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Question Number 175244 by infinityaction last updated on 24/Aug/22

$$provethat$$$${\int }_0^a{\int }_0^{\sqrt{a^2-x^2}}\frac{dxdy}{(1+e^y)\sqrt{a^2-x^2-y^2}}=\frac{\pi }{2}\log \frac{2e^a}{1+e^a}$$

Answered by Ar Brandon last updated on 24/Aug/22

$$I={\int }_0^a{\int }_0^{\sqrt{a^2-{\mathrm{y}}^2}}\frac{dxd\mathrm{y}}{(1+e^{\mathrm{y}})\sqrt{a^2-x^2-{\mathrm{y}}^2}}$$$$={\int }_0^a\frac{1}{1+e^{\mathrm{y}}}{\int }_0^{\sqrt{a^2-{\mathrm{y}}^2}}\frac{dx}{\sqrt{(a^2-{\mathrm{y}}^2)-x^2}}d\mathrm{y}$$$$={\int }_0^a\frac{1}{1+e^{\mathrm{y}}}{\left[\arcsin \left(\frac{x}{\sqrt{a^2-{\mathrm{y}}^2}}\right)\right]}_0^{\sqrt{a^2-{\mathrm{y}}^2}}d\mathrm{y}$$$$=\frac{\pi }{2}{\int }_0^a\frac{1}{1+e^{\mathrm{y}}}d\mathrm{y}=\frac{\pi }{2}{\int }_0^a\frac{e^{-\mathrm{y}}}{e^{-\mathrm{y}}+1}d\mathrm{y}$$$$=-\frac{\pi }{2}{\left[\ln (e^{-\mathrm{y}}+1)\right]}_0^a=\frac{\pi }{2}(\ln 2-\ln (e^{-a}+1))$$$$=\frac{\pi }{2}\ln \left(\frac{2}{e^{-a}+1}\right)=\frac{\pi }{2}\ln \left(\frac{2e^a}{1+e^a}\right)$$

Commented by infinityaction last updated on 24/Aug/22

$$thankssir$$

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