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Question Number 116112 by Lordose last updated on 01/Oct/20

$$\mathbf{show}\mathbf{that}$$$${\int }_0^{\mathrm{\infty }}\frac{\mathbf{lnx}}{1+{\mathbf{x}}^2}\mathbf{dx}=0$$$$$$

Answered by MJS_new last updated on 01/Oct/20

$$\underset{0}{\stackrel{1}{\int }}\frac{\ln x}{1+x^2}dx=$$$$[t=\frac{1}{x}\to dx=-x^{2}dt=-\frac{dt}{t^2}]$$$$=\underset{\mathrm{\infty }}{\stackrel{1}{\int }}\frac{\ln \frac{1}{t}}{1+\frac{1}{t^2}}\times -\frac{dt}{t^2}=$$$$[\ln \frac{1}{t}=-\ln t\wedge -\underset{a}{\stackrel{b}{\int }}f\left(t\right)dt=\underset{b}{\stackrel{a}{\int }}f\left(t\right)dt]$$$$=-\underset{1}{\stackrel{\mathrm{\infty }}{\int }}\frac{\ln t}{1+t^2}dt$$$$\Rightarrow$$$$\underset{0}{\stackrel{1}{\int }}\frac{\ln x}{1+x^2}=-\underset{1}{\stackrel{\mathrm{\infty }}{\int }}\frac{\ln x}{1+x^2}dx$$$$\underset{0}{\stackrel{1}{\int }}\frac{\ln x}{1+x^2}+\underset{1}{\stackrel{\mathrm{\infty }}{\int }}\frac{\ln x}{1+x^2}dx=0\iff \underset{0}{\stackrel{\mathrm{\infty }}{\int }}\frac{\ln x}{1+x^2}dx=0$$$$[\underset{a}{\stackrel{b}{\int }}f\left(x\right)dx+\underset{b}{\stackrel{c}{\int }}f\left(x\right)dx=\underset{a}{\stackrel{c}{\int }}f\left(x\right)dx]$$

Answered by Bird last updated on 02/Oct/20

$$aeazywaybych.x=tan\theta \Rightarrow$$$${\int }_0^{\mathrm{\infty }}\frac{lnx}{1+x^2}dx={\int }_0^{\frac{\pi }{2}}\frac{ln\left(tan\theta \right)}{1+{tan}^2\theta }(1+{tan}^2\theta )d\theta$$$$={\int }_0^{\frac{\pi }{2}}ln\left(sin\theta \right)-{\int }_0^{\frac{\pi }{2}}ln\left(cos\theta \right)d\theta =0$$$$becausetheintegralsareequals$$

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