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Question Number 112697 by mnjuly1970 last updated on 09/Sep/20

$$....calculus...$$$$pleaseprove:$$$$$$$$\mathrm{\Omega }={\int }_0^{\mathrm{\infty }}\frac{1}{{(1+x^{\phi })}^{\phi }}dx=1$$$$\phi ::goldenratio...$$

Commented by mathdave last updated on 09/Sep/20

$$thisissimplenah$$

Commented by MJS_new last updated on 09/Sep/20

$$\mathrm{I}{\mathrm{haven}}^{\prime }\mathrm{t}\mathrm{found}\mathrm{it}\mathrm{yet}\mathrm{but}\mathrm{we}\mathrm{have}\mathrm{to}\mathrm{play}$$$$\mathrm{around}\mathrm{with}\mathrm{the}\mathrm{fact}$$$$\frac{1}{\phi }=\phi -1\mathrm{and}{\phi }^2=\phi +1$$

Commented by mnjuly1970 last updated on 09/Sep/20

$$youareright..$$

Answered by MJS_new last updated on 09/Sep/20

$$\mathrm{trying}‘‘\mathrm{reverse}{\mathrm{method}}^{″}$$$$\frac{d}{dx}\left[f\left(x\right){(1+x^{\phi })}^{1-\phi }\right]=$$$$=f^{\prime }\left(x\right){(1+x^{\phi })}^{1-\phi }+(\phi -{\phi }^2)f\left(x\right)\frac{x^{\phi -1}}{{(1+x^{\phi })}^{\phi }}=$$$$=\frac{f^{\prime }\left(x\right)(1+x^{\phi })+(\phi -{\phi }^2)f\left(x\right)x^{\phi -1}}{{(1+x^{\phi })}^{\phi }}$$$$\mathrm{putting}f\left(x\right)=x\mathrm{we}\mathrm{get}$$$$\frac{(1+\phi -{\phi }^2)x^{\phi }+1}{{(1+x^{\phi })}^{\phi }}$$$$\mathrm{but}1+\phi -{\phi }^2=0$$$$\Rightarrow$$$$\int \frac{dx}{{(1+x^{\phi })}^{\phi }}=\frac{x}{{(1+x^{\phi })}^{\phi -1}}+C=\frac{x}{{(1+x^{\phi })}^{1/\phi }}+C$$$$\underset{0}{\stackrel{\mathrm{\infty }}{\int }}\frac{dx}{{(1+x^{\phi })}^{\phi }}=\underset{x\to \mathrm{\infty }}{\lim }\frac{x}{{(1+x^{\phi })}^{1/\phi }}=1$$

Commented by mnjuly1970 last updated on 09/Sep/20

$$nicemethod..$$

Commented by MJS_new last updated on 09/Sep/20

$$\mathrm{it}\mathrm{works}\mathrm{only}\mathrm{sometimes}...$$

Answered by ajfour last updated on 09/Sep/20

$$I=\int \frac{dx}{{[1+x\left(x^{1/\phi }\right)]}^{\phi }}$$$$=\int \frac{dx}{x{(x^{-1/\phi }+x)}^{\phi }}$$$$letx=e^t\Rightarrow \frac{dx}{x}=dt$$$$I={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{dt}{{(e^{-t\left(\frac{1}{\phi }\right)}+e^t)}^{\phi }}$$$$but\frac{1}{\phi }=\phi -1,so$$$$I=\int \frac{dt}{{(\frac{e^t}{e^{\phi t}}+e^t)}^{\phi }}={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{e^{-\phi t}dt}{{(e^{-\phi t}+1)}^{\phi }}$$$$let{e}^{-\phi t}+1=z$$$$\Rightarrow e^{-\phi t}dt=-\frac{dz}{\phi }$$$$I=-\frac{1}{\phi }{\int }_{\mathrm{\infty }}^1{z}^{-\phi }dz$$$$I=\frac{1}{\phi (\phi -1)}{\left[z^{-\phi +1}\right]}_{\mathrm{\infty }}^1$$$$I=\left(\frac{\phi -1}{\phi -1}\right)(1-0)=1$$$$★.................................★$$$$$$

Commented by mnjuly1970 last updated on 09/Sep/20

$$niceverynicemrajfour.$$$$thankyouverymuch.$$

Answered by mnjuly1970 last updated on 09/Sep/20

$$solution.$$$$weknowthat::$$$${\int }_0^{\mathrm{\infty }}\frac{x^{p-1}}{{(1+x)}^{p+q}}dx=\frac{\mathrm{\Gamma }\left(p\right)\mathrm{\Gamma }\left(q\right)}{\mathrm{\Gamma }(p+q)}$$$$x^{\phi }=t\Rightarrow dx=\frac{1}{\phi }{t}^{\frac{1}{\phi }-1}$$$$\mathrm{\Omega }=\frac{1}{\phi }{\int }_0^{\mathrm{\infty }}\frac{t^{\frac{1}{\phi }-1}}{{(1+t)}^{\phi }}dt=\frac{1}{\phi }\left[\frac{\mathrm{\Gamma }\left(\frac{1}{\phi }\right)\mathrm{\Gamma }(\phi -\frac{1}{\phi })}{\mathrm{\Gamma }\left(\phi \right)}\right]$$$$weknowthat:{\phi }^2=\phi +1\Rightarrow \phi =1+\frac{1}{\phi }$$$$\therefore \mathrm{\Omega }=\frac{1}{\phi }\left(\frac{\mathrm{\Gamma }\left(\frac{1}{\phi }\right)}{\mathrm{\Gamma }\left(\phi \right)}\right)=\frac{\mathrm{\Gamma }(1+\frac{1}{\phi })}{\mathrm{\Gamma }\left(\phi \right)}=\frac{\mathrm{\Gamma }\left(\phi \right)}{\mathrm{\Gamma }\left(\phi \right)}$$$$=1✓$$$$...m.n...$$$$$$

Commented by Tawa11 last updated on 06/Sep/21

$$\mathrm{great}\mathrm{sir}$$

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