i am solving the follwong integral using Cauchy principal value $$ \int_{-\infty }^\infty { y\over \sqrt{(x-a)^2 + (y-b)^2}} dy $$ the integral of this comes out to be $$ \ln \left ( \left | \sqrt{(x-a)^2 + (y-b)^2} + y -b \right | \right ) + \sqrt {(x-a)^2 + (y-b)^2} + C $$
the first of integral becomes 0 and applying limits second part becomes $$ \sqrt {(L-a)^2 - (x-b)^2} - \sqrt {(-L-a)^2 - (x-b)^2}$$
Can i consider the second part equal to 0 because if L is very large number then L+a and L-a will be approximately same and whole expression will become 0 .
Will this assumption is correct or total blunder . thanks for any help
No, you definitely can't do that. While $L+a$ and $L-a$ are "approximately the same" in the relative sense that $\frac{L-a}{L+a}\approx 1$, here we're taking a difference, not a quotient. Indeed, for any $L$, $(L-a)-(L+a)=-2a$, so $$\lim_{L\to\infty}((L-a)-(L+a))=\lim_{L\to\infty}-2a=-2a.$$