What is a method for solving principal value integral of $\frac{1}{\pi}\int_{-B}^{B} \frac{x \sqrt{B^2-x^2}}{x-y}\mathrm{d} x$?

Question

Question:

I am trying to solve a principal value integral involving a square root. Using Mathematica I can get an answer but I would like to know a general approach to obtain them by hand. To be clear I am interested in a clear explanation of the method not just the solution.

The principal value integral is of the form: $$ I_{B}\left(y\right)=\frac{1}{\pi} \int_0^B J\left(x\right)\left(\frac{\mathcal{P}}{x+y}+\frac{\mathcal{P}}{x-y}\right)\mathrm{d}x,$$ where $\mathcal{P}$ denotes a principal value integral and $0\le y\le B$ (we only consider real numbers). I would be interested in a method of solution for $J\left(x\right)$ given by: $$J\left(x\right)=x\sqrt{B^2-x^2}.$$

Using Mathematica I have ascertained that $I\left(y\right)=\frac{B^2}{2}-y^2$ and since $J\left(x\right)$ is odd we have: $$ I_{B}\left(y\right)=\frac{1}{\pi} \int_{-B}^B J\left(x\right)\frac{\mathcal{P}}{x-y}\mathrm{d}x,$$ so I wonder if it could be solved with something akin to contour integration but I am at a loss as to how to proceed further.

Context:

The integral is required to map spectral densities of Hamiltonians used in Open Quantum Systems.

Bonus:

I will accept any answer that provides a step-by-step (analytic) method of solution for $I$ but I would also be interested in methods of solution for a couple of other integrals. I can post these as a separate question if people prefer.

Firstly: $$ L_{\left(C,D\right)}\left(y\right)=\frac{1}{\pi} \int_{C}^{D} J_2\left(x\right)\frac{\mathcal{P}}{x-y}\mathrm{d}x,$$ with $0\le C<y<D$, all positive real numbers and: $$J_2\left(x\right)=\sqrt{\left(D-x\right)\left(x-C\right)}.$$ Unfortunately, I have been unable to compute $L$ with Mathematica, but I have used alternative methods to ascertain $L_{\left(C,D\right)}\left(y\right)=\frac{C+D}{2}-y$.

And also: $$ K_{\left(A,B\right)}\left(y\right)=\frac{1}{\pi} \int_{A}^{B} J_1\left(x\right)\left(\frac{\mathcal{P}}{x+y}+\frac{\mathcal{P}}{x-y}\right)\mathrm{d}x,$$ with $0\le A<y<B$ all positive real numbers and: $$J_1\left(x\right)=\sqrt{\left(B^2-x^2\right)\left(x^2-A^2\right)}.$$ Since: $$ \frac{1}{\pi} \int_{A}^{B} J_1\left(x\right)\left(\frac{1}{x+y}+\frac{1}{x-y}\right)\mathrm{d}x=\frac{1}{\pi} \int_{A}^{B} J_1\left(x\right)\frac{2x}{x^2-y^2}\mathrm{d}x=\frac{1}{\pi} \int_{A^2}^{B^2} J_1\left(\sqrt{w}\right)\frac{1}{w-y^2}\mathrm{d}x=L\left(y^2\right)=\frac{A^2+B^2}{2}-y^2,$$ where we have used the substitution $w=x^2$ and defined $C=A^2$ and $D=B^2$.

It is probably also worth noting that: $$ L_{\left(C,D\right)}\left(y\right)=\frac{1}{\pi} \int_{0}^{D-C} \sqrt{w}\sqrt{D-C-w}\frac{\mathcal{P}}{w-(y-C)}\mathrm{d}w,$$ which can be found using the substitution $w=x-C$.

Answer 1

The evaluation of the Cauchy principal value integral via contour integration is relatively straightforward. To begin, consider the contour integral

$$\oint_C dz \, \frac{z \sqrt{z^2-B^2}}{z-y} $$

where $C$ is the following contour:

There are semicircular detours of radius $\epsilon$ around the branch points at $z=\pm B$ and the pole at $z=y$. Also, the large circle has a radius $R$. The pieces of the contour integral as labeled in the figure are as follows. (Yes, there are a lot of pieces, but as you will see, most will vanish or cancel.)

