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Question Number 163400 by mnjuly1970 last updated on 06/Jan/22

$$$$$$prove$$$$\mathrm{\Omega }={\int }_0^{\mathrm{\infty }}{cot}^{-1}(1+x^2)=\left(\sqrt{\frac{1}{\sqrt{2}}-\frac{1}{2}}\right)\pi$$$$$$

Answered by Ar Brandon last updated on 06/Jan/22

$$\mathrm{\Omega }=\int {\cot }^{-1}(1+x^2)=x{\cot }^{-1}(1+x^2)+\int \frac{2x^2}{x^4+2x^2+1}dx$$$$=x{\cot }^{-1}(1+x^2)+\int \frac{(x^2+1)+(x^2-1)}{x^4+2x^2+1}dx$$$$=x{\cot }^{-1}(1+x^2)+\int \frac{1+\frac{1}{x^2}}{{(x-\frac{1}{x})}^2+4}dx+\int \frac{1-\frac{1}{x^2}}{{(x+\frac{1}{x})}^2}dx$$$$=x{\cot }^{-1}(1+x^2)+\frac{1}{2}{\tan }^{-1}\left(\frac{x^2-1}{2x}\right)-\frac{x}{x^2+1}+C$$

Answered by Kamel last updated on 07/Jan/22

$$prove$$$$\mathrm{\Omega }={\int }_0^{\mathrm{\infty }}{cot}^{-1}(1+x^2)=\left(\sqrt{\frac{1}{\sqrt{2}}-\frac{1}{2}}\right)\pi$$$$=2{\int }_0^{+\mathrm{\infty }}\frac{x^{2}dx}{1+{(1+x^2)}^2}\stackrel{t=x^2}{=}{\int }_0^{+\mathrm{\infty }}\frac{\sqrt{t}dt}{1+{(1+t)}^2}$$$$=\frac{\sqrt{\pi }}{2}{\int }_0^{+\mathrm{\infty }}\frac{sin\left(t\right)e^{-t}}{t^{\frac{3}{2}}}dt=\frac{\sqrt{\pi }}{2}{\int }_1^{+\mathrm{\infty }}{\int }_0^{+\mathrm{\infty }}\frac{sin\left(t\right)}{\sqrt{t}}{e}^{-at}dtda$$$$=\sqrt{\pi }{\int }_1^{+\mathrm{\infty }}{\int }_0^{+\mathrm{\infty }}sin\left(u^2\right)e^{-{au}^2}duda$$$$=\frac{\pi }{4i}{\int }_1^{+\mathrm{\infty }}(\frac{1}{\sqrt{a-i}}-\frac{1}{\sqrt{a+i}})da=\frac{\pi }{2i}(\sqrt{1+i}-\sqrt{1-i})=\pi \sqrt{\sqrt{2}}sin\left(\frac{\pi }{8}\right)$$$$sin\left(\frac{\pi }{8}\right)=\sqrt{\frac{1-cos\left(\frac{\pi }{4}\right)}{2}}=\sqrt{\frac{\sqrt{2}-1}{2\sqrt{2}}}=\frac{1}{\sqrt{\sqrt{2}}}\sqrt{\frac{1}{\sqrt{2}}-\frac{1}{2}}$$$$\therefore \mathrm{\Omega }=\pi \sqrt{\frac{1}{\sqrt{2}}-\frac{1}{2}}$$

Commented by mnjuly1970 last updated on 07/Jan/22

$$bravo...$$

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