Question Number 219357 by SdC355 last updated on 23/Apr/25
$$\int_{\mathrm{0}} ^{\:\infty} \:\:\:\frac{\mathrm{ln}\left({z}\right)}{{z}^{\mathrm{2}} +\mathrm{1}}\:\mathrm{d}{z} \\ $$
Answered by breniam last updated on 23/Apr/25
$$=\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\frac{\mathrm{ln}\:\left({z}\right)}{{z}^{\mathrm{2}} +\mathrm{1}}\mathrm{d}{z}+\underset{\mathrm{1}} {\overset{\infty} {\int}}\frac{\mathrm{ln}\left({z}\right)}{{z}^{\mathrm{2}} +\mathrm{1}}\mathrm{d}{z} \\ $$$$\mathrm{After}\:\mathrm{we}\:\mathrm{apply}\:\overset{−} {{z}}=\frac{\mathrm{1}}{{z}}\:\mathrm{substitution}\:\mathrm{to}\:\mathrm{the}\:\mathrm{right}\:\mathrm{integral}\:\mathrm{we}\:\mathrm{obtain} \\ $$$$\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\frac{\mathrm{ln}\left({z}\right)}{{z}^{\mathrm{2}} +\mathrm{1}}\mathrm{d}{z}−\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\frac{\mathrm{ln}\left({z}\right)}{{z}^{\mathrm{2}} +\mathrm{1}}\mathrm{d}{z}=\mathrm{0} \\ $$
Commented by Nicholas666 last updated on 23/Apr/25
$${good} \\ $$