$$\int_{AB} dz \frac{z \sqrt{z^2-B^2}}{z-y} = \int_{-R}^{-B-\epsilon} dx \frac{x \sqrt{x^2-B^2}}{x-y}$$

$$\int_{BC} dz \frac{z \sqrt{z^2-B^2}}{z-y} = i \epsilon \int_{\pi}^0 d\phi \, e^{i \phi} \, \frac{(-B+\epsilon e^{i \phi})\sqrt{(-B+\epsilon e^{i \phi})^2-B^2}}{-B+\epsilon e^{i \phi}-y} $$

$$\int_{CD} dz \frac{z \sqrt{z^2-B^2}}{z-y} = \int_{-B+\epsilon}^{y-\epsilon} dx \frac{x \sqrt{x^2-B^2}}{x-y} = \int_{-B+\epsilon}^{y-\epsilon} dx \frac{x i \sqrt{B^2-x^2}}{x-y}$$

$$\int_{DE} dz \frac{z \sqrt{z^2-B^2}}{z-y} = i \epsilon \int_{\pi}^0 d\phi \, e^{i \phi} \, \frac{(y+\epsilon e^{i \phi})\sqrt{(y+\epsilon e^{i \phi})^2-B^2}}{\epsilon e^{i \phi}} \\ = i \epsilon \int_{\pi}^0 d\phi \, e^{i \phi} \, \frac{(y+\epsilon e^{i \phi})i \sqrt{B^2-(y+\epsilon e^{i \phi})^2}}{\epsilon e^{i \phi}}$$

$$\int_{EF} dz \frac{z \sqrt{z^2-B^2}}{z-y} = \int_{y+\epsilon}^{B-\epsilon} dx \frac{x \sqrt{x^2-B^2}}{x-y} = \int_{y+\epsilon}^{B-\epsilon} dx \frac{x i \sqrt{B^2-x^2}}{x-y}$$

$$\int_{FG} dz \frac{z \sqrt{z^2-B^2}}{z-y} = i \epsilon \int_{\pi}^{-\pi} d\phi \, e^{i \phi} \, \frac{(B+\epsilon e^{i \phi}) \sqrt{(B+\epsilon e^{i \phi})^2-B^2}}{B+\epsilon e^{i \phi}-y} $$

$$\int_{GH} dz \frac{z \sqrt{z^2-B^2}}{z-y} = \int_{B-\epsilon}^{y+\epsilon} dx \frac{x \sqrt{x^2-B^2}}{x-y} = \int_{B-\epsilon}^{y+\epsilon} dx \frac{x (-i) \sqrt{B^2-x^2}}{x-y}$$

$$\int_{HI} dz \frac{z \sqrt{z^2-B^2}}{z-y} = i \epsilon \int_{0}^{-\pi} d\phi \, e^{i \phi} \, \frac{(y+\epsilon e^{i \phi})\sqrt{(y+\epsilon e^{i \phi})^2-B^2}}{\epsilon e^{i \phi}} \\ = i \epsilon \int_{0}^{-\pi} d\phi \, e^{i \phi} \, \frac{(y+\epsilon e^{i \phi})(-i) \sqrt{B^2-(y+\epsilon e^{i \phi})^2}}{\epsilon e^{i \phi}} $$

$$\int_{IJ} dz \frac{z \sqrt{z^2-B^2}}{z-y} = \int_{y-\epsilon}^{-B+\epsilon} dx \frac{x \sqrt{x^2-B^2}}{x-y} = \int_{y-\epsilon}^{-B+\epsilon} dx \frac{x (-i) \sqrt{B^2-x^2}}{x-y}$$

$$\int_{JK} dz \frac{z \sqrt{z^2-B^2}}{z-y} = i \epsilon \int_{0}^{-\pi} d\phi \, e^{i \phi} \, \frac{(-B+\epsilon e^{i \phi})\sqrt{(-B+\epsilon e^{i \phi})^2-B^2}}{-B+\epsilon e^{i \phi}-y} $$

$$\int_{KL} dz \frac{z \sqrt{z^2-B^2}}{z-y} = \int_{-B-\epsilon}^{-R} dx \frac{x \sqrt{x^2-B^2}}{x-y}$$

$$\int_{LA} dz \frac{z \sqrt{z^2-B^2}}{z-y} = i R \int_{-\pi}^{\pi} d\theta \, e^{i \theta} \frac{R e^{i \theta} \sqrt{R^2 e^{i 2 \theta}-B^2}}{R e^{i \theta}-y} $$

Note that, on the branch above the real axis, $-1=e^{i \pi}$ and on the branch below the real axis, $-1=e^{-i \pi}$. Thus, the sign of $i$ in front of the square root when $|x| \lt B$ is positive above the real axis and negative below the real axis.

First, note that the integrals over $AB$ and $KL$ cancel because the square root does not introduce any phase change there.

Second, as $\epsilon \to 0$, the integrals over $BC$, $FG$, and $JK$ vanish.

Third, as $\epsilon \to 0$, the integrals over $DE$ and $HI$ cancel. In this case, the phase difference over the branch cut due to the square root turns what is normally a constructive interference (i.e., the contributions usually add) into a destructive interference (i.e., they cancel.)

