Chapter 14.7, Problem 66P

In equation (7.18), let $u(x)$ be an even function and $v(x)$ be an odd function.

(a) If $f(x)=u(x)+iv(x),$ show that these conditions are equivalent to the equation $f^{*}(x)=f(-x).$

(b) Show that

$\pi u(a)=PV{\displaystyle {\int }_0^{\infty }\frac{2xv(x)}{x^2-a^2}}dx,\pi v(a)=-PV{\displaystyle {\int }_0^{\infty }\frac{2au(x)}{x^2-a^2}}dx.$

These are the Kramers-Kronig relations. Hint: To find $u(a),$ write the integral for $u(a)$ in (7.18) as an integral from $-\infty$ to 0 plus an integral from 0 to $\infty .$ Then in the $-\infty$ to 0 integral, replace $x$ by $-x$ to get an integral from 0 to $\infty ,$ and use $v(-x)=-v(x).$ Add the two 0 to $\infty$ integrals and simplify. Similarly find $v(a)$.