overline{AB}+\overline{BC}+\overline{CA}}P_0(u,v)\mathrm{d}u+Q_0(u,v)\mathrm{d}v\\ \end{split} } \end{align*}\ [![image 2][2]][2] \begin{align*} I_1&=\lim\limits_{\varepsilon\to0}\kern{6pt}\bigcirc\kern{-12pt}\int\limits_{\overline{AB}}P_0(u,v)\mathrm{d}u+Q_0(u,v)\mathrm{d}v\\ I_2&=\lim\limits_{\varepsilon\to0}\kern{6pt}\bigcirc\kern{-12pt}\int\limits_{\overline{BC}}P_0(u,v)\mathrm{d}u+Q_0(u,v)\mathrm{d}v\\ I_3&=\lim\limits_{\varepsilon\to0}\kern{6pt}\bigcirc\kern{-12pt}\int\limits_{\overline{CA}}P_0(u,v)\mathrm{d}u+Q_0(u,v)\mathrm{d}v\\ \end{align*}\

Calculate curve integralI_1I_1:Since the line segment\overset{,}{\underset{,}{\overline{AB}}}\overset{,}{\underset{,}{\overline{AB}}}’sparametric equationis\begin{align*} \left\{ \begin{split} u&=t\\ v&=0 \end{split} \right.\quad\left(\tfrac{\varepsilon}{2}\leqslant\,t\leqslant\tfrac{\pi}{2}\right) \end{align*}\\begin{align*} \left\{ \begin{split} u&=t\\ v&=0 \end{split} \right.\quad\left(\tfrac{\varepsilon}{2}\leqslant\ ,t\leqslant\tfrac{\pi}{2}\right) \end{align*}\ but\begin{align*} I_1&=\lim\limits_{\varepsilon\to0}\kern{6pt}\bigcirc\kern{-12pt}\int\limits_{\overline{AB}}P_0(u,v)\mathrm{d}u+Q_0(u,v)\mathrm{d}v\\ &=\lim\limits_{\varepsilon\to0}\left[\int_{\frac{\varepsilon}{2}}^{\frac{\pi}{2}}P_0(t,0)\mathrm{d}t+\int_{\frac{\varepsilon}{2}}^{\frac{\pi}{2}}Q_0(t,0)\mathrm{d}(0)\right]\\ &=0 \end{align*}\\begin{align*} I_1&=\lim\limits_{\varepsilon\to0}\kern{6pt}\bigcirc\kern{-12pt}\int\limits_{\overline{AB}}P_0(u,v)\ mathrm{d}u+Q_0(u,v)\mathrm{d}v\\ &=\lim\limits_{\varepsilon\to0}\left[\int_{\frac{\varepsilon}{2}}^ {\frac{\pi}{2}}P_0(t,0)\mathrm{d}t+\int_{\frac{\varepsilon}{2}}^{\frac{\pi}{2}}Q_0( t,0)\mathrm{d}(0)\right]\\ &=0 \end{align*}\ ② Calculate curve integralI_2I_2:Since the line segment\overset{,}{\underset{,}{\overline{BC}}}The parametric equation of \overset{,}{\underset{,}{\overline{BC}}} is\begin{align*} \left\{ \begin{split} u&=t\\ v&=\dfrac{\pi}{2}-t \end{split} \right.\quad\left(\tfrac{\pi+\varepsilon}{4}\leqslant\,t\leqslant\tfrac{\pi}{2}\right) \end{align*}\\begin{align*} \left\{ \begin{split} u&=t\\ v&=\dfrac{\pi}{2}-t \end{split} \right.