Fourth, as $\epsilon \to 0$, the integrals over $CD$ and $EF$ combine to form

$$i PV \int_{-B}^B dx \frac{x \sqrt{B^2-x^2}}{x-y}$$

and the integrals over $GH$ and $IJ$ combine to form

$$-i PV \int_{B}^{-B} dx \frac{x \sqrt{B^2-x^2}}{x-y}$$

so together, the contribution to the contour integral over these four intervals is

$$i 2 PV \int_{-B}^B dx \frac{x \sqrt{B^2-x^2}}{x-y}$$

Fifth, the final contribution to the contour integral is the integral over $LA$ in the limit as $R \to \infty$. That limit is evaluated as follows:

$$\begin{align} \int_{LA} dz \frac{z \sqrt{z^2-B^2}}{z-y} &= i R \int_{-\pi}^{\pi} d\theta \, e^{i \theta} \frac{R e^{i \theta} \sqrt{R^2 e^{i 2 \theta}-B^2}}{R e^{i \theta}-y} \\ &= i R^2 \int_{-\pi}^{\pi} d\theta \, e^{i 2 \theta} \left (1-\frac{B^2}{R^2 e^{i 2 \theta}} \right )^{1/2} \left (1-\frac{y}{R e^{i \theta}} \right )^{-1} \\ &= i R^2 \int_{-\pi}^{\pi} d\theta \, e^{i 2 \theta} \left [1+y \frac{1}{R e^{i \theta}} -\left (\frac{B^2}{2} - y^2 \right ) \frac1{R^2 e^{i 2 \theta}} + O \left ( \frac1{R^3} \right ) \right ]\end{align}$$

After integration, the first two contributions inside the brackets vanish. Further, as $R \to \infty$, all terms $O(1/R^3)$ will also vanish. This simply leaves the $1/R^2$ term in the integrand, and we may finally write an expression for the contour integral:

$$\oint_C dz \, \frac{z \sqrt{z^2-B^2}}{z-y} = i 2 PV \int_{-B}^B dx \frac{x \sqrt{B^2-x^2}}{x-y} - i 2 \pi \left (\frac{B^2}{2} - y^2 \right )$$

By Cauchy's theorem, the contour integral is equal to zero. Therefore, when $y \in (-B,B)$

$$\frac1{\pi} PV \int_{-B}^B dx \frac{x \sqrt{B^2-x^2}}{x-y} = \frac{B^2}{2} - y^2$$

as asserted by the OP.

ADDENDUM

The case $|y| \gt B$ is a straightforward application of the residue theorem and the result is

$$\frac1{\pi} \int_{-B}^B dx \frac{x \sqrt{B^2-x^2}}{x-y} = \sqrt{y^2-B^2} \left (y - \sqrt{y^2-B^2} \right ) - \frac{B^2}{2}$$

Answer 2

Preliminary definitions and equivalences:

Firstly we define (as per the question): \begin{equation} \begin{split} I_{B}\left(y\right)=&\frac{1}{\pi} \int_{0}^{B} x\sqrt{B^2-x^2}\left(\frac{\mathcal{P}_{x+y}}{x+y}+\frac{\mathcal{P}_{x-y}}{x-y}\right)\mathrm{d}x,\\ K_{\left(A,B\right)}\left(y\right)=&\frac{1}{\pi} \int_{A}^{B} \sqrt{\left(B^2-x^2\right)\left(x^2-A^2\right)}\left(\frac{\mathcal{P}_{x+y}}{x+y}+\frac{\mathcal{P}_{x-y}}{x-y}\right)\mathrm{d}x,\\ L_{\left(C,D\right)}\left(y\right)=&\frac{1}{\pi} \int_{C}^{D} \sqrt{\left(D-x\right)\left(x-C\right)}\frac{\mathcal{P}_{x-y}}{x-y}\mathrm{d}x,\\ M_{E}\left(y\right)=&\frac{1}{\pi} \int_{0}^{E} \sqrt{x}\sqrt{E-x}\frac{\mathcal{P}_{x-y}}{x-y}\mathrm{d}x,\\ N\left(y\right)=&\frac{1}{\pi} \int_{0}^{1} \sqrt{x}\sqrt{1-x}\frac{\mathcal{P}_{x-y}}{x-y}\mathrm{d}x. \end{split} \end{equation} where $\frac{\mathcal{P}_{x}}{x}$ is a principal value distribution centred at $x=0$: \begin{equation} \begin{split} \frac{\mathcal{P}_{x}}{x}=\lim_{\varepsilon\rightarrow 0}\left[\frac{x}{x^2+\varepsilon^2}\right]\underset{\mathcal{D}}{=}\begin{cases} \frac{1}{x}, &\text{ for } x\ne 0\\ 0, &\text{ for } x= 0\\ \end{cases} \end{split} \end{equation} where the limit should be taken after integration, we will also define a distributional equality: $=_{\mathcal{D}}$. If two distributions are distributionally equal this means they will have the same value when integrated over any domain (assuming there are no poles in the integrand other than those that appear explicitly in the distributional equality). If two distributions are distributionally equal this means they differ by a finite amount (in fact they can differ by an infinite amount as long as the infinity diverges slower than $1/x$ does, as $x\rightarrow 0$) at a finite (or countably infinite) number of points within their domain. Distributional equality holds in such a case because (unless a distribution is infinitely valued at such a point) the contribution of a single point in an integration domain to the full integral is infinitesimal (zero). A commonly used distributionally equality is: \begin{equation} \begin{split} x\frac{\mathcal{P}_{x}}{x}\underset{\mathcal{D}}{=}1, \end{split} \end{equation} since the two only differ at a single point $x=0$, where the former is zero and the later one (so they differ by a finite amount (1) at a finite number (1) of points).

