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Question Number 122882 by mnjuly1970 last updated on 20/Nov/20

$$...nicecalculus...$$$$provethat:::$$$$\mathrm{\Omega }={\int }_0^1{\left(\frac{x^{\phi }-1}{ln\left(x\right)}\right)}^{2}dx=\sqrt{5}ln\left(\phi \right)$$$$.m.n.$$

Answered by TANMAY PANACEA last updated on 20/Nov/20

$$f\left(\phi \right)={\int }_0^1\frac{x^{\phi }-1}{lnx}dx$$$$\frac{df}{d\phi }={\int }_0^1\frac{x^{\phi }lnx}{lnx}dx=\mid \frac{x^{\phi +1}}{\phi +1}{\mid }_0^1=\frac{1}{\phi +1}$$$$df=\frac{d\phi }{\phi +1}$$$$f=ln(\phi +1)+C$$$$when\phi =0f\left(\phi \right)=0C=0$$$$f\left(\phi \right)=ln(\phi +1)$$$$wait...$$

Commented by mnjuly1970 last updated on 21/Nov/20

$$thankyoumrtanmayforyour$$$$effort$$$$withyourpermission$$$$ipresentthesolution.$$$$note:ln\left(N\right)={\int }_0^1\frac{x^{N-1}-1}{ln\left(x\right)}dx$$$$f\left(a\right)={\int }_0^1\frac{{(x^a-1)}^2}{{ln}^2\left(x\right)}dx$$$$f{}^{\prime }\left(a\right)={\int }_0^1\frac{\partial }{\partial a}\left[\frac{{(x^a-1)}^2}{{ln}^2\left(x\right)}\right]$$$$={\int }_0^1\frac{2(x^a-1)x^{a}ln\left(x\right)}{{ln}^2\left(x\right)}dx=2{\int }_0^1\frac{x^{2a}-x^a}{ln\left(x\right)}dx$$$$=2{\int }_0^1\frac{x^{2a}-1}{ln\left(x\right)}+2{\int }_0^1\frac{1-x^a}{ln\left(x\right)}dx$$$$=2ln(2a+1)-2ln(a+1)$$$$f\left(a\right)=2{\int }^{}ln(2a+1)da-2\int ln(a+1)da+C$$$$=(2a+1)ln(2a+1)-(2a+1)$$$$-2(a+1)ln(a+1)+2(a+1)+C$$$$f\left(0\right)=1+C\Rightarrow C=-1$$$$f\left(a\right)=(2a+1)ln(2a+1)-2(a+1)ln(a+1)+1-1$$$$$$$$\mathrm{\Omega }=f\left(\phi \right)=(2\phi +1)ln(2\phi +1)-2(\phi +1)ln(\phi +1)$$$$\phi :isgoldenratio$$$$andweknow:{\phi }^2=\phi +1$$$$=({\phi }^2+\phi )ln({\phi }^2+\phi )-2{\phi }^{2}ln\left({\phi }^2\right)$$$$={\phi }^{3}ln\left({\phi }^3\right)-4{\phi }^{2}ln\left(\phi \right)$$$$=(3{\phi }^3-4{\phi }^2)ln\left(\phi \right)=(3{\phi }^2+3\phi -4{\phi }^2)ln\left(\phi \right)$$$$=(-{\phi }^2+3\phi )ln\left(\phi \right)=(-\phi -1+3\phi )ln\left(\phi \right)$$$$=(2\phi -1)ln\left(\phi \right)=(2\left(\frac{1+\sqrt{5}}{2}\right)-1)ln\left(\phi \right)$$$$=\sqrt{5}ln\left(\phi \right)✓✓✓$$

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