\quad\left(\tfrac {\pi+\varepsilon}{4}\leqslant\,t\leqslant\tfrac{\pi}{2}\right) \end{align*}\ but\begin{align*} I_2&=\lim\limits_{\varepsilon\to0}\kern{6pt}\bigcirc\kern{-12pt}\int\limits_{\overline{BC}}P_0(u,v)\mathrm{d}u+Q_0(u,v)\mathrm{d}v\\ &=\lim\limits_{\varepsilon\to0}\left[\int_{\frac{\pi}{2}}^{\frac{\pi+\varepsilon}{4}}P_0\left(t,\tfrac{\pi}{2}-t\right)\mathrm{d}t\right.\\ &\qquad\qquad\qquad\,+\left.\int_{\frac{\pi}{2}}^{\frac{\pi+\varepsilon}{4}}Q_0\left(t,\tfrac{\pi}{2}-t\right)\mathrm{d}\left(\tfrac{\pi}{2}-t\right)\right]\\ &=\int_{\frac{\pi}{2}}^{\frac{\pi}{4}}P_0\left(t,\tfrac{\pi}{2}-t\right)\mathrm{d}t\\ &\qquad\qquad\qquad\,+\int_{\frac{\pi}{2}}^{\frac{\pi}{4}}Q_0\left(t,\tfrac{\pi}{2}-t\right)\mathrm{d}\left(\tfrac{\pi}{2}-t\right)\\ &=-\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}P_0\left(t,\tfrac{\pi}{2}-t\right)\mathrm{d}t+\int_{0}^{\frac{\pi}{4}}Q_0\left(\tfrac{\pi}{2}-\tau,\tau\right)\mathrm{d}\tau\\ &=-\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\dfrac{t}{2\sin^2t}\ln\left|\dfrac{\sin(t)-\cos(t)}{\sin(t)+\cos(t)}\right|\mathrm{d}t\\ &\qquad\qquad+\int_{0}^{\frac{\pi}{4}}\left[-\dfrac{\tau}{2\sin^2\tau}\ln\left|\dfrac{\sin(\tau)-\cos(\tau)}{\sin(\tau)+\cos(\tau)}\right|\,\right]\mathrm{d}\tau\\ &=-\dfrac{1}{2}\int_{0}^{\frac{\pi}{2}}\dfrac{t}{\sin^2t}\ln\left|\dfrac{\sin\,\!t-\cos\,\!t}{\sin\,\!t+\cos\,\!t}\right|\mathrm{d}t\\ &=\int_{0}^{\frac{\pi}{2}}\dfrac{t}{2\sin^2t}\ln\left|\cot\left(t-\dfrac{\pi}{4}\right)\right|\mathrm{d}t\\ \end{align*}\\begin{align*} I_2&=\lim\limits_{\varepsilon\to0}\kern{6pt}\bigcirc\kern{-12pt}\int\limits_{\overline{BC}}P_0(u,v)\ mathrm{d}u+Q_0(u,v)\mathrm{d}v\\ &=\lim\limits_{\varepsilon\to0}\left[\int_{\frac{\pi}{2}}^ {\frac{\pi+\varepsilon}{4}}P_0\left(t,\tfrac{\pi}{2}-t\right)\mathrm{d}t\right.\\ &\qquad\qquad \qquad\,+\left.