In all the integrals above we assume that $y$ is within the integration domain (though it can be at a boundary), e.g. for $I_{B}\left(y\right)$ we have $0\le y\le B$ and for $K_{\left(A,B\right)}\left(y\right)$ we have $A\le y\le B$. Now we have $I_{B}\left(y\right)=K_{\left(0,B\right)}\left(y\right)$ by definition; \begin{equation} \begin{split} K_{\left(A,B\right)}\left(y\right)=&\frac{1}{\pi} \int_{A}^{B} \sqrt{\left(B^2-x^2\right)\left(x^2-A^2\right)}\left(\frac{\mathcal{P}_{x+y}}{x+y}+\frac{\mathcal{P}_{x-y}}{x-y}\right)\mathrm{d}x\\ =&\frac{1}{\pi} \int_{A}^{B} 2x\sqrt{\left(B^2-x^2\right)\left(x^2-A^2\right)}\frac{\mathcal{P}_{x^2-y^2}}{x^2-y^2}\mathrm{d}x\\ =&\frac{1}{\pi} \int_{A^2}^{B^2} \sqrt{\left(B^2-w\right)\left(w-A^2\right)}\frac{\mathcal{P}_{w-y^2}}{w-y^2}\mathrm{d}w=L_{\left(A^2,B^2\right)}\left(y^2\right), \end{split} \end{equation} using the distributional equality [1]: \begin{equation} \begin{split} \frac{\mathcal{P}_{x+y}}{x+y}+\frac{\mathcal{P}_{x-y}}{x-y}\underset{\mathcal{D}}{=}2x\frac{\mathcal{P}_{x^2-y^2}}{x^2-y^2}, \end{split} \end{equation} and the substitution $w=x^2$ (so $2x\mathrm{d} x=\mathrm{d} w$); we also have: \begin{equation} \begin{split} L_{\left(C,D\right)}\left(y\right)=&\frac{1}{\pi} \int_{C}^{D} \sqrt{\left(D-x\right)\left(x-C\right)}\frac{\mathcal{P}_{x-y}}{x-y}\mathrm{d}x\\ =&\frac{1}{\pi} \int_{0}^{D-C} \sqrt{\left(D-C-w\right)w}\frac{\mathcal{P}_{w-\left(y-C\right)}}{w-\left(y-C\right)}\mathrm{d}w=M_{D-C}\left(y-C\right),\\ \end{split} \end{equation} where we use the substitution $w=x-C$; and lastly [2]: \begin{equation} \begin{split} M_{E}\left(y\right)=&\frac{1}{\pi} \int_{0}^{E} \sqrt{x}\sqrt{E-x}\frac{\mathcal{P}_{x-y}}{x-y}\mathrm{d}x=\frac{1}{\pi} \int_{0}^{1} \sqrt{Ew}\sqrt{E-Ew}\frac{\mathcal{P}_{Ew-y}}{Ew-y}E\mathrm{d}w\\ =&E\frac{1}{\pi} \int_{0}^{1} \sqrt{w}\sqrt{1-w}\frac{\mathcal{P}_{w-\frac{y}{E}}}{w-\frac{y}{E}}\mathrm{d}w=E N\left(\frac{y}{E}\right),\\ \end{split} \end{equation} where we have used the substitution $w=\frac{x}{E}$.

