Evaluating Cauchy Principal value in Tricomi's formula

Question

How to evaluate the Cauchy principal value of the following integral which appears in Tricomi's formula,

$$P \int _a^b dt\, \frac{\sqrt {(t - a)(b - t)}}{\pi (x - t)} t$$

$x$ takes values in $[a, b]$ and $P$ stands for Cauchy Principal value.

Answer 1

$$I(a,b,x)=P.V. \int _a^b \, \frac{\sqrt {(s - a)(b - s)}}{\pi (x - s)} s\,ds$$ Making change $s=a+(b-a)t$ and introducing $c=\frac{x-a}{b-a}$ $$I(a,b)=P.V. \frac{b-a}{\pi}\int _0^1\frac{\sqrt {t(1 - t)}}{(c - t)} (a+(b-a)t)\,dt$$ $$=P.V. \frac{a(b-a)}{\pi}\int _0^1\frac{\sqrt {t(1 - t)}}{(c - t)}dt+P.V. \frac{(b-a)^2c}{\pi}\int _0^1\frac{\sqrt {t(1 - t)}}{(c - t)}dt$$ $$-\,\frac{(b-a)^2}{\pi}\int _0^1\sqrt {t(1 - t)}dt$$ $$=(b-a)\frac{a+(b-a)c}{\pi}I_1-\frac{(b-a)^2}{\pi}I_2=\frac{(b-a)x}{\pi}I_1(c)-\,\frac{(b-a)^2}{\pi}I_2$$ Where $$I_1(c)=P.V.\int _0^1\frac{\sqrt {t(1 - t)}}{(c - t)}dt; \,c\in(0,1)$$ $$I_2=\int _0^1\sqrt {t(1 - t)}dt=B(3/2;3/2)=\frac{\pi}{8}$$

To evaluate $I_1$ we go in the complex plane and integrate along the following contour, going clockwise:

To make the contour closed we have to add two small semi-circles of radius $r$ around $t=c$ on the upper and lower banks of the cut (in both cases we go clockwise), $I_{c_+}, I_{c_+}$, and two small circles around $x=0$ and $x=1$. It is easy to seet that last two integrals (around $x=0$ and $x=1$) $\to 0$ as $r\to0$, and $$\oint=I_1 +I_{c_+}-(-I_1)+I_{c_-}\Rightarrow \,2I_1=\oint-I_{c_+}-I_{c_-}$$

$$I_{c_+}=\sqrt{c(1-c)}\int_{\pi}^0\frac {ire^{i\phi}}{-re^{i\phi}}d\phi=\pi i\sqrt{c(1-c)}$$ $$I_{c_-}=-\pi i\sqrt{c(1-c)}$$ so the last two integrals cancel each other, and we are left with $I_1=\frac{1}{2}\oint$.

Because there are no singularities outside the contour, we can deform it making a big circle of radius $R\to\infty$ $$\oint=\int_{2\pi}^0\frac{\sqrt{Re^{i\phi}(1+Re^{-\pi i}e^{i\phi})}}{c-Re^{i\phi}}iRe^{i\phi}d\phi=i\sqrt{e^{-\pi i}}\int_0^{2\pi}\frac{(Re^{i\phi})^2\sqrt{1-\frac{1}{Re^{i\phi}}}}{Re^{i\phi}(1-\frac{c}{Re^{i\phi}})}d\phi$$ $$=\int_0^{2\pi}Re^{i\phi}\Bigl(1-\frac{1}{2Re^{i\phi}}+...\Bigr)\Bigl(1+\frac{c}{Re^{i\phi}}+..\Bigr)d\phi=2\pi\Bigl(c-\frac{1}{2}\Bigr)$$ Therefore, $$I_1(c)=\pi\bigl(c-\frac{1}{2}\bigr)$$