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Question Number 219357 by SdC355 last updated on 23/Apr/25

∫_0 ^( ∞) ((ln(z))/(z^2 +1)) dz

$$\int_{\mathrm{0}} ^{\:\infty} \:\:\:\frac{\mathrm{ln}\left({z}\right)}{{z}^{\mathrm{2}} +\mathrm{1}}\:\mathrm{d}{z} \\ $$

Answered by breniam last updated on 23/Apr/25

=∫_0 ^1 ((ln (z))/(z^2 +1))dz+∫_1 ^∞ ((ln(z))/(z^2 +1))dz After we apply z^− =(1/z) substitution to the right integral we obtain ∫_0 ^1 ((ln(z))/(z^2 +1))dz−∫_0 ^1 ((ln(z))/(z^2 +1))dz=0

$$=\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\frac{\mathrm{ln}\:\left({z}\right)}{{z}^{\mathrm{2}} +\mathrm{1}}\mathrm{d}{z}+\underset{\mathrm{1}} {\overset{\infty} {\int}}\frac{\mathrm{ln}\left({z}\right)}{{z}^{\mathrm{2}} +\mathrm{1}}\mathrm{d}{z} \\ $$$$\mathrm{After}\:\mathrm{we}\:\mathrm{apply}\:\overset{−} {{z}}=\frac{\mathrm{1}}{{z}}\:\mathrm{substitution}\:\mathrm{to}\:\mathrm{the}\:\mathrm{right}\:\mathrm{integral}\:\mathrm{we}\:\mathrm{obtain} \\ $$$$\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\frac{\mathrm{ln}\left({z}\right)}{{z}^{\mathrm{2}} +\mathrm{1}}\mathrm{d}{z}−\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\frac{\mathrm{ln}\left({z}\right)}{{z}^{\mathrm{2}} +\mathrm{1}}\mathrm{d}{z}=\mathrm{0} \\ $$

Commented by Nicholas666 last updated on 23/Apr/25

good

$${good} \\ $$

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