So we have: \begin{equation} \begin{split} M_{E}\left(y\right)=&E N\left(\frac{y}{E}\right),\\ L_{\left(C,D\right)}\left(y\right)=&M_{D-C}\left(y-C\right)=\left(D-C\right) N\left(\frac{y-C}{D-C}\right)\\ K_{\left(A,B\right)}\left(y\right)=&L_{\left(A^2,B^2\right)}\left(y^2\right)=M_{B^2-A^2}\left(y^2-A^2\right)=\left(B^2-A^2\right) N\left(\frac{y^2-A^2}{B^2-A^2}\right),\\ I_{B}\left(y\right)=&K_{\left(0,B\right)}\left(y\right)=B^2 N\left(\frac{y^2}{B^2}\right), \end{split} \end{equation} and only need to determine a method to integrate $N\left(y\right)$ to answer all the original questions.

Integrations:

If we consider $N\left(y\right)$ for $0<y<1$ we find: \begin{equation} \begin{split} N\left(y\right)=&\frac{1}{\pi} \int_{0}^{1} \sqrt{x}\sqrt{1-x}\frac{\mathcal{P}_{x-y}}{x-y}\mathrm{d}x=\frac{2}{\pi} \int_{0}^{1} w^2\sqrt{1-w^2}\frac{\mathcal{P}_{w^2-y}}{w^2-y}\mathrm{d}w\\ =&\frac{2}{\pi} \int_{0}^{1} \left(w^2-y+y\right)\sqrt{1-w^2}\frac{\mathcal{P}_{w^2-y}}{w^2-y}\mathrm{d}w\\ =&\frac{2}{\pi} \int_{0}^{1} \sqrt{1-w^2}\mathrm{d}w+\frac{2y}{\pi} \int_{0}^{1} \sqrt{1-w^2}\frac{\mathcal{P}_{w^2-y}}{w^2-y}\mathrm{d}w\\ =&\frac{1}{\pi} \int_{0}^{\frac{\pi}{2}} \left(1+\cos\left(2\theta\right)\right)\mathrm{d}\theta+\frac{2y}{\pi} \int_{0}^{\frac{\pi}{2}} \left(1-\sin^2\theta+y-y\right)\frac{\mathcal{P}_{\sin^2\theta-y}}{\sin^2\theta-y}\mathrm{d}\theta\\ =&\frac{1}{\pi} \left[\theta+\frac{1}{2}\sin\left(2\theta\right)\right]_{0}^{\frac{\pi}{2}}-\frac{2y}{\pi} \int_{0}^{\frac{\pi}{2}} \left(\sin^2\theta-y\right)\frac{\mathcal{P}_{\sin^2\theta-y}}{\sin^2\theta-y}\mathrm{d}\theta\\ &+\frac{2y\left(1-y\right)}{\pi} \int_{0}^{\frac{\pi}{2}} \frac{\mathcal{P}_{\sin^2\theta-y}}{\sin^2\theta-y}\mathrm{d}\theta\\ =&\frac{1}{2}-y, \end{split} \end{equation} where we have used the substitution $w=\sqrt{x}$ (so $\mathrm{d}x=2w\mathrm{d}w$) and then $w= \sin\theta$ (so $\mathrm{d}w=\cos\theta\mathrm{d}\theta$), we have also used the distributional equality: \begin{equation} \begin{split} \left(x-y\right)\frac{\mathcal{P}_{x-y}}{x-y}&\underset{\mathcal{D}}{=}1,\\ \end{split} \end{equation} the trigonometric identity: \begin{equation} \begin{split} 1-\sin^2\theta&=\cos^2\theta=\frac{1}{2}\left(1+\cos\left(2\theta\right)\right), \end{split} \end{equation} and the fact [3]: \begin{equation} \begin{split} \int_{0}^{\frac{\pi}{2}} \frac{\mathcal{P}}{\sin^2\theta-y}\mathrm{d}\theta=0. \end{split} \end{equation}