\int_{\frac{\pi}{2}}^{\frac{\pi+\varepsilon}{4}}Q_0\left(t,\tfrac{\pi}{2} -t\right)\mathrm{d}\left(\tfrac{\pi}{2}-t\right)\right]\\ &=\int_{\frac{\pi}{2}}^{ \frac{\pi}{4}}P_0\left(t,\tfrac{\pi}{2}-t\right)\mathrm{d}t\\ &\qquad\qquad\qquad\,+\ int_{\frac{\pi}{2}}^{\frac{\pi}{4}}Q_0\left(t,\tfrac{\pi}{2}-t\right)\mathrm{d}\ left(\tfrac{\pi}{2}-t\right)\\ &=-\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}P_0\left (t,\tfrac{\pi}{2}-t\right)\mathrm{d}t+\int_{0}^{\frac{\pi}{4}}Q_0\left(\tfrac{\pi} {2}-\tau,\tau\right)\mathrm{d}\tau\\ &=-\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \dfrac{t}{2\sin^2t}\ln\left|\dfrac{\sin(t)-\cos(t)}{\sin(t)+\cos(t)}\right|\mathrm {d}t\\ &\qquad\qquad+\int_{0}^{\frac{\pi}{4}}\left[-\dfrac{\tau}{2\sin^2\tau}\ln \left|\dfrac{\sin(\tau)-\cos(\tau)}{\sin(\tau)+\cos(\tau)}\right|\,\right]\mathrm{d}\tau \\ &=-\dfrac{1}{2}\int_{0}^{\frac{\pi}{2}}\dfrac{t}{\sin^2t}\ln\left|\dfrac{ \sin\,\!t-\cos\,\!t}{\sin\,\!t+\cos\,\!t}\right|\mathrm{d}t\\ &=\int_{0 }^{\frac{\pi}{2}}\dfrac{t}{2\sin^2t}\ln\left|\cot\left(t-\dfrac{\pi}{4}\right)\ right|\mathrm{d}t\\ \end{align*}\ Looking forward to it, looking forward to it, I have encountered the problem of finding a function of one variabledefective integralUse the substitution method (not easy to think of) to make transformations w=\tan(\tfrac{\pi}{4}-t)w=\tan(\tfrac{\pi}{4}-t) , then\begin{alignat*}{4} &&w&=\tan\left(\tfrac{\pi}{4}-t\right)\\ \Rightarrow\quad&& w&=\frac{1-\tan\,\!t}{1+\tan\,\!t}\\ \Rightarrow\quad&& t&=\tfrac{\pi}{4}-\arctan\,\!w\\ \Rightarrow\quad&& \tan\,\!t&=\frac{1-w}{1+w}\\ \Rightarrow\quad&& {\mathrm{d}}t&=-\dfrac{1}{1+w^2}{\mathrm{d}}w\\ \end{alignat*}\\begin{alignat*}{4} &&w&=\tan\left(\tfrac{\pi}{4}-t\right)\\ \Rightarrow\quad&& w&=\frac{ 1-\tan\,\!t}{1+\tan\,\!t}\\ \Rightarrow\quad&& t&=\tfrac{\pi}{4}-\arctan\,\!w \\ \Rightarrow\quad&& \tan\,\!t&=\frac{1-w}{1+w}\\ \Rightarrow\quad&& {\mathrm{d}}t&=- \dfrac{1}{1+w^2}{\mathrm{d}}w\\ \end{alignat*}\ According tomultiple angle formula, universal formula, we have< /span>\begin{align*} \dfrac{1}{2\sin^2t}&=\dfrac{1}{1-\cos(2t)}=\dfrac{1}{1-\frac{1-\tan^2t}{1+\tan^2t}}\\ &=\dfrac{1+\tan^2t}{2\tan^2t}=\dfrac{1+\left(\frac{1-w}{1+w}\right)^2}{2\left(\frac{1-w}{1+w}\right)^2}\\ &=\frac{1+w^{2}}{\left(1-w\right)^{2}} \end{align*}

