Question

In equation (7.18), let u (x) be an even function and $\mathit{\upsilon }\left(x\right)$be an odd function.

1. If $\mathit{f}\left(x\right)\mathbf{=}\mathit{u}\left(x\right)\mathbf{+}\mathit{i}\mathit{\upsilon }\left(x\right)$, show that these conditions are equivalent to the equation${\mathit{f}}^{\mathbf{*}}\left(x\right)\mathbf{=}\mathit{f}\left(-x\right)$ .
2. Show that

$\mathit{\pi }\mathit{u}\left(a\right)\mathbf{=}\mathit{P}\mathit{V}\underset{\mathbf{0}}{\overset{\mathbf{\infty }}{\mathbf{\int }}}\frac{\mathbf{2}\mathbf{x}\mathbf{\upsilon }\left(x\right)}{{\mathbf{x}}^{\mathbf{2}}\mathbf{-}{\mathbf{a}}^{\mathbf{2}}}\mathbf{ }\mathit{d}\mathit{x}\mathbf{,}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathit{\pi }\mathit{\upsilon }\left(a\right)\mathbf{=}\mathbf{-}\mathit{P}\mathit{V}\underset{\mathbf{0}}{\overset{\mathbf{\infty }}{\mathbf{\int }}}\frac{\mathbf{2}\mathbf{a}\mathbf{u}\left(x\right)}{{\mathbf{x}}^{\mathbf{2}}\mathbf{-}{\mathbf{a}}^{\mathbf{2}}}\mathbf{ }\mathit{d}\mathit{x}$

These are Kramers-Kroning relations. Hint: To find u(a), write the integral for u(a) in (7.18) as an integral from $\mathbf{-}\mathbf{\infty }$to 0 plus an integral from 0 to $\mathbf{\infty }$. Then in the to integral $\mathbf{-}\mathbf{\infty }$to 0, replace x by -x to get an integral from 0 to$\mathbf{\infty }$ , and userole="math" localid="1664350095623" $\mathit{\upsilon }\left(-x\right)\mathbf{=}\mathbf{-}\mathit{\upsilon }\left(x\right)$ . Add the two to integrals and simplify. Similarly findrole="math" localid="1664350005594" $\mathit{\upsilon }\mathbf{\left(}\mathbf{a}\mathbf{\right)}$ .

Short answer

(a)It is proved that, if $f\left(x\right)=u\left(x\right)+i\upsilon \left(x\right)$, then the conditions are equivalent to $f^{*}\left(x\right)=f\left(-x\right)$.

(b) For u(x) even, we have,

$PV\underset{-\infty }{\overset{\infty }{\int }}\frac{u\left(x\right)}{x-a} dx= PV\underset{0}{\overset{\infty }{\int }}\frac{2 au\left(x\right)}{x^2-a^2} dx=-\pi \upsilon \left(a\right)$

for $\upsilon \left(x\right)$odd, we have,

$PV\underset{-\infty }{\overset{\infty }{\int }}\frac{\upsilon \left(x\right)}{x-a} dx= PV\underset{0}{\overset{\infty }{\int }}\frac{2 x\upsilon \left(x\right)}{x^2-a^2} dx=\pi u\left(a\right)$

Step-by-step solution

(a) To show that f(x)=u(x)+iυ(x) is equivalent to the equation f*(x)=f(-x)

Let u(x) be an even function and $\upsilon \left(x\right)$be an odd function.

We have given that $f\left(x\right)=u\left(x\right)+i\upsilon \left(x\right) \cdots \left(1\right)$.

Therefore, .$f^{*}\left(x\right)=u\left(x\right)-i\upsilon \left(x\right) \cdots \left(2\right)$

Since u(x) is an even function, we have, .$u\left(-x\right)=u\left(x\right)$

Also, since $\upsilon \left(x\right)$is an odd function, we have, .$\upsilon \left(-x\right)=-\upsilon \left(x\right)$

Thus, replacing x by -x in (1) and using above definitions of an even and an odd function, we get,

$\begin{array}{rcl}f\left(-x\right)& =& u\left(-x\right)+i\upsilon \left(-x\right)\\ & =& u\left(x\right)-i\upsilon \left(x\right)\\ & =& f^{*}\left(x\right) \cdots \left(\because by \left(2\right)\right)\end{array}$

Therefore$f\left(-x\right)=f^{*}\left(x\right)$

Hence, .$f^{*}\left(x\right)=f\left(-x\right)$

Thus, if $f\left(x\right)=u\left(x\right)+i\upsilon \left(x\right)$, then the conditions are equivalent to .$f^{*}\left(x\right)=f\left(-x\right)$

(b)To show that πu(a)=PV∫0∞2xυ(x)x2-a2 dx,      πυ(a)=-PV∫0∞2au(x)x2-a2 dx

Let u(x) be an even function and $\upsilon \left(x\right)$be an odd function.

