Prove that \(i\log\left(\dfrac{x-i}{x+i}\right)=\pi-2\tan^{-1}x\) for \(x>0\) (principal value).

Answer: Proved; \(\log\frac{x-i}{x+i}=-2i\tan^{-1}\frac1x\), so \(i\log\frac{x-i}{x+i}=2\tan^{-1}\frac1x=\pi-2\tan^{-1}x\).

Use \(\log(a+ib)=\frac12\log(a^2+b^2)+i\tan^{-1}\frac ba\) (principal value, \(a>0\)).

\(\log(x-i)=\frac12\log(x^2+1)-i\tan^{-1}\frac1x\), \(\log(x+i)=\frac12\log(x^2+1)+i\tan^{-1}\frac1x\).

\[\log\frac{x-i}{x+i}=\log(x-i)-\log(x+i)=-2i\tan^{-1}\frac1x.\]

Multiply by \(i\): \(i\log\frac{x-i}{x+i}=-2i^2\tan^{-1}\frac1x=2\tan^{-1}\frac1x\).

For \(x>0\), \(\tan^{-1}\frac1x=\frac\pi2-\tan^{-1}x\). Hence

\[i\log\left(\frac{x-i}{x+i}\right)=2\left(\frac\pi2-\tan^{-1}x\right)=\pi-2\tan^{-1}x.\]