\begin{align*} \dfrac{1}{2\sin^2t}&=\dfrac{1}{1-\cos(2t)}=\dfrac{1}{1-\frac{1-\ tan^2t}{1+\tan^2t}}\\ &=\dfrac{1+\tan^2t}{2\tan^2t}=\dfrac{1+\left(\frac{1-w }{1+w}\right)^2}{2\left(\frac{1-w}{1+w}\right)^2}\\ &=\frac{1+w^{2} }{\left(1-w\right)^{2}} \end{align*}\ therefore\begin{align*} I_2 &=\int_{0}^{\frac{\pi}{2}}\dfrac{t}{2\sin^2t}\ln\left|\cot\left(t-\dfrac{\pi}{4}\right)\right|\mathrm{d}t\\ &=\int_{1}^{-1}\dfrac{\left(\frac{\pi}{4}-\arctan\,\!w\right)\left(1+w^2\right)\ln\left|\frac{1}{w}\right|}{(1-w)^2}\left(\dfrac{-1}{1+w^2}\right)\mathrm{d}w\\ &=-\int_{-1}^{1}\dfrac{\frac{\pi}{4}-\arctan\,\!w}{(1-w)^2}\ln\left|w\right|\mathrm{d}w\\ &=-\int_{0}^{1}\dfrac{\left(\frac{\pi}{4}-\arctan\,\!w\right)\ln(w)}{(1-w)^2}\mathrm{d}w\\ &\qquad\qquad\qquad-\int_{-1}^0\dfrac{\left(\frac{\pi}{4}-\arctan\,\!w\right)\ln|w|}{(1-w)^2}\mathrm{d}w\\ &=-\int_{0}^{1}\dfrac{\left(\frac{\pi}{4}-\arctan\,\!w\right)\ln(w)}{(1-w)^2}\mathrm{d}w\\ &\qquad\quad-\int_{1}^0\dfrac{\left[\frac{\pi}{4}-\arctan(-\omega)\right]\ln|-\omega|}{(1+\omega)^2}\mathrm{d}(-\omega)\\ &=-\int_{0}^{1}\left[\dfrac{\left(\frac{\pi}{4}-\arctan\,\!w\right)\ln(w)}{(1-w)^2}\right.\\ &\qquad\qquad\qquad\quad\left.+\dfrac{\left(\frac{\pi}{4}+\arctan\,\!w\right)\ln(w)}{(1+w)^2}\right]\mathrm{d}w\\ \end{align*}\\begin{align*} I_2 &=\int_{0}^{\frac{\pi}{2}}\dfrac{t}{2\sin^2t}\ln\left|\cot\left(t -\dfrac{\pi}{4}\right)\right|\mathrm{d}t\\ &=\int_{1}^{-1}\dfrac{\left(\frac{\pi}{ 4}-\arctan\,\!w\right)\left(1+w^2\right)\ln\left|\frac{1}{w}\right|}{(1-w)^2} \left(\dfrac{-1}{1+w^2}\right)\mathrm{d}w\\ &=-\int_{-1}^{1}\dfrac{\frac{\pi} {4}-\arctan\,\!w}{(1-w)^2}\ln\left|w\right|\mathrm{d}w\\ &=-\int_{0}^{1 }\dfrac{\left(\frac{\pi}{4}-\arctan\,\!w\right)\ln(w)}{(1-w)^2}\mathrm{d}w\\ &\qquad\qquad\qquad-\int_{-1}^0\dfrac{\left(\frac{\pi}{4}-\arctan\,\!w\right)\ln|w|}{ (1-w)^2}\mathrm{d}w\\ &=-\int_{0}^{1}\dfrac{\left(\frac{\pi}{4}-\arctan\,\ !w\right)\ln(w)}{(1-w)^2}\mathrm{d}w\\ &\qquad\quad-\int_{1}^0\dfrac{\left[\frac {\pi}{4}-\arctan(-\omega)\right]\ln|-\omega|}{(1+\omega)^2}\mathrm{d}(-\omega)\\ & =-\int_{0}^{1}\left[\dfrac{\left(\frac{\pi}{4}-\arctan\,\!w\right)\ln(w)}{(1- w)^2}\right.\\ &\qquad\qquad\qquad\quad\left.+\dfrac{\left(\frac{\pi}{4}+\arctan\,\!w\right) \ln(w)}{(1+w)^2}\right]\mathrm{d}w\\ \end{align*}\ because \begin{align*} W_1&=\dfrac{(k-\arctan\,\!w)\ln\,\!w}{(1-w)^2}+\dfrac{(k+\arctan\, \!w)\ln\,\!w}{(1+w)^2}\\ &=\dfrac{2\left(kw^2+k-2w\arctan\,\!w\right) \ln\,\!w}{(1-w)^2(1+w)^2}\\ &=\dfrac{2\left(kw^2-2kw+k+2kw-2w\arctan\ ,\!w\right)\ln\,\!w}{(1-w)^2(1+w)^2}\\ &=\dfrac{2k\ln\,\!w}{( 1+w)^2}+\dfrac{4\left(k-\arctan\,\!w\right)w\ln\,\!w}{(1-w)^2(1+w)^ 2}\\ \end{align*}

