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Question Number 124201 by mnjuly1970 last updated on 01/Dec/20

$$:::nicecalculus:::$$$$pleaseprove:::$$$$\mathrm{\Omega }={\int }_0^{\mathrm{\infty }}\frac{x^{\frac{1}{2}}}{x^2+2x+5}dx=\frac{\pi }{\sqrt{\phi }}$$$$where\phi isGoldenratio...$$

Answered by ajfour last updated on 02/Dec/20

$$I=\int \frac{\sqrt{x}dx}{{(x+1)}^2+{(2\phi -1)}^2}$$$$let\sqrt{x}=t\Rightarrow \sqrt{x}dx=2t^{2}dt$$$$I=\int \frac{2t^{2}dt}{{(t^2+1)}^2+{(2\phi -1)}^2}$$$$lett=\frac{1}{z}\Rightarrow dt=-\frac{dz}{z^2}$$$$I=\int \frac{2dz}{{(1+z^2)}^2+{(2\phi -1)}^2{z}^4}$$$$=\int \frac{2dz}{6z^4+2z^2+1}$$$$=\frac{1}{3}\int \frac{dz}{{(z^2+\frac{1}{6})}^2+{\left(\frac{\sqrt{5}}{6}\right)}^2}$$$$....$$

Commented by mindispower last updated on 02/Dec/20

$$sirajfouryourmethode$$$$worckssince6z^4+2z^2+1canbeefactorised$$$$inpolynomsofdegree2$$$$and\int \frac{dx}{x^2+ax+b}canbeesolved$$$$$$$$$$

Commented by mnjuly1970 last updated on 03/Dec/20

$$thankingforyoureffort$$$$sirajfor..$$

Commented by mnjuly1970 last updated on 03/Dec/20

Answered by mindispower last updated on 02/Dec/20

$$\mathrm{\Delta }={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{z^{\frac{1}{2}}}{z^2+2z+5}dz={\int }_{-\mathrm{\infty }}^0\frac{z^{\frac{1}{2}}}{z^2+2z+5}dz+\mathrm{\Omega }$$$$=i{\int }_0^{\mathrm{\infty }}\frac{z^{\frac{1}{2}}}{z^2-2z+5}dz+\mathrm{\Omega }=\mathrm{\Omega }+it,\mathrm{\Omega },t\in \mathbb{R}$$$$\mathrm{\Delta }=2i\pi Res(\frac{z^{\frac{1}{2}}}{z^2+2z+5},z=-1+2i)$$$$.=2i\pi \frac{{(-1+2i)}^{\frac{1}{2}}}{2(-1+2i)+2}=\frac{\pi }{2}{(-1+2i)}^{\frac{1}{2}}$$$${(-1+2i)}^{\frac{1}{2}}=x+iy$$$$\Rightarrow xy=1,x^2+y^2=\sqrt{5},x^2-y^2=-1$$$$x={\left(\frac{\sqrt{5}-1}{2}\right)}^{\frac{1}{2}},y=\frac{2}{\sqrt{5}-1}={\left(\frac{\sqrt{5}+1}{2}\right)}^{\frac{1}{2}}$$$$x=\sqrt{\phi -1},y=\sqrt{\phi }$$$$\mathrm{\Delta }=\frac{\pi }{2}\sqrt{\phi -1}+\frac{\pi }{2}i\sqrt{\phi }$$$$\mathrm{\Omega }=Re\left(\mathrm{\Delta }\right)=\frac{\pi }{2}\sqrt{\phi -1}$$$${\phi }^2=\phi +1\Rightarrow \phi (\phi -1)=1$$$$\Rightarrow \sqrt{\phi -1}=\frac{1}{\sqrt{\phi }}$$$$\mathrm{\Omega }=\frac{\pi }{2}\sqrt{\phi -1}=\frac{\pi }{2\sqrt{\phi }}$$$$$$$$$$

Commented by mnjuly1970 last updated on 02/Dec/20

$$excellent$$

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