Question Number 175244 by infinityaction last updated on 24/Aug/22
$$\:{prove}\:{that} \\ $$$$\:\int_{\mathrm{0}} ^{{a}} \int_{\mathrm{0}} ^{\sqrt{{a}^{\mathrm{2}} −{x}^{\mathrm{2}} }} \frac{{dx}\:{dy}}{\left(\mathrm{1}+{e}^{{y}} \right)\sqrt{{a}^{\mathrm{2}} −{x}^{\mathrm{2}} −{y}^{\mathrm{2}} }}\:=\:\frac{\pi}{\mathrm{2}}\mathrm{log}\:\frac{\mathrm{2}{e}^{{a}} }{\mathrm{1}+{e}^{{a}} } \\ $$
Answered by Ar Brandon last updated on 24/Aug/22
![I=∫_0 ^a ∫_0 ^(√(a^2 −y^2 )) ((dxdy)/((1+e^y )(√(a^2 −x^2 −y^2 )))) =∫_0 ^a (1/(1+e^y ))∫_0 ^(√(a^2 −y^2 )) (dx/( (√((a^2 −y^2 )−x^2 ))))dy =∫_0 ^a (1/(1+e^y ))[arcsin((x/( (√(a^2 −y^2 )))))]_0 ^(√(a^2 −y^2 )) dy =(π/2)∫_0 ^a (1/(1+e^y ))dy=(π/2)∫_0 ^a (e^(−y) /(e^(−y) +1))dy =−(π/2)[ln(e^(−y) +1)]_0 ^a =(π/2)(ln2−ln(e^(−a) +1)) =(π/2)ln((2/(e^(−a) +1)))=(π/2)ln(((2e^a )/(1+e^a )))](Q175245.png)
$${I}=\int_{\mathrm{0}} ^{{a}} \int_{\mathrm{0}} ^{\sqrt{{a}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} }} \frac{{dxd}\mathrm{y}}{\left(\mathrm{1}+{e}^{\mathrm{y}} \right)\sqrt{{a}^{\mathrm{2}} −{x}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} }} \\ $$$$\:\:=\int_{\mathrm{0}} ^{{a}} \frac{\mathrm{1}}{\mathrm{1}+{e}^{\mathrm{y}} }\int_{\mathrm{0}} ^{\sqrt{{a}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} }} \frac{{dx}}{\:\sqrt{\left({a}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} \right)−{x}^{\mathrm{2}} }}{d}\mathrm{y} \\ $$$$\:\:=\int_{\mathrm{0}} ^{{a}} \frac{\mathrm{1}}{\mathrm{1}+{e}^{\mathrm{y}} }\left[\mathrm{arcsin}\left(\frac{{x}}{\:\sqrt{{a}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} }}\right)\right]_{\mathrm{0}} ^{\sqrt{{a}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} }} {d}\mathrm{y} \\ $$$$\:\:=\frac{\pi}{\mathrm{2}}\int_{\mathrm{0}} ^{{a}} \frac{\mathrm{1}}{\mathrm{1}+{e}^{\mathrm{y}} }{d}\mathrm{y}=\frac{\pi}{\mathrm{2}}\int_{\mathrm{0}} ^{{a}} \frac{{e}^{−\mathrm{y}} }{{e}^{−\mathrm{y}} +\mathrm{1}}{d}\mathrm{y} \\ $$$$\:\:=−\frac{\pi}{\mathrm{2}}\left[\mathrm{ln}\left({e}^{−\mathrm{y}} +\mathrm{1}\right)\right]_{\mathrm{0}} ^{{a}} =\frac{\pi}{\mathrm{2}}\left(\mathrm{ln2}−\mathrm{ln}\left({e}^{−{a}} +\mathrm{1}\right)\right) \\ $$$$\:\:=\frac{\pi}{\mathrm{2}}\mathrm{ln}\left(\frac{\mathrm{2}}{{e}^{−{a}} +\mathrm{1}}\right)=\frac{\pi}{\mathrm{2}}\mathrm{ln}\left(\frac{\mathrm{2}{e}^{{a}} }{\mathrm{1}+{e}^{{a}} }\right) \\ $$
Commented by infinityaction last updated on 24/Aug/22
$${thanks}\:{sir} \\ $$