So, using the alternative indefinite integral formula, we have

\begin{align*} I_2 &=-\int_{0}^{1}\left[\dfrac{\left(\frac{\pi}{4}-\arctan\,\!w\right)\ ln(w)}{(1-w)^2}\right.\\ &\qquad\qquad\qquad\quad\left.+\dfrac{\left(\frac{\pi}{4}+\ arctan\,\!w\right)\ln(w)}{(1+w)^2}\right]\mathrm{d}w\\ &=-\dfrac{\pi}{2}\int_ {0}^{1}\dfrac{\ln\,\!w}{(1+w)^2}\mathrm{d}w\\ &\qquad\qquad\quad-4\int_{0} ^{1}\dfrac{\left(\frac{\pi}{4}-\arctan\,\!w\right)w\ln\,\!w}{(1-w)^2(1+ w)^2}\mathrm{d}w\\ &=\dfrac{\pi}{2}\ln(2)-4\int_{0}^{1}\dfrac{\left(\frac{ \pi}{4}-\arctan\,\!w\right)w\ln\,\!w}{(1-w^2)^2}\mathrm{d}w\\ &=\dfrac {\pi}{2}\ln(2)-4J\\ \end{align*}

Usingintegration by parts method, we have

\begin{align*} J &=\int_{0}^{1}\dfrac{\left(\frac{\pi}{4}-\arctan\,\!w\right)w\ln\, \!w}{(1-w^2)^2}\mathrm{d}w\\ &=\int_{0}^{1}\left(\frac{\pi}{4}-\arctan \,\!w\right)\mathrm{d}\left[\frac{w^{2}\ln\,\!w}{2\left(1-w^2\right)}+\frac{ \ln(1-w^2)}{4}\right]\\ &=\left.\left\{\left(\frac{\pi}{4}-\arctan\,\!w\right )\left[\frac{w^{2}\ln\,\!w}{2\left(1-w^2\right)}+\frac{\ln(1-w^2)}{4 }\right]\right\}\right|_{0}^{1}\\ &\qquad\quad-\int_{0}^{1}\left[\frac{w^{2}\ln \,\!w}{2\left(1-w^2\right)}+\frac{\ln(1-w^2)}{4}\right]\mathrm{d}\left(\frac {\pi}{4}-\arctan\,\!w\right)\\ &=\dfrac{1}{2}\int_{0}^{1}\left[\frac{w^{2 }\ln\,\!w}{\left(1-w^2\right)(1+w^2)}+\frac{\ln(1-w^2)}{2(1+w^ 2)}\right]\mathrm{d}w \end{align*}

And because of

\begin{align*} W_2&=\frac{w^{2}\ln\,\!w}{\left(1-w^2\right)(1+w^2)}+\frac{\ ln(1-w^2)}{2(1+w^2)}\\ &=\dfrac{2w^2\ln(w)+(1-w^2)\ln(1-w^ 2)}{2(1-w^4)}\\ &=\dfrac{(1+w^2)\ln(w)-(1-w^2)\ln(w)+(1- w^2)\ln(1-w^2)}{2(1-w^2)(1+w^2)}\\ &=\dfrac{\ln\,\!w}{2( 1-w^2)}+\dfrac{\ln(\frac{1-w^2}{w})}{2(1+w^2)} \end{align*}

Furthermore, there are

\begin{align*} 4J &=2\int_{0}^{1}\left[\frac{w^{2}\ln\,\!w}{\left(1-w^2\right )(1+w^2)}+\frac{\ln(1-w^2)}{2(1+w^2)}\right]\mathrm{d}w\\ &=\int_{ 0}^{1}\left[\dfrac{\ln\,\!w}{1-w^2}+\dfrac{\ln\left(\frac{1-w^2}{w}\right )}{1+w^2}\right]\mathrm{d}w\\ &=\int_{0}^{1}\dfrac{\ln\,\!w}{1-w^2} \mathrm{d}w+\int_{0}^{1}\dfrac{\ln\left(\frac{1-w^2}{w}\right)}{1+w^2}\mathrm{d }w\\ &=-\dfrac{\pi^2}{8}+\int_{0}^{1}\dfrac{\ln\left(\frac{1-w^2}{w}\ right)}{1+w^2}\mathrm{d}w\\ &=-\dfrac{\pi^2}{8}+J_1\\ \end{align*}

Use substitution again (not easy to think of) for transformation $w=\dfrac{1-\xi}{1+\xi}$ , then

\begin{align*} J_1 &=\int_{0}^{1}\dfrac{\ln\left(\frac{1-w^2}{w}\right)}{1+w^2} \mathrm{d}w\\ &=\int_{1}^{0}\dfrac{\ln\left(\frac{4\xi}{1-\xi^2}\right)}{1+ \left(\frac{1-\xi}{1+\xi}\right)^2}\mathrm{d}\left(\frac{1-\xi}{1+\xi}\right)\\ &=\int_{1}^{0}\dfrac{\ln\left(\frac{1-\xi^2}{4\xi}\right)}{1+\xi^2}\mathrm{ d}\xi\\ &=\int_{0}^{1}\dfrac{2\ln2}{1+\xi^2}\mathrm{d}\xi-\int_{0}^{1} \dfrac{\ln\left(\frac{1-\xi^2}{\xi}\right)}{1+\xi^2}\mathrm{d}\xi\\ &=\int_{0 }^{1}\dfrac{2\ln2}{1+\xi^2}\mathrm{d}\xi-J_1\\ &=\int_{0}^{1}\dfrac{\ln2}{ 1+\xi^2}\mathrm{d}\xi=\dfrac{\pi}{4}\ln2 \end{align*}

