Question Number 116112 by Lordose last updated on 01/Oct/20
$$\boldsymbol{\mathrm{show}}\:\boldsymbol{\mathrm{that}}\: \\ $$$$\int_{\:\mathrm{0}} ^{\:\infty} \frac{\boldsymbol{\mathrm{lnx}}}{\mathrm{1}+\boldsymbol{\mathrm{x}}^{\mathrm{2}} }\boldsymbol{\mathrm{dx}}\:=\:\mathrm{0} \\ $$$$ \\ $$
Answered by MJS_new last updated on 01/Oct/20
![∫_0 ^1 ((ln x)/(1+x^2 ))dx= [t=(1/x) → dx=−x^2 dt=−(dt/t^2 )] =∫_∞ ^1 ((ln (1/t))/(1+(1/t^2 )))×−(dt/t^2 )= [ln (1/t) =−ln t ∧ −∫_a ^b f(t)dt=∫_b ^a f(t)dt] =−∫_1 ^∞ ((ln t)/(1+t^2 ))dt ⇒ ∫_0 ^1 ((ln x)/(1+x^2 ))=−∫_1 ^∞ ((ln x)/(1+x^2 ))dx ∫_0 ^1 ((ln x)/(1+x^2 ))+∫_1 ^∞ ((ln x)/(1+x^2 ))dx=0 ⇔ ∫_0 ^∞ ((ln x)/(1+x^2 ))dx=0 [∫_a ^b f(x)dx+∫_b ^c f(x)dx=∫_a ^c f(x)dx]](https://www.tinkutara.com/question/Q116121.png)
$$\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\frac{\mathrm{ln}\:{x}}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\frac{\mathrm{1}}{{x}}\:\rightarrow\:{dx}=−{x}^{\mathrm{2}} {dt}=−\frac{{dt}}{{t}^{\mathrm{2}} }\right] \\ $$$$=\underset{\infty} {\overset{\mathrm{1}} {\int}}\frac{\mathrm{ln}\:\frac{\mathrm{1}}{{t}}}{\mathrm{1}+\frac{\mathrm{1}}{{t}^{\mathrm{2}} }}×−\frac{{dt}}{{t}^{\mathrm{2}} }= \\ $$$$\:\:\:\:\:\left[\mathrm{ln}\:\frac{\mathrm{1}}{{t}}\:=−\mathrm{ln}\:{t}\:\wedge\:−\underset{{a}} {\overset{{b}} {\int}}{f}\left({t}\right){dt}=\underset{{b}} {\overset{{a}} {\int}}{f}\left({t}\right){dt}\right] \\ $$$$=−\underset{\mathrm{1}} {\overset{\infty} {\int}}\frac{\mathrm{ln}\:{t}}{\mathrm{1}+{t}^{\mathrm{2}} }{dt} \\ $$$$\Rightarrow \\ $$$$\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\frac{\mathrm{ln}\:{x}}{\mathrm{1}+{x}^{\mathrm{2}} }=−\underset{\mathrm{1}} {\overset{\infty} {\int}}\frac{\mathrm{ln}\:{x}}{\mathrm{1}+{x}^{\mathrm{2}} }{dx} \\ $$$$\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\frac{\mathrm{ln}\:{x}}{\mathrm{1}+{x}^{\mathrm{2}} }+\underset{\mathrm{1}} {\overset{\infty} {\int}}\frac{\mathrm{ln}\:{x}}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}=\mathrm{0}\:\Leftrightarrow\:\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{\mathrm{ln}\:{x}}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}=\mathrm{0} \\ $$$$\:\:\:\:\:\left[\underset{{a}} {\overset{{b}} {\int}}{f}\left({x}\right){dx}+\underset{{b}} {\overset{{c}} {\int}}{f}\left({x}\right){dx}=\underset{{a}} {\overset{{c}} {\int}}{f}\left({x}\right){dx}\right] \\ $$
Answered by Bird last updated on 02/Oct/20
$${a}\:{eazy}\:{way}\:{by}\:{ch}.{x}\:={tan}\theta\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\frac{{lnx}}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{ln}\left({tan}\theta\right)}{\mathrm{1}+{tan}^{\mathrm{2}} \theta}\left(\mathrm{1}+{tan}^{\mathrm{2}} \theta\right){d}\theta \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left({sin}\theta\right)−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left({cos}\theta\right){d}\theta=\mathrm{0} \\ $$$${because}\:{the}\:{integrals}\:{are}\:{equals} \\ $$