91Ó°ÊÓ

Let u(x) be an even function and $\mathrm{Ï\dots }\left(x\right)$be an odd function.

We have given that $f\left(x\right)=u\left(x\right)+i\mathrm{Ï\dots }\left(x\right)    ⋯\left(1\right)$.

Therefore, .$f^{*}\left(x\right)=u\left(x\right)-i\mathrm{Ï\dots }\left(x\right)     ⋯\left(2\right)$

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

Also, since $\mathrm{Ï\dots }\left(x\right)$is an odd function, we have, .$\mathrm{Ï\dots }\left(-x\right)=-\mathrm{Ï\dots }\left(x\right)$

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

$\begin{array}{rcl}f\left(-x\right)& =& u\left(-x\right)+i\mathrm{Ï\dots }\left(-x\right)\\ & =& u\left(x\right)-i\mathrm{Ï\dots }\left(x\right)\\ & =& f^{*}\left(x\right)    ⋯\left(âˆ\mu   by  \left(2\right)\right)\end{array}$

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

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

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

Let u(x) be an even function and $\mathrm{Ï\dots }\left(x\right)$be an odd function.

We have $PV\underset{-∞}{\overset{∞}{∫}}\frac{u\left(x\right)}{x-a} dx=-\mathrm{Ï€}\mathrm{Ï\dots }\left(a\right),  u\left(-x\right)=u\left(x\right)    ⋯\left(1\right)$

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

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

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

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

$ PV\underset{-∞}{\overset{∞}{∫}}\frac{u\left(x\right)}{x-a} dx= PV\underset{0}{\overset{∞}{∫}}\frac{2 au\left(x\right)}{x^2-a^2} dx=-\mathrm{Ï€}\mathrm{Ï\dots }\left(a\right)$

.

Next, we have $PV\underset{-∞}{\overset{∞}{∫}}\frac{\mathrm{Ï\dots }\left(x\right)}{x-a} dx=\mathrm{Ï€}u\left(a\right),  \mathrm{Ï\dots }\left(-x\right)=-\mathrm{Ï\dots }\left(x\right)    ⋯\left(3\right)$

Therefore,$PV\underset{-∞}{\overset{∞}{∫}}\frac{\mathrm{Ï\dots }\left(x\right)}{x-a} dx= PV\underset{-∞}{\overset{0}{∫}}\frac{\mathrm{Ï\dots }\left(x\right)}{x-a} dx+PV\underset{0}{\overset{∞}{∫}}\frac{\mathrm{Ï\dots }\left(x\right)}{x-a} dx     ⋯\left(4\right)$

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

$\begin{array}{rcl}PV\underset{-∞}{\overset{∞}{∫}}\frac{\mathrm{Ï\dots }\left(x\right)}{x-a} dx& =&  -PV\underset{0}{\overset{∞}{∫}}\frac{\mathrm{Ï\dots }\left(-x\right)}{-x-a} \left(-dx\right)+PV\underset{0}{\overset{∞}{∫}}\frac{\mathrm{Ï\dots }\left(x\right)}{x-a} dx \\ PV\underset{-∞}{\overset{∞}{∫}}\frac{\mathrm{Ï\dots }\left(x\right)}{x-a} dx& =&  -PV\underset{0}{\overset{∞}{∫}}\frac{-\mathrm{Ï\dots }\left(x\right)}{\left(x+a\right)} dx+PV\underset{0}{\overset{∞}{∫}}\frac{\mathrm{Ï\dots }\left(x\right)}{x-a} dx\\ PV\underset{-∞}{\overset{∞}{∫}}\frac{\mathrm{Ï\dots }\left(x\right)}{x-a} dx& =&  PV\underset{0}{\overset{∞}{∫}}\frac{\mathrm{Ï\dots }\left(x\right)\left(x-a\right)+\mathrm{Ï\dots }\left(x\right)\left(x+a\right)}{x^2-a^2} dx\\ PV\underset{-∞}{\overset{∞}{∫}}\frac{\mathrm{Ï\dots }\left(x\right)}{x-a} dx& =&  PV\underset{0}{\overset{∞}{∫}}\frac{2 x\mathrm{Ï\dots }\left(x\right)}{x^2-a^2} dx\end{array}$

Hence, for $\mathrm{Ï\dots }\left(x\right)$odd, we have, .$PV\underset{-∞}{\overset{∞}{∫}}\frac{\mathrm{Ï\dots }\left(x\right)}{x-a} dx= PV\underset{0}{\overset{∞}{∫}}\frac{2 x\mathrm{Ï\dots }\left(x\right)}{x^2-a^2} dx=\mathrm{Ï€}u\left(a\right)$