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Question Number 97007 by bagjamath last updated on 06/Jun/20

Commented by bagjamath last updated on 06/Jun/20

$$WhatisthetheormSir?$$$${\int }_0^{1}ydx={\int }_{\mathrm{\infty }}^{0}xdy??$$

Commented by PRITHWISH SEN 2 last updated on 06/Jun/20

$$\mathrm{Itdoes}\mathrm{not}\mathrm{make}\mathrm{any}\mathrm{sense}.$$

Commented by PRITHWISH SEN 2 last updated on 06/Jun/20

$$\mathrm{Instead}\mathrm{it}\mathrm{can}\mathrm{be}\mathrm{written}\mathrm{as}$$$${\int }_0^1\mathrm{ydx}={\int }_0^1\sqrt{-1+\sqrt{\frac{4}{\mathrm{x}}-3}}\mathrm{dx}$$

Commented by bagjamath last updated on 06/Jun/20

Commented by bagjamath last updated on 06/Jun/20

$$Thisisquestion$$

Commented by PRITHWISH SEN 2 last updated on 06/Jun/20

$$\mathrm{if}\mathrm{x}=\frac{4}{3+{(1+{\mathrm{y}}^2)}^2}\mathrm{then}$$$$\mathrm{dx}=\frac{-16({\mathrm{y}}^3+\mathrm{y})}{{({\mathrm{y}}^4+{2\mathrm{y}}^2+4)}^2}\mathrm{dy}$$$$\mathrm{then}\mathrm{the}\mathrm{integration}\mathrm{turns}\mathrm{to}$$$${\int }_0^{\mathrm{\infty }}\mathrm{y}.\frac{16({\mathrm{y}}^3+\mathrm{y})}{{({\mathrm{y}}^4+{2\mathrm{y}}^2+4)}^2}\mathrm{dy}$$

Commented by mr W last updated on 06/Jun/20

$${it}^{\prime }snotatheorem,butonlyatechnique.$$$$techniqueseeQ76146$$

Commented by mr W last updated on 06/Jun/20

$$seealsoQ76680$$

Commented by mr W last updated on 06/Jun/20

Commented by mr W last updated on 06/Jun/20

$$thetechniqueisfollowing:$$$${\int }_0^{1}ydx={\int }_0^{1}f\left(x\right)dxistheareaunder$$$$thecurve\left(shadedarea\right).$$$$thisareacanalsobecalculatedas$$$${\int }_0^{\mathrm{\infty }}xdy={\int }_0^{\mathrm{\infty }}{f}^{-1}\left(x\right)dx.$$$$inourcasethecalculationof$$$${\int }_0^{\mathrm{\infty }}{f}^{-1}\left(x\right)dxismucheasierthan$$$${\int }_0^{1}f\left(x\right)dx.$$

Commented by PRITHWISH SEN 2 last updated on 06/Jun/20

$$\mathrm{yes}\mathrm{sir}\mathrm{it}\mathrm{is}\mathrm{the}\mathrm{concept}\mathrm{of}\mathrm{inverse}\mathrm{function}.$$$$\mathrm{Thank}\mathrm{you}.\mathrm{But}\mathrm{I}\mathrm{think}\mathrm{in}\mathrm{the}\mathrm{question}\mathrm{it}\mathrm{should}$$$$\mathrm{be}\mathrm{mentioned}\mathrm{somewhere}\mathrm{as}\mathrm{x}\mathrm{is}\mathrm{a}\mathrm{function}\mathrm{of}$$$$\mathrm{y}.\mathrm{Then}\mathrm{this}\mathrm{misunderstood}\mathrm{can}\mathrm{be}\mathrm{avoided}.$$

Commented by bagjamath last updated on 06/Jun/20

$$ThankYouSir,Nowiunderstand$$

Commented by bagjamath last updated on 06/Jun/20

$$Whatapplicationdoyouuse,Sir?$$

Commented by PRITHWISH SEN 2 last updated on 06/Jun/20

$$\mathrm{you}\mathrm{can}\mathrm{use}\mathrm{GeoGebra}\mathrm{or}\mathrm{Desmos}$$