For $y=1$ we have: \begin{equation} \begin{split} N\left(1\right)=&\frac{1}{\pi} \int_{0}^{1} \sqrt{x}\sqrt{1-x}\frac{\mathcal{P}_{x-1}}{x-1}\mathrm{d}x=-\frac{1}{\pi} \int_{0}^{1} \sqrt{\frac{x}{1-x}}\mathrm{d}x\\ =&-\frac{2}{\pi} \int_{-\frac{\pi}{2}}^{0}\cos^2\theta\mathrm{d}\theta\\ =&-\frac{1}{\pi} \int_{-\frac{\pi}{2}}^{0}\left(1+\cos\left(2\theta\right)\right)\mathrm{d}\theta=-\frac{1}{\pi} \left[\theta+\frac{1}{2}\sin\left(2\theta\right)\right]_{-\frac{\pi}{2}}^{0}=-\frac{1}{2}, \end{split} \end{equation} where we used the distributional equality: \begin{equation} \begin{split} \sqrt{1-x}\frac{\mathcal{P}_{x-1}}{x-1}\underset{\mathcal{D}}{=}-\frac{1}{\sqrt{1-x}}, \end{split} \end{equation} we have also used substitutions $x=\cos^2 \theta$ (so $\mathrm{d}x=\left|-2\cos\theta\sin\theta\mathrm{d}\theta\right|$) (where we have preserved the orientation of our integral so we require the modulus sign). For $y=0$ we have: \begin{equation} \begin{split} N\left(0\right)=&\frac{1}{\pi} \int_{0}^{1} \sqrt{x}\sqrt{1-x}\frac{\mathcal{P}_{x}}{x}\mathrm{d}x=\frac{1}{\pi} \int_{0}^{1} \sqrt{\frac{1-x}{x}}\mathrm{d}x=\frac{2}{\pi} \int_{-\frac{\pi}{2}}^{0}\sin^2\theta\mathrm{d}\theta\\ =&\frac{1}{\pi} \int_{-\frac{\pi}{2}}^{0}\left(1-\cos\left(2\theta\right)\right)\mathrm{d}\theta=\frac{1}{\pi} \left[\theta-\frac{1}{2}\sin\left(2\theta\right)\right]_{-\frac{\pi}{2}}^{0}=\frac{1}{2}, \end{split} \end{equation} where we used the distributional equality: \begin{equation} \begin{split} \sqrt{x}\frac{\mathcal{P}_{x}}{x}\underset{\mathcal{D}}{=}\frac{1}{\sqrt{x}}, \end{split} \end{equation} we have also used substitutions $x=\cos^2 \theta$ (so $\left|\mathrm{d}x\right|=\left|2\cos\theta\sin\theta\mathrm{d}\theta\right|$).

Bringing it all together we find: \begin{equation} \begin{split} N\left(y\right)=&\frac{1}{2}-y, \end{split} \end{equation} for $0\le y\le 1$.

Conclusion:

Since $N\left(y\right)=\frac{1}{2}-y$ for $0\le y\le 1$ (as derived above) and: \begin{equation} \begin{split} M_{E}\left(y\right)=&E N\left(\frac{y}{E}\right),\\ L_{\left(C,D\right)}\left(y\right)=&\left(D-C\right) N\left(\frac{y-C}{D-C}\right)\\ K_{\left(A,B\right)}\left(y\right)=&\left(B^2-A^2\right) N\left(\frac{y^2-A^2}{B^2-A^2}\right),\\ I_{B}\left(y\right)=&B^2 N\left(\frac{y^2}{B^2}\right), \end{split} \end{equation} (as derived in the section before last) we find: \begin{equation} \begin{split} M_{E}\left(y\right)=&E \left(\frac{1}{2}-\frac{y}{E}\right)=\frac{E}{2}-y,\\ L_{\left(C,D\right)}\left(y\right)=&\left(D-C\right) \left(\frac{1}{2}-\frac{y-C}{D-C}\right)=\frac{D+C}{2}-y\\ K_{\left(A,B\right)}\left(y\right)=&\left(B^2-A^2\right) \left(\frac{1}{2}-\frac{y^2-A^2}{B^2-A^2}\right)= \frac{B^2+A^2}{2}-y^2,\\ I_{B}\left(y\right)=&B^2 \left(\frac{1}{2}-\frac{y^2}{B^2}\right)=\frac{B^2}{2}-y^2, \end{split} \end{equation} for $0\le y\le E$, $C\le y\le D$, $A\le y\le B$, and $0\le y\le B$ respectively. This completes the derivation.