Therefore

\begin{align*} I_2 &=\dfrac{\pi}{2}\ln(2)-4J\ &=\dfrac{\pi}{2}\ln(2)-\left(- \dfrac{\pi^2}{8}+J_1\right)\ &=\dfrac{\pi}{2}\ln(2)-\left[-\dfrac{\pi^2}{8 }+\dfrac{\pi}{4}\ln(2)\right]\ &=\dfrac{\pi^2}{8}+\dfrac{\pi}{4}\ln2 \end{ align*}

Calculate curve integral $I_3$:

Becauseline segment The parametric equation of $\overset{\,}{\underset{\,}{\overline{CA}}}$ is

\begin{align*} \left\{ \begin{split} u&=t\\ v&=t-\dfrac{\varepsilon}{2} \end{split} \right.\quad\left(\tfrac {\epsilon}{2}\leqslant\,t\leqslant\tfrac{\pi+\varepsilon}{4}\right) \end{align*}

but

\begin{align*} I_3&=\lim\limits_{\varepsilon\to0}\kern{6pt}\bigcirc\kern{-12pt}\int\limits_{\overline{CA}}P_0(u,v)\ mathrm{d}u+Q_0(u,v)\mathrm{d}v\\ &=\lim\limits_{\varepsilon\to0}\left[\int_{\frac{\pi+\varepsilon}{4} }^{\frac{\varepsilon}{2}}P_0\left(t,t-\tfrac{\varepsilon}{2}\right)\mathrm{d}t\right.\\ &\qquad\qquad \qquad\,+\left.\int_{\frac{\pi+\varepsilon}{4}}^{\frac{\varepsilon}{2}}Q_0\left(t,t-\tfrac{\varepsilon}{ 2}\right)\mathrm{d}\left(t-\tfrac{\varepsilon}{2}\right)\right]\\ &=\lim\limits_{\varepsilon\to0}\left[\int_ {\frac{\pi}{4}}^{\frac{\varepsilon}{4}}P_0\left(\lambda+\tfrac{\varepsilon}{4},\lambda-\tfrac{\varepsilon}{4 }\right)\mathrm{d}\left(\lambda+\tfrac{\varepsilon}{4}\right)\right.\\ &\qquad\qquad\qquad\,+\left.\int_{\frac {\pi}{4}}^{\frac{\varepsilon}{4}}Q_0\left(\lambda+\tfrac{\varepsilon}{4},\lambda-\tfrac{\varepsilon}{4}\right )\mathrm{d}\left(\lambda-\tfrac{\varepsilon}{4}\right)\right]\\ &=-\lim\limits_{\varepsilon\to0}\left\{\int_{ 1}^{\frac{\pi}{\varepsilon}}P_0\left[\tfrac{(\mu+1)\varepsilon}{4},\tfrac{(\mu-1)\varepsilon}{4} \right]\mathrm{d}\left[\tfrac{(\mu+1)\varepsilon}{4}\right]\right.\\ &\qquad\qquad\qquad\,+\left.\int_ {1}^{\frac{\pi}{\varepsilon}}Q_0\left[\tfrac{(\mu+1)\varepsilon}{4},\tfrac{(\mu-1)\varepsilon}{4 }\right]\mathrm{d}\left[\tfrac{(\mu-1)\varepsilon}{4}\right]\right\}\\ &=-\dfrac{1}{2}\lim \limits_{\varepsilon\to0}\left\{\int_{1}^{\frac{\pi}{\varepsilon}}\dfrac{\frac{(\mu+1)\varepsilon}{4}}{ \sin^2\left[\frac{(\mu+1)\varepsilon}{4}\right]}\ln\left|\dfrac{\cot\left(\frac{\mu\varepsilon}{4} \right)}{\cot\left(\frac{\varepsilon}{4}\right)}\right|\mathrm{d}\left(\frac{\mu\varepsilon}{4}\right)\right .\\ &\qquad\qquad\qquad\,+\left.\int_{1}^{\frac{\pi}{\varepsilon}}\dfrac{-\,\frac{(\mu-1) \varepsilon}{4}}{\sin^2\left[\frac{(\mu-1)\varepsilon}{4}\right]}\ln\left|\dfrac{\cot\left(\frac{ \mu\varepsilon}{4}\right)}{\cot\left(\frac{\varepsilon}{4}\right)}\right|\mathrm{d}\left(\frac{\mu\varepsilon} {4}\right)\right\}\\ &=-\dfrac{1}{2}\lim\limits_{\varepsilon\to0}\int_{1}^{\frac{\pi}{\varepsilon }}\left\{\dfrac{(\mu+1)\left(\frac{\varepsilon}{4}\right)^2}{\sin^2\left[\frac{(\mu+1) \varepsilon}{4}\right]}\ln\left|\dfrac{\cot\left(\frac{\mu\varepsilon}{4}\right)}{\cot\left(\frac{\varepsilon} {4}\right)}\right|\right.\\ &\qquad\qquad\qquad\qquad\qquad\,-\left.\dfrac{(\mu-1)\left(\frac{\varepsilon }{4}\right)^2}{\sin^2\left[\frac{(\mu-1)\varepsilon}{4}\right]}\ln\left|\dfrac{\cot\left( \frac{\mu\varepsilon}{4}\right)}{\cot\left(\frac{\varepsilon}{4}\right)}\right|\right\}\mathrm{d}\mu\\ &=-\dfrac{1}{2}\lim\limits_{\delta\to0}\int_{1}^{\frac{\pi}{4\delta}}\left\{\dfrac{(\ mu+1)\,{\delta}^2}{\sin^2\left[(\mu+1){\delta}\right]}\ln\left|\dfrac{\cot\left(\mu \delta\right)}{\cot\left(\delta\right)}\right|\right.\\ &\qquad\qquad\qquad\qquad\qquad\,-\left.\dfrac{(\mu -1)\,{\delta}^2}{\sin^2\left[(\mu-1){\delta}\right]}\ln\left|\dfrac{\cot\left(\mu\delta\right)} {\cot\left(\delta\right)}\right|\right\}\mathrm{d}\mu\\ &=-\dfrac{1}{2}\int_{1}^{+\infty }\left(\dfrac{1}{\mu-1}-\dfrac{1}{\mu+1}\right)\ln\mu\,\mathrm{d}\mu\\ \\ &= -\dfrac{\pi^2}{8} \end{align*}

