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Question Number 136497 by mnjuly1970 last updated on 22/Mar/21

$$......nicecalculus.....$$$$prove::$$$${\int }_0^1\frac{1}{1+{ln}^2\left(x\right)}dx={\int }_0^{\mathrm{\infty }}\frac{sin\left(x\right)}{1+x}dx$$$$$$

Commented by Dwaipayan Shikari last updated on 22/Mar/21

$${\int }_0^{\mathrm{\infty }}\frac{sin\left(x\right)}{1+x}dx$$$$={\int }_0^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}{e}^{-(x+1)t}sin\left(x\right)dtdx$$$$=\frac{1}{2i}{\int }_0^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}{e}^{-t}{e}^{-(t-i)x}-e^{-t}{e}^{-(t+i)x}dxdt$$$$=\frac{1}{2i}{\int }_0^{\mathrm{\infty }}\frac{e^{-t}}{t-i}-\frac{e^{-t}}{t+i}dt$$$$={\int }_0^{\mathrm{\infty }}\frac{e^{-t}}{t^2+1}dt-t=log\left(u\right)\Rightarrow 1=-\frac{1}{u}.\frac{du}{dt}$$$$={\int }_0^1\frac{1}{1+{log}^2\left(u\right)}du$$

Commented by mnjuly1970 last updated on 22/Mar/21

$$gratefulmrpayan...$$

Answered by mindispower last updated on 22/Mar/21

$$t=-ln\left(x\right)$$$$={\int }_0^{\mathrm{\infty }}\frac{e^{-t}}{1+t^2}dt={\int }_0^{\mathrm{\infty }}\frac{(t+i)-(t-i)}{2i(t+i)(t-i)}{e}^{-t}$$$$=\frac{1}{2i}({\int }_0^{\mathrm{\infty }}\frac{e^{-t}}{t-i}dt-{\int }_0^{\mathrm{\infty }}\frac{e^{-t}}{t+i})$$$$=Im{\int }_0^{\mathrm{\infty }}\frac{e^{-t}}{t-i}dt$$$$letz=it$$$$=Im{\int }_0^{i\mathrm{\infty }}\frac{e^{iz}}{i(t+1)}.\frac{dz}{i}=Im(-{\int }_0^{i\mathrm{\infty }}\frac{e^{iz}}{t+1}dz)...2$$$$let{C}_R=[0,R]\cup {({Re}^{i\theta },\theta \in [0,\frac{\pi }{2}])}_D\cup [iR,0]$$$${\int }_{C_R}\frac{e^{iz}}{z+1}dz=0..1,z\to \frac{e^{iz}}{z+1}holomorphicwithoutpols$$$$over{C}_R$$$${\int }_D\frac{e^{iz}}{z+1}=0\Rightarrow ..2\left(1\right)\mathrm{\&}\left(2\right)\Rightarrow$$$${\int }_0^{\mathrm{\infty }}\frac{e^{iz}}{z+1}dz=-{\int }_0^{i\mathrm{\infty }}\frac{e^{iz}}{z+1}dz$$$$wegetIm{\int }_0^{\mathrm{\infty }}\frac{e^{iz}}{z+1}dz={\int }_0^{\mathrm{\infty }}\frac{sin\left(z\right)}{z+1}dz$$$$$$$$$$$$$$

Commented by mnjuly1970 last updated on 22/Mar/21

$$thankyousomuchmrpower...$$

Commented by mindispower last updated on 22/Mar/21

$$pleasur$$

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