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Footnote [1]: We were a bit cavalier with our principal value equating: \begin{equation} \begin{split} \frac{\mathcal{P}_{x+y}}{x+y}+\frac{\mathcal{P}_{x-y}}{x-y}\underset{\mathcal{D}}{=}2x\frac{\mathcal{P}_{x^2-y^2}}{x^2-y^2}, \end{split} \end{equation} so we shall verify our procedure was valid here: \begin{equation} \begin{split} &\frac{\mathcal{P}_{x+y}}{x+y}+\frac{\mathcal{P}_{x-y}}{x-y}=\lim_{\varepsilon_1,\varepsilon_2\rightarrow 0}\left[\frac{x+y}{\left(x+y\right)^2+\varepsilon_1^2}+\frac{x-y}{\left(x-y\right)^2+\varepsilon_2^2}\right]\\ =&\lim_{\varepsilon_1,\varepsilon_2\rightarrow 0}\left[\frac{\left(x^2-y^2\right)\left(x-y\right)+\varepsilon_2^2\left(x+y\right)+\left(x^2-y^2\right)\left(x+y\right)+\varepsilon_1^2\left(x-y\right)}{\left(\left(x+y\right)^2+\varepsilon_1^2\right)\left(\left(x-y\right)^2+\varepsilon_2^2\right)}\right]\\ =&\lim_{\varepsilon_1,\varepsilon_2\rightarrow 0}\left[\frac{2x\left(x^2-y^2\right)+\varepsilon_2^2\left(x+y\right)+\varepsilon_1^2\left(x-y\right)}{\left(x^2-y^2\right)^2+\varepsilon_1^2\left(x-y\right)^2+\varepsilon_2^2\left(x+y\right)^2+\varepsilon_1^2\varepsilon_2^2}\right]\\ =&\begin{cases} \frac{2x}{x^2-y^2}, &\text{ for } x^2\ne y^2\\ \lim_{\varepsilon_1,\varepsilon_2\rightarrow 0}\left[\frac{\left(x+y\right)}{\left(x+y\right)^2+\varepsilon_1^2}\right]=\frac{\mathcal{P}_{x+y}}{x+y}=\frac{1}{2y}, &\text{ for } x= y\ne 0\\ \lim_{\varepsilon_1,\varepsilon_2\rightarrow 0}\left[\frac{\left(x-y\right)}{\left(x-y\right)^2+\varepsilon_2^2}\right]=\frac{\mathcal{P}_{x-y}}{x-y}=-\frac{1}{2y}, &\text{ for } x= -y\ne 0\\ \lim_{\varepsilon_1,\varepsilon_2\rightarrow 0}\left[\frac{2x\left(x^2+\varepsilon_2^2+\varepsilon_1^2\right)}{x^4+\left(\varepsilon_1^2 +\varepsilon_2^2\right)x^2+\varepsilon_1^2\varepsilon_2^2}\right]_{x=0}=0, &\text{ for } x= \pm y= 0\\ \end{cases} \end{split} \end{equation} where we use the definition of principal value distributions above, if we compare this to: \begin{equation} \begin{split} &2x\frac{\mathcal{P}_{x^2-y^2}}{x^2-y^2}=\lim_{\varepsilon\rightarrow 0}\left[2x\frac{x^2-y^2}{\left(x^2-y^2\right)^2+\varepsilon^2}\right]\\ =&\begin{cases} \frac{2x}{x^2-y^2}, &\text{ for } x^2\ne y^2\\ 0, &\text{ for } x= y\ne 0\\ 0, &\text{ for } x= -y\ne 0\\ \lim_{\varepsilon\rightarrow 0}\left[2x\frac{x^2}{x^4+\varepsilon^2}\right]_{x=0}=0, &\text{ for } x= \pm y= 0\\ \end{cases} \end{split} \end{equation} we see that the distributions are equal everywhere except at two points $x=y$ and $x=-y$ (where $y\ne 0$), but here they only differ by a finite amount ($\pm\frac{1}{2y}$) at a finite number (2) of points (unless $y=0$ in which case they do not differ at any points) and thus they can be considered to be distributionally equal.

Footnote [2]: We leave proof of the distributional equality: \begin{equation} \begin{split} E\frac{\mathcal{P}_{Ew-y}}{Ew-y}\underset{\mathcal{D}}{=} \frac{\mathcal{P}_{w-\frac{y}{E}}}{w-\frac{y}{E}}, \end{split} \end{equation} as an exercise for the reader.