So, there is \begin{align*} \bbox[#EFF,15px,border:2px solid blue]{ \begin{split} I &= \int_0^{\frac{\pi}{2}} \int_0^{\frac{\pi}{2}} \dfrac{x\cot x - y\cot y}{\cos x - \cos y} \, \mathrm{d}x \, \mathrm{d}y \\ &= 4(I_1 + I_2 + I_3) = 4\left[ 0 + \dfrac{\pi^2}{8} + \dfrac{\pi}{4}\ln(2) + \left(-\dfrac{\pi^2}{8}\right) \right] \\ &= \pi\ln2 \end{split} } \end{align*}

Remaining Problem The author use the sandwich criterion and controlled convergence to prove the theoremThe following limit is equal to an infinite integral, but the process is extremely cumbersome.

\begin{align*} L &= \lim\limits_{\delta \to 0} \int_{1}^{\frac{\pi}{4\delta}} \left\{ \frac{(\mu+1) \, \delta^2}{\sin^2\left[(\mu+1)\delta\right]} \ln\left| \frac{\cot\left(\mu\delta\right)}{\cot\left(\delta\right)} \right| - \frac{(\mu-1) \, \delta^2}{\sin^2\left[(\mu-1)\delta\right]} \ln\left| \frac{\cot\left(\mu\delta\right)}{\cot\left(\delta\right)} \right| \right\} \mathrm{d}\mu \\ &= \int_{1}^{+\infty} \left( \frac{1}{\mu-1} - \frac{1}{\mu+1} \right) \ln\mu \, \mathrm{d}\mu \end{align*}

\begin{align*} L &= \lim\limits_{\delta \to 0} \int_{1}^{\frac{\pi}{4\delta}} \left\{ \frac{(\mu+1) \, \delta^2}{\sin^2\left[(\mu+1)\delta\right]} \ln\left| \frac{\cot\left(\mu\delta\right)}{\cot\left(\delta\right)} \right| - \frac{(\mu-1) \, \delta^2}{\sin^2\left[(\mu-1)\delta\right]} \ln\left| \frac{\cot\left(\mu\delta\right)}{\cot\left(\delta\right)} \right| \right\} \mathrm{d}\mu \\ &= \int_{1}^{+\infty} \left( \frac{1}{\mu-1} - \frac{1}{\mu+1} \right) \ln\mu \, \mathrm{d}\mu \end{align*}

Therefore, I don’t know if there is a faster and more clever method. I hope readers can tell me. Thank you in advance. [1]: https://i.sstatic.net/NCEUX.png [2]: https://i.sstatic.net/ID2UA.png