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Question Number 123961 by mnjuly1970 last updated on 29/Nov/20

$$...nicecalculus...$$$$provethat::$$$$$$$$\mathrm{\Omega }={\int }_0^{\mathrm{\infty }}\left(\frac{x^{\phi }}{(x+1)\sqrt{x^{4\phi }+x^4}}\right)dx$$$$\stackrel{???}{=}\frac{\phi {\mathrm{\Gamma }}^2\left(\frac{1}{4}\right)}{\sqrt{\pi }}$$$$\phi isGoldenratio...$$

Answered by mindispower last updated on 29/Nov/20

$$2\mathrm{\Omega }={\int }_0^{\mathrm{\infty }}\frac{x^{\phi }}{(1+x)\sqrt{x^{4\phi }+x^4}}+\frac{1}{x^{\phi }}.\frac{dx}{x^2(1+\frac{1}{x})\sqrt{\frac{x^{4\phi }+x^4}{x^{4\phi +4}}}}$$$$={\int }_0^{\mathrm{\infty }}\frac{x^{\phi }+x^{\phi +1}}{(x+1)\sqrt{x^4+x^{4\phi }}}dx=\int \frac{x^{\phi }}{\sqrt{x^4+x^{4\phi }}}$$$$={\int }_0^{\mathrm{\infty }}\frac{x^{\phi -2}}{\sqrt{1+x^{4(\phi -1)}}}dx$$$$let{x}^{(\phi -1)}=t\Rightarrow x^{\phi -2}dx=\frac{dt}{\phi -1}$$$$={\int }_0^{\mathrm{\infty }}\frac{dt}{(\phi -1)\sqrt{1+t^4}}$$$$t=\sqrt{tg\left(r\right)}\Rightarrow dt=\frac{dx}{2{cos}^{\frac{3}{2}}\left(x\right){sin}^{\frac{1}{2}}}$$$$\mathrm{\Omega }{=}_{}\frac{1}{2(\phi -1)}{\int }_0^{\frac{\pi }{2}}\frac{cos\left(x\right)dx}{{cos}^{\frac{3}{2}}\left(x\right){sin}^{\frac{1}{2}}\left(x\right)}$$$$\mathrm{\Omega }=\frac{1}{4}.2{\int }_0^{\frac{\pi }{2}}{cos}^{2\left(\frac{1}{4}\right)-1}\left(x\right){sin}^{2\left(\frac{1}{4}\right)-1}\left(x\right)dx$$$$=\frac{1}{4(\phi -1)}\beta (\frac{1}{4},\frac{1}{4})=\frac{1}{4(\phi -1)}\frac{{\mathrm{\Gamma }}^2\left(\frac{1}{4}\right)}{\mathrm{\Gamma }\left(\frac{1}{2}\right)}=\frac{{\mathrm{\Gamma }}^2\left(\frac{1}{4}\right)}{4\sqrt{\pi }(\phi -1)}$$$${\phi }^2-\phi -1=0=\phi (\phi -1)-1\Rightarrow \frac{1}{\phi -1}=\phi$$$$2\mathrm{\Omega }=\frac{\phi {\mathrm{\Gamma }}^2\left(\frac{1}{4}\right)}{4\sqrt{\pi }},\mathrm{\Omega }=\frac{\phi }{8\sqrt{\pi }}{\mathrm{\Gamma }}^2\left(\frac{1}{4}\right)$$$$$$

Commented by mnjuly1970 last updated on 29/Nov/20

$$thankyoumastermindspower$$$$excellentasalways..$$

Commented by mindispower last updated on 02/Dec/20

$$witheplessursirhaveaniceday$$

Answered by Dwaipayan Shikari last updated on 29/Nov/20

$${\int }_0^{\mathrm{\infty }}\left(\frac{x^{\phi }}{(x+1)\sqrt{x^{4\phi }+x^4}}\right)x=\frac{1}{t}$$$$={\int }_0^{\mathrm{\infty }}\frac{1}{t}.\frac{t^{-\phi }}{(1+t)\sqrt{\frac{1}{t^{4\phi }}+\frac{1}{t^4}}}dt$$$$={\int }_0^{\mathrm{\infty }}\frac{t^{\phi +1}}{(1+t)\sqrt{t^4+t^{4\phi }}}dt=I$$$$2I={\int }_0^{\mathrm{\infty }}\frac{t^{\phi +1}}{(1+t)\sqrt{t^{4\phi }+t^4}}+\frac{t^{\phi }}{(1+t)\sqrt{t^{4\phi }+t^4}}dt$$$$={\int }_0^{\mathrm{\infty }}\frac{t^{\phi -2}}{\sqrt{t^{4\phi -4}+1}}dt{t}^{\phi -1}=u\Rightarrow (\phi -1)t^{\phi -2}=\frac{du}{dt}$$$$=\frac{1}{2(\phi -1)}{\int }_0^{\frac{\pi }{2}}\frac{sec\theta }{u}d\theta u^2=tan\theta \Rightarrow 2u={sec}^2\theta \frac{d\theta }{du}$$$$=\frac{1}{2(\phi -1)}{\int }_0^{\frac{\pi }{2}}{sin}^{-\frac{1}{2}}\theta {cos}^{-\frac{1}{2}}\theta d\theta {sin}^2\theta =p$$$$=\frac{1}{4(\phi -1)}{\int }_0^1{p}^{-\frac{1}{4}-\frac{1}{2}}{(1-p)}^{-\frac{1}{4}-\frac{1}{2}}d\theta \phi -1=\frac{-1+\sqrt{5}}{4}=\frac{1}{\phi }$$$$=\frac{1}{4(\phi -1)}.\frac{{\mathrm{\Gamma }}^2\left(\frac{1}{4}\right)}{\mathrm{\Gamma }\left(\frac{1}{2}\right)}=\frac{{\mathrm{\Gamma }}^2\left(\frac{1}{4}\right)}{\sqrt{\pi }}.\frac{\phi }{4}=2I$$$$I=\frac{{\mathrm{\Gamma }}^2\left(\frac{1}{4}\right)(\sqrt{5}+1)}{32\sqrt{\pi }}$$

Commented by mnjuly1970 last updated on 29/Nov/20

$$verynicesolution$$$$bravobravo$$$$sirpayan.grateful...$$

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