We have $PV\underset{-\infty }{\overset{\infty }{\int }}\frac{u\left(x\right)}{x-a} dx=-\pi \upsilon \left(a\right), u\left(-x\right)=u\left(x\right) \cdots \left(1\right)$

Therefore, .$PV\underset{-\infty }{\overset{\infty }{\int }}\frac{u\left(x\right)}{x-a} dx= PV\underset{-\infty }{\overset{0}{\int }}\frac{u\left(x\right)}{x-a} dx+PV\underset{0}{\overset{\infty }{\int }}\frac{u\left(x\right)}{x-a} dx \cdots \left(2\right)$

Now, replacing x = -x gives dx = -dx, in the first integral of (2), we get,

$\begin{array}{rcl}PV\underset{-\infty }{\overset{\infty }{\int }}\frac{u\left(x\right)}{x-a} dx& =& -PV\underset{0}{\overset{\infty }{\int }}\frac{u\left(-x\right)}{-x-a} \left(-dx\right)+PV\underset{0}{\overset{\infty }{\int }}\frac{u\left(x\right)}{x-a} dx \\ PV\underset{-\infty }{\overset{\infty }{\int }}\frac{u\left(x\right)}{x-a} dx& =& -PV\underset{0}{\overset{\infty }{\int }}\frac{u\left(x\right)}{\left(x+a\right)} dx+PV\underset{0}{\overset{\infty }{\int }}\frac{u\left(x\right)}{x-a} dx\\ PV\underset{-\infty }{\overset{\infty }{\int }}\frac{u\left(x\right)}{x-a} dx& =& PV\underset{0}{\overset{\infty }{\int }}\frac{-u\left(x\right)\left(x-a\right)+u\left(x\right)\left(x+a\right)}{x^2-a^2} dx\\ PV\underset{-\infty }{\overset{\infty }{\int }}\frac{u\left(x\right)}{x-a} dx& =& PV\underset{0}{\overset{\infty }{\int }}\frac{2 au\left(x\right)}{x^2-a^2} dx\end{array}$

Hence, for u(x) even, we have,

$PV\underset{-\infty }{\overset{\infty }{\int }}\frac{u\left(x\right)}{x-a} dx= PV\underset{0}{\overset{\infty }{\int }}\frac{2 au\left(x\right)}{x^2-a^2} dx=-\pi \upsilon \left(a\right)$

.

Next, we have $PV\underset{-\infty }{\overset{\infty }{\int }}\frac{\upsilon \left(x\right)}{x-a} dx=\pi u\left(a\right), \upsilon \left(-x\right)=-\upsilon \left(x\right) \cdots \left(3\right)$

Therefore,$PV\underset{-\infty }{\overset{\infty }{\int }}\frac{\upsilon \left(x\right)}{x-a} dx= PV\underset{-\infty }{\overset{0}{\int }}\frac{\upsilon \left(x\right)}{x-a} dx+PV\underset{0}{\overset{\infty }{\int }}\frac{\upsilon \left(x\right)}{x-a} dx \cdots \left(4\right)$

Now, replacing x = -x gives dx = -dx, in the first integral of (4) , we get,

$\begin{array}{rcl}PV\underset{-\infty }{\overset{\infty }{\int }}\frac{\upsilon \left(x\right)}{x-a} dx& =& -PV\underset{0}{\overset{\infty }{\int }}\frac{\upsilon \left(-x\right)}{-x-a} \left(-dx\right)+PV\underset{0}{\overset{\infty }{\int }}\frac{\upsilon \left(x\right)}{x-a} dx \\ PV\underset{-\infty }{\overset{\infty }{\int }}\frac{\upsilon \left(x\right)}{x-a} dx& =& -PV\underset{0}{\overset{\infty }{\int }}\frac{-\upsilon \left(x\right)}{\left(x+a\right)} dx+PV\underset{0}{\overset{\infty }{\int }}\frac{\upsilon \left(x\right)}{x-a} dx\\ PV\underset{-\infty }{\overset{\infty }{\int }}\frac{\upsilon \left(x\right)}{x-a} dx& =& PV\underset{0}{\overset{\infty }{\int }}\frac{\upsilon \left(x\right)\left(x-a\right)+\upsilon \left(x\right)\left(x+a\right)}{x^2-a^2} dx\\ PV\underset{-\infty }{\overset{\infty }{\int }}\frac{\upsilon \left(x\right)}{x-a} dx& =& PV\underset{0}{\overset{\infty }{\int }}\frac{2 x\upsilon \left(x\right)}{x^2-a^2} dx\end{array}$

Hence, for $\upsilon \left(x\right)$odd, we have, .$PV\underset{-\infty }{\overset{\infty }{\int }}\frac{\upsilon \left(x\right)}{x-a} dx= PV\underset{0}{\overset{\infty }{\int }}\frac{2 x\upsilon \left(x\right)}{x^2-a^2} dx=\pi u\left(a\right)$