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Question Number 141914 by mnjuly1970 last updated on 24/May/21

$$$$$$provethat::$$$$\varphi :={\int }_0^{\pi }\frac{x}{{\left(sin\left(x\right)\right)}^{\frac{1}{2}}}dx=\sqrt{\frac{\pi }{8}}{\mathrm{\Gamma }}^2\left(\frac{1}{4}\right)...✓$$$$.........$$

Answered by qaz last updated on 24/May/21

$$\varphi ={\int }_0^{\pi }\frac{\mathrm{x}}{\sqrt{\sin \mathrm{x}}}\mathrm{dx}....................\left(1\right)$$$$={\int }_0^{\pi }\frac{\pi -\mathrm{x}}{\sqrt{\sin \mathrm{x}}}\mathrm{dx}$$$$=\pi {\int }_0^{\pi }\frac{\mathrm{dx}}{\sqrt{\sin \mathrm{x}}}-{\int }_0^{\pi }\frac{\mathrm{x}}{\sqrt{\sin \mathrm{x}}}\mathrm{dx}.......\left(2\right)$$$$\left(1\right)+\left(2\right)$$$$\Rightarrow \varphi =\frac{\pi }{2}{\int }_0^{\pi }\frac{\mathrm{dx}}{\sqrt{\sin \mathrm{x}}}=\pi {\int }_0^{\pi /2}\frac{\mathrm{dx}}{\sqrt{\sin \mathrm{x}}}$$$$=\frac{\pi }{2}\frac{\mathrm{\Gamma }\left(\frac{1}{4}\right)\mathrm{\Gamma }\left(\frac{1}{2}\right)}{\mathrm{\Gamma }\left(\frac{3}{4}\right)}$$$$=\frac{\pi \sqrt{\pi }}{2}\frac{{\mathrm{\Gamma }}^2\left(\frac{1}{4}\right)}{\pi }\sin \frac{\pi }{4}$$$$=\sqrt{\frac{\pi }{8}}{\mathrm{\Gamma }}^2\left(\frac{1}{4}\right)$$

Answered by Dwaipayan Shikari last updated on 24/May/21

$${\int }_0^{\pi }\frac{x}{\sqrt{sinx}}dx={\int }_0^{\pi }\frac{\pi -x}{\sqrt{sinx}}dx=\frac{\pi }{2}{\int }_0^{\pi }\frac{1}{\sqrt{sin\left(x\right)}}dx=\pi .\frac{\mathrm{\Gamma }\left(\frac{1}{4}\right)\mathrm{\Gamma }\left(\frac{1}{2}\right)}{2\mathrm{\Gamma }\left(\frac{3}{4}\right)}$$$$=\sqrt{\frac{\pi }{2}}\frac{{\mathrm{\Gamma }}^2\left(\frac{1}{4}\right)}{2}=\sqrt{\frac{\pi }{8}}{\mathrm{\Gamma }}^2\left(\frac{1}{4}\right)$$$$$$