Footnote [3]: It would be nice to prove $\int_{0}^{\frac{\pi}{2}} \frac{\mathcal{P}_{\sin^2\theta-y}}{\sin^2\theta-y}\mathrm{d}\theta=0 $ using symmetry but we can show it in a somewhat clumsy way for $0<y<1$. We have, where we use the second part of the principle value definition (that removes the pole $y=\sin\theta$, i.e. $\theta=\phi$, from the integration range by taking explicit limits), the result: \begin{equation} \begin{split} &\int_{0}^{\frac{\pi}{2}} \frac{\mathcal{P}_{\sin^2\theta-y}}{\sin^2\theta-y}\mathrm{d}\theta=\int_{0}^{\frac{\pi}{2}} \frac{\mathcal{P}_{\sin\left(\theta+\phi\right)\sin\left(\theta-\phi\right)}}{\sin\left(\theta+\phi\right)\sin\left(\theta-\phi\right)}\mathrm{d}\theta\\ =&\lim_{\varepsilon\rightarrow 0}\left[\int_{0}^{\phi-\varepsilon} \frac{1}{\sin\left(\theta+\phi\right)\sin\left(\theta-\phi\right)}\mathrm{d}\theta+\int_{\phi+\varepsilon}^{\frac{\pi}{2}} \frac{1}{\sin\left(\theta+\phi\right)\sin\left(\theta-\phi\right)}\mathrm{d}\theta\right]\\ =&\lim_{\varepsilon\rightarrow 0}\left[\left[\frac{1}{\sin \left(2\phi\right)}\ln\left|\frac{\sin\left(\theta-\phi\right)}{\sin\left(\theta+\phi\right)}\right|\right]_{0}^{\phi-\varepsilon}+\left[\frac{1}{\sin \left(2\phi\right)}\ln\left|\frac{\sin\left(\theta-\phi\right)}{\sin\left(\theta+\phi\right)}\right|\right]_{\phi+\varepsilon}^{\frac{\pi}{2}} \right]\\ =&\frac{1}{\sin \left(2\phi\right)}\lim_{\varepsilon\rightarrow 0}\left[\ln\left|\frac{\sin\left(-\varepsilon\right)}{\sin\left(2\phi-\varepsilon\right)}\right|-\ln\left|\frac{\sin\left(-\phi\right)}{\sin\left(\phi\right)}\right|\right.\\ &~~~~~~~~~~~~~~~~~~~~~~~~~\left.+\ln\left|\frac{\sin\left(\frac{\pi}{2}-\phi\right)}{\sin\left(\frac{\pi}{2}+\phi\right)}\right|-\ln\left|\frac{\sin\varepsilon}{\sin\left(2\phi+\varepsilon\right)}\right| \right]\\ =&\frac{1}{\sin \left(2\phi\right)}\lim_{\varepsilon\rightarrow 0}\left[\ln\left|\frac{\sin\varepsilon}{\sin\left(2\phi\right)\cos\varepsilon-\cos\left(2\phi\right)\sin\varepsilon}\right|\right.\\ &~~~~~~~~~~~~~~~~~~~~~~~~~\left.-\ln\left|\frac{\sin\varepsilon}{\sin\left(2\phi\right)\cos\varepsilon+\cos\left(2\phi\right)\sin\varepsilon}\right| \right]\\ =&\frac{1}{\sin \left(2\phi\right)}\lim_{\varepsilon\rightarrow 0}\left[\ln\left|\frac{1+\cot\left(2\phi\right)\varepsilon}{1-\cot\left(2\phi\right)\varepsilon}+\mathcal{O}\left(\varepsilon^2\right)\right| \right]\\ =&\frac{1}{\sin \left(2\phi\right)}\lim_{\varepsilon\rightarrow 0}\left[\ln\left|1+2\cot\left(2\phi\right)\varepsilon+\mathcal{O}\left(\varepsilon^2\right)\right| \right]\\ =&\lim_{\varepsilon\rightarrow 0}\left[\frac{2\cot\left(2\phi\right)}{\sin \left(2\phi\right)}\varepsilon+\mathcal{O}\left(\varepsilon^2\right)\right]=0, \end{split} \end{equation} where we have used the substitution $\sin\phi=y$ (valid since $0<y<1$, it means $0<\phi<\frac{\pi}{2}$) and the trigonometric identities $\sin^2\theta-\sin^2\phi=\sin\left(\theta+\phi\right)\sin\left(\theta-\phi\right)$ and $\sin\left(x+y\right)=\sin x\cos y +\cos x\sin y$ as well as the Taylor expansions $\sin x=x+\mathcal{O}\left(x^3\right)$, $\cos x=1+\mathcal{O}\left(x^2\right)$, and $\ln\left(1+x\right)=x+\mathcal{O}\left(x^2\right)$. We have also used the fact that: \begin{equation} \begin{split} &\frac{\mathrm{d}}{\mathrm{d}\theta}\left[\frac{1}{\sin \left(2\phi\right)}\ln\left|\frac{\sin\left(\theta-\phi\right)}{\sin\left(\theta+\phi\right)}\right|\right]\\ =&\frac{1}{\sin \left(2\phi\right)}\frac{\cos\left(\theta-\phi\right)\sin\left(\theta+\phi\right)-\sin\left(\theta-\phi\right)\cos\left(\theta+\phi\right)}{\sin\left(\theta-\phi\right)\sin\left(\theta+\phi\right)}\\ =&\frac{1}{\sin\left(\theta-\phi\right)\sin\left(\theta+\phi\right)}, \end{split} \end{equation} where we have used the trigonometric identity: $\sin\left(2\theta\right)=\sin\left(\theta+\phi\right)\cos\left(\theta-\phi\right)-\cos\left(\theta+\phi\right)\sin\left(\theta-\phi\right)$. We were able to substitute the antiderivative because neither of the integrals we considered featured poles (the poles were at $\theta =\phi$).