Commented by abdomathmax last updated on 07/Jun/20

$$\mathrm{I}={\int }_0^1\sqrt{-1+\sqrt{\frac{4}{\mathrm{x}}-3}}\mathrm{dx}\mathrm{we}\mathrm{do}\mathrm{the}\mathrm{changement}$$$$\sqrt{-1+\sqrt{\frac{4}{\mathrm{x}}-3}}=\mathrm{t}\Rightarrow \sqrt{\frac{4}{\mathrm{x}}-3}={\mathrm{t}}^2+1\Rightarrow$$$$\frac{4}{\mathrm{x}}-3={({\mathrm{t}}^2+1)}^2\Rightarrow \frac{4}{\mathrm{x}}={({\mathrm{t}}^2+1)}^2+3\Rightarrow$$$$\frac{\mathrm{x}}{4}=\frac{1}{{({\mathrm{t}}^2+1)}^2+3}\Rightarrow \mathrm{x}=\frac{4}{{({\mathrm{t}}^2+1)}^2+3}\Rightarrow$$$$\mathrm{dx}=-4\times \frac{2\left(2\mathrm{t}\right)({\mathrm{t}}^2+1)}{{({\mathrm{t}}^2+1)}^2+3}=-16\times \frac{{\mathrm{t}}^3+\mathrm{t}}{{({\mathrm{t}}^2+1)}^2+3}\Rightarrow$$$$\mathrm{I}=16{\int }_0^{\mathrm{\infty }}\mathrm{t}\times \frac{{\mathrm{t}}^3+\mathrm{t}}{{({\mathrm{t}}^2+1)}^2+3}\mathrm{dt}$$$$=16{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{t}}^4+{\mathrm{t}}^2}{{({\mathrm{t}}^2+1)}^2+3}\mathrm{dt}$$$$=8{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{{\mathrm{t}}^4+{\mathrm{t}}^2}{{({\mathrm{t}}^2+1)}^2+3}\mathrm{dt}\mathrm{let}\phi \left(\mathrm{z}\right)=\frac{{\mathrm{z}}^4+{\mathrm{z}}^2}{{({\mathrm{z}}^2+1)}^2+3}$$$$\mathrm{poles}\mathrm{of}\phi ?$$$$\phi \left(\mathrm{z}\right)=\frac{{\mathrm{z}}^4+{\mathrm{z}}^2}{({\mathrm{z}}^2+1-\mathrm{i}\sqrt{3})({\mathrm{z}}^2+1+\mathrm{i}\sqrt{3})}$$$$-1+\mathrm{i}\sqrt{3}=2(-\frac{1}{2}+\mathrm{i}\frac{\sqrt{3}}{2})={2\mathrm{e}}^{\frac{2\mathrm{i}\pi }{3}}\mathrm{and}-1-\mathrm{i}\sqrt{3}={2\mathrm{e}}^{-\frac{2\mathrm{i}\pi }{3}}$$$$\phi \left(\mathrm{z}\right)=\frac{{\mathrm{z}}^4+{\mathrm{z}}^2}{({\mathrm{z}}^2-{2\mathrm{e}}^{\frac{2\mathrm{i}\pi }{3}})({\mathrm{z}}^2-{2\mathrm{e}}^{-\frac{\mathrm{i}2\pi }{3}})}$$$$=\frac{{\mathrm{z}}^4+{\mathrm{z}}^2}{(\mathrm{z}-\sqrt{2}{\mathrm{e}}^{\frac{\mathrm{i}\pi }{3}})(\mathrm{z}+\sqrt{2}{\mathrm{e}}^{\frac{\mathrm{i}\pi }{3}})(\mathrm{z}-\sqrt{2}{\mathrm{e}}^{-\frac{\mathrm{i}\pi }{3}})(\mathrm{z}+\sqrt{2}{\mathrm{e}}^{-\frac{\mathrm{i}\pi }{3}})}$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(\mathrm{z}\right)\mathrm{dz}=2\mathrm{i}\pi \{\mathrm{Res}(\phi ,\sqrt{2}{\mathrm{e}}^{\frac{\mathrm{i}\pi }{3}})+\mathrm{Res}(\phi ,-\sqrt{2}{\mathrm{e}}^{\frac{-\mathrm{i}\pi }{3}})\}$$$$\mathrm{Res}(\phi ,\sqrt{2}{\mathrm{e}}^{\frac{\mathrm{i}\pi }{3}})=\frac{4{\mathrm{e}}^{\frac{4\mathrm{i}\pi }{3}}+2{\mathrm{e}}^{\frac{\mathrm{i}2\pi }{3}}}{4\sqrt{2}{\mathrm{e}}^{\frac{\mathrm{i}\pi }{3}}\left(2\mathrm{i}\sin \left(\frac{2\pi }{3}\right)\right)}$$$$=\frac{2{\mathrm{e}}^{\frac{4\mathrm{i}\pi }{3}}+{\mathrm{e}}^{\frac{\mathrm{i}2\pi }{3}}}{4\sqrt{2}\left(\mathrm{i}\frac{\sqrt{3}}{2}\right)}\times {\mathrm{e}}^{-\frac{\mathrm{i}\pi }{3}}=\frac{2{\mathrm{e}}^{\mathrm{i}\pi }+{\mathrm{e}}^{\frac{\mathrm{i}\pi }{3}}}{2\mathrm{i}\sqrt{2}\sqrt{3}}=\frac{-2+{\mathrm{e}}^{\frac{\mathrm{i}\pi }{3}}}{2\mathrm{i}\sqrt{6}}$$$$\mathrm{Res}(\phi ,-\sqrt{2}{\mathrm{e}}^{\frac{-\mathrm{i}\pi }{3}})=\frac{4{\mathrm{e}}^{\frac{-4\mathrm{i}\pi }{3}}+2{\mathrm{e}}^{-\frac{2\mathrm{i}\pi }{3}}}{-2\sqrt{2}{\mathrm{e}}^{-\frac{\mathrm{i}\pi }{3}}({2\mathrm{e}}^{-\frac{2\mathrm{i}\pi }{3}}-{2\mathrm{e}}^{\frac{2\mathrm{i}\pi }{3}})}$$$$=\frac{4{\mathrm{e}}^{-\frac{4\mathrm{i}\pi }{3}}+2{\mathrm{e}}^{-\frac{2\mathrm{i}\pi }{3}}}{4\sqrt{2}\left(2\mathrm{isin}\left(\frac{2\pi }{3}\right)\right)}\times {\mathrm{e}}^{\frac{\mathrm{i}\pi }{3}}=\frac{2{\mathrm{e}}^{-\mathrm{i}\pi }+{\mathrm{e}}^{-\frac{\mathrm{i}\pi }{3}}}{2\sqrt{3}\left(2\mathrm{i}\times \frac{\sqrt{3}}{2}\right)}$$$$=\frac{-2+{\mathrm{e}}^{-\frac{\mathrm{i}\pi }{3}}}{2\mathrm{i}\sqrt{6}}....\mathrm{be}\mathrm{continued}....$$$$$$$$$$

Commented by mathmax by abdo last updated on 07/Jun/20

$$\mathrm{error}\mathrm{in}\mathrm{dx}\mathrm{i}\mathrm{will}\mathrm{delete}\mathrm{the}\mathrm{post}!$$

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