Answered by mathmax by abdo last updated on 24/May/21

$$\mathrm{\Phi }={\int }_0^{\pi }\frac{\mathrm{x}}{\sqrt{\mathrm{sinx}}}\mathrm{dx}\Rightarrow \mathrm{\Phi }{=}_{\mathrm{x}=\pi -\mathrm{t}}-{\int }_0^{\pi }\frac{\pi -\mathrm{t}}{\sqrt{\mathrm{sint}}}(-\mathrm{dt})={\int }_0^{\pi }\frac{\pi \mathrm{dt}}{\sqrt{\mathrm{sint}}}-\mathrm{\Phi }\Rightarrow$$$$2\mathrm{\Phi }=\pi {\int }_0^{\pi }\frac{\mathrm{dt}}{\sqrt{\mathrm{sint}}}\Rightarrow \mathrm{\Phi }=\frac{\pi }{2}{\int }_0^{\pi }\frac{\mathrm{dt}}{\sqrt{\mathrm{sint}}}$$$$\mathrm{we}\mathrm{have}{\int }_0^{\pi }\frac{\mathrm{dt}}{\sqrt{\mathrm{sint}}}={\int }_0^{\frac{\pi }{2}}\frac{\mathrm{dt}}{\sqrt{\mathrm{sint}}}+{\int }_{\frac{\pi }{2}}^{\pi }\frac{\mathrm{dt}}{\sqrt{\mathrm{sint}}}(\to \mathrm{t}=\frac{\pi }{2}+\mathrm{u})$$$$={\int }_0^{\frac{\pi }{2}}\frac{\mathrm{dt}}{\sqrt{\mathrm{sint}}}+{\int }_0^{\frac{\pi }{2}}\frac{\mathrm{du}}{\sqrt{\mathrm{cosu}}}\mathrm{but}$$$${\int }_0^{\frac{\pi }{2}}\frac{\mathrm{dt}}{\sqrt{\mathrm{sint}}}={\int }_0^{\frac{\pi }{2}}{\left(\mathrm{cost}\right)}^0{\left(\mathrm{sint}\right)}^{-\frac{1}{2}}\mathrm{dt}$$$$2\mathrm{p}-1=0\Rightarrow \mathrm{p}=\frac{1}{2}\mathrm{and}2\mathrm{q}-1=-\frac{1}{2}\Rightarrow 2\mathrm{q}=\frac{1}{2}\Rightarrow \mathrm{q}=\frac{1}{4}\Rightarrow$$$${\int }_0^{\frac{\pi }{2}}\frac{\mathrm{dt}}{\sqrt{\mathrm{sint}}}={\int }_0^{\frac{\pi }{2}}{\left(\mathrm{cost}\right)}^{2.\frac{1}{2}-1}.{\left(\mathrm{sint}\right)}^{2.\frac{1}{4}-1}\mathrm{dt}=\frac{1}{2}\mathrm{B}(\frac{1}{2},\frac{1}{4})$$$$=\frac{1}{2}\frac{\mathrm{\Gamma }\left(\frac{1}{2}\right).\mathrm{\Gamma }\left(\frac{1}{4}\right)}{\mathrm{\Gamma }\left(\frac{3}{4}\right)}$$$$\mathrm{\Gamma }\left(\frac{1}{4}\right).\mathrm{\Gamma }(1-\frac{1}{4})=\frac{\pi }{\sin \left(\frac{\pi }{4}\right)}=\pi \sqrt{2}\Rightarrow \mathrm{\Gamma }\left(\frac{3}{4}\right)=\frac{\pi \sqrt{2}}{\mathrm{\Gamma }\left(\frac{1}{4}\right)}\Rightarrow$$$${\int }_0^{\frac{\pi }{2}}\frac{\mathrm{dt}}{\sqrt{\mathrm{sint}}}=\frac{\sqrt{\pi }}{2}\times \pi \sqrt{2}.{\mathrm{\Gamma }}^2\left(\frac{1}{4}\right)=\frac{\pi \sqrt{\pi }}{\sqrt{2}}{\mathrm{\Gamma }}^2\left(\frac{1}{4}\right)\mathrm{also}$$$${\int }_0^{\frac{\pi }{2}}\frac{\mathrm{dt}}{\sqrt{\mathrm{cost}}}=\frac{\pi \sqrt{\pi }}{\sqrt{2}}{\mathrm{\Gamma }}^2\left(\frac{1}{4}\right)\Rightarrow {\int }_0^{\pi }\frac{\mathrm{dt}}{\sqrt{\mathrm{sint}}}=\frac{2\pi \sqrt{\pi }}{\sqrt{2}}{\mathrm{\Gamma }}^2\left(\frac{1}{4}\right)\Rightarrow$$$$\mathrm{\Phi }=\pi \sqrt{2\pi }.{\mathrm{\Gamma }}^2\left(\frac{1}{4}\right)$$

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