Question Number 126238 by mohammad17 last updated on 18/Dec/20 | ||
Answered by Olaf last updated on 18/Dec/20 | ||
$$\Omega\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \int_{{y}^{\mathrm{2}} } ^{\mathrm{1}} \frac{{e}^{{x}} }{\:\sqrt{{x}}}{dxdy} \\ $$$$\mathrm{Let}\:{u}\:=\:\sqrt{{x}} \\ $$$$\Omega\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \int_{{y}} ^{\mathrm{1}} \frac{{e}^{{u}^{\mathrm{2}} } }{\:{u}}\left(\mathrm{2}{udu}\right){dy} \\ $$$$\Omega\:=\:\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \int_{{y}} ^{\mathrm{1}} {e}^{{u}^{\mathrm{2}} } {dudy} \\ $$$$\Omega\:=\:\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \left[\frac{\sqrt{\pi}}{\mathrm{2}}\mathrm{erfi}\left({u}\right)\right]_{{y}} ^{\mathrm{1}} {dy} \\ $$$$\Omega\:=\:\sqrt{\pi}\int_{\mathrm{0}} ^{\mathrm{1}} \left[\mathrm{erfi}\left(\mathrm{1}\right)−\mathrm{erfi}\left({y}\right)\right]{dy} \\ $$$$\mathrm{erfi}\left({z}\right)\:=\:\frac{\mathrm{2}}{\:\sqrt{\pi}}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right){n}!}{z}^{\mathrm{2}{n}+\mathrm{1}} \\ $$$$\Omega\:=\:\mathrm{2}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right){n}!}\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−{y}^{\mathrm{2}{n}+\mathrm{1}} \right){dy} \\ $$$$\Omega\:=\:\mathrm{2}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right){n}!}\left[{y}−\frac{{y}^{\mathrm{2}{n}+\mathrm{2}} }{\mathrm{2}{n}+\mathrm{2}}\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$\Omega\:=\:\mathrm{2}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right){n}!}\left[\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{2}}\right] \\ $$$$\Omega\:=\:\mathrm{2}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right){n}!}\left[\frac{\mathrm{2}{n}+\mathrm{1}}{\mathrm{2}{n}+\mathrm{2}}\right] \\ $$$$\Omega\:=\:\mathrm{2}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{2}\right){n}!} \\ $$$$\Omega\:=\:\mathrm{2}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{2}\left({n}+\mathrm{1}\right){n}!} \\ $$$$\Omega\:=\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)!} \\ $$$$\Omega\:=\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}!}−\mathrm{1}\:=\:{e}−\mathrm{1} \\ $$ | ||
Answered by Olaf last updated on 18/Dec/20 | ||
$$\Omega\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \int_{\mathrm{0}} ^{\sqrt{{y}−{y}^{\mathrm{2}} }} \frac{{dxdy}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} −{y}^{\mathrm{2}} }} \\ $$$$\Omega\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \int_{\mathrm{0}} ^{\sqrt{{y}−{y}^{\mathrm{2}} }} \left(\frac{{dx}}{\:\sqrt{\left(\mathrm{1}−{y}^{\mathrm{2}} \right)−{x}^{\mathrm{2}} }}\right){dy} \\ $$$$\int\frac{{dx}}{\:\sqrt{{a}^{\mathrm{2}} −{x}^{\mathrm{2}} }}\:=\:\mathrm{arctan}\left(\frac{{x}}{\:\sqrt{{a}^{\mathrm{2}} −{x}^{\mathrm{2}} }}\right) \\ $$$$\mathrm{here}\:{a}^{\mathrm{2}} \:=\:\mathrm{1}−{y}^{\mathrm{2}} \\ $$$$\Omega\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \left[\mathrm{arctan}\left(\frac{{x}}{\:\sqrt{\left(\mathrm{1}−{y}^{\mathrm{2}} \right)−{x}^{\mathrm{2}} }}\right)\right]_{\mathrm{0}} ^{\sqrt{{y}−{y}^{\mathrm{2}} }} {dy} \\ $$$$\Omega\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{arctan}\left(\frac{\sqrt{{y}−{y}^{\mathrm{2}} }}{\:\sqrt{\left(\mathrm{1}−{y}^{\mathrm{2}} \right)−\left({y}−{y}^{\mathrm{2}} \right)}}\right){dy} \\ $$$$\Omega\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{arctan}\left(\frac{\sqrt{{y}−{y}^{\mathrm{2}} }}{\:\sqrt{\mathrm{1}−{y}}}\right){dy} \\ $$$$\Omega\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{arctan}\left(\sqrt{{y}}\right){dy} \\ $$$$\mathrm{Let}\:{u}\:=\:\sqrt{{y}} \\ $$$$\Omega\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{arctan}{u}\left(\mathrm{2}{udu}\right) \\ $$$$\Omega\:=\:\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} {u}.\mathrm{arctan}{u}.{du} \\ $$$$\Omega\:=\:\mathrm{2}\left\{\left[\frac{{u}^{\mathrm{2}} }{\mathrm{2}}\mathrm{arctan}{u}\right]_{\mathrm{0}} ^{\mathrm{1}} −\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{u}^{\mathrm{2}} }{\mathrm{2}}.\frac{{du}}{\mathrm{1}+{u}^{\mathrm{2}} }\right\} \\ $$$$\Omega\:=\:\frac{\pi}{\mathrm{4}}−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{u}^{\mathrm{2}} }{\mathrm{1}+{u}^{\mathrm{2}} }{du} \\ $$$$\Omega\:=\:\frac{\pi}{\mathrm{4}}−\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{1}+{u}^{\mathrm{2}} }\right){du} \\ $$$$\Omega\:=\:\frac{\pi}{\mathrm{4}}−\left[{u}−\mathrm{arctan}{u}\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$\Omega\:=\:\frac{\pi}{\mathrm{4}}−\left[\mathrm{1}−\frac{\pi}{\mathrm{4}}\right] \\ $$$$\Omega\:=\:\frac{\pi}{\mathrm{2}}−\mathrm{1} \\ $$ | ||
Answered by Ar Brandon last updated on 19/Dec/20 | ||
$$\mathcal{I}=\int_{\mathrm{0}} ^{\mathrm{1}} \int_{\mathrm{0}} ^{\sqrt{\mathrm{y}−\mathrm{y}^{\mathrm{2}} }} \frac{\mathrm{dxdy}}{\:\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} }}=\int_{\mathrm{0}} ^{\mathrm{1}} \int_{\mathrm{0}} ^{\sqrt{\mathrm{y}−\mathrm{y}^{\mathrm{2}} }} \frac{\mathrm{dxdy}}{\:\sqrt{\left(\mathrm{1}−\mathrm{y}^{\mathrm{2}} \right)−\mathrm{x}^{\mathrm{2}} }} \\ $$$$\:\:\:=\int_{\mathrm{0}} ^{\mathrm{1}} \left[\mathrm{Arcsin}\left(\frac{\mathrm{x}}{\:\sqrt{\mathrm{1}−\mathrm{y}^{\mathrm{2}} }}\right)\right]_{\mathrm{0}} ^{\sqrt{\mathrm{y}−\mathrm{y}^{\mathrm{2}} }} \mathrm{dy}=\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{Arcsin}\left(\sqrt{\frac{\mathrm{y}−\mathrm{y}^{\mathrm{2}} }{\mathrm{1}−\mathrm{y}^{\mathrm{2}} }}\right)\mathrm{dy} \\ $$$$\:\:\:=\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{Arcsin}\left(\sqrt{\frac{\mathrm{y}}{\mathrm{y}+\mathrm{1}}}\right)\mathrm{dy} \\ $$$$\mathrm{By}\:\mathrm{part} \\ $$$$\begin{cases}{\mathrm{u}\left(\mathrm{y}\right)=\mathrm{Arcsin}\left(\sqrt{\frac{\mathrm{y}}{\mathrm{y}+\mathrm{1}}}\right)}\\{\mathrm{v}'\left(\mathrm{y}\right)=\mathrm{1}}\end{cases}\Rightarrow\begin{cases}{\mathrm{u}'\left(\mathrm{y}\right)=\frac{\mathrm{1}}{\left(\mathrm{y}+\mathrm{1}\right)^{\mathrm{2}} }\centerdot\frac{\mathrm{1}}{\mathrm{2}}\centerdot\sqrt{\frac{\mathrm{y}+\mathrm{1}}{\mathrm{y}}}\centerdot\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−\frac{\mathrm{y}}{\mathrm{y}+\mathrm{1}}}}}\\{\mathrm{v}\left(\mathrm{y}\right)=\mathrm{y}}\end{cases} \\ $$$$\Rightarrow\begin{cases}{\mathrm{u}'\left(\mathrm{y}\right)=\frac{\mathrm{1}}{\left(\mathrm{y}+\mathrm{1}\right)^{\mathrm{2}} }\centerdot\frac{\mathrm{1}}{\mathrm{2}}\centerdot\sqrt{\frac{\mathrm{y}+\mathrm{1}}{\mathrm{y}}}\centerdot\sqrt{\frac{\mathrm{y}+\mathrm{1}}{\mathrm{1}}}}\\{\mathrm{v}\left(\mathrm{y}\right)=\mathrm{y}}\end{cases}\Rightarrow\begin{cases}{\mathrm{u}'\left(\mathrm{y}\right)=\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{y}+\mathrm{1}\right)\sqrt{\mathrm{y}}}}\\{\mathrm{v}\left(\mathrm{y}\right)=\mathrm{y}}\end{cases} \\ $$$$\mathcal{I}=\left\{\mathrm{yArcsin}\left(\sqrt{\frac{\mathrm{y}}{\mathrm{y}+\mathrm{1}}}\right)−\int\frac{\mathrm{y}}{\mathrm{2}\left(\mathrm{y}+\mathrm{1}\right)\sqrt{\mathrm{y}}}\mathrm{dy}\right\}_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$\:\:\:=\mathrm{Arcsin}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\sqrt{\mathrm{y}}}{\mathrm{y}+\mathrm{1}}\mathrm{dy} \\ $$$$\mathrm{Let}\:\mathrm{y}=\mathrm{t}^{\mathrm{2}} \:\Rightarrow\:\mathrm{dy}=\mathrm{2tdt} \\ $$$$\mathcal{I}=\frac{\pi}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{t}}{\mathrm{t}^{\mathrm{2}} +\mathrm{1}}\centerdot\mathrm{2tdt}=\frac{\pi}{\mathrm{4}}−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{t}^{\mathrm{2}} }{\mathrm{t}^{\mathrm{2}} +\mathrm{1}}\mathrm{dt} \\ $$$$\:\:\:=\frac{\pi}{\mathrm{4}}−\int_{\mathrm{0}} ^{\mathrm{1}} \left\{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{t}^{\mathrm{2}} +\mathrm{1}}\right\}\mathrm{dt}=\frac{\pi}{\mathrm{4}}−\left[\mathrm{t}−\mathrm{Arctan}\left(\mathrm{t}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$\:\:\:=\frac{\pi}{\mathrm{4}}−\mathrm{1}+\frac{\pi}{\mathrm{4}}=\frac{\pi}{\mathrm{2}}−\mathrm{1} \\ $$ | ||