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Question Number 172953 by Mathspace last updated on 03/Jul/22

$$find{\int }_0^1\sqrt{1-x^4}lnxdx$$

Answered by Ar Brandon last updated on 03/Jul/22

$$\mathrm{\Omega }={\int }_0^1\sqrt{1-x^4}\ln xdx,x=t^{\frac{1}{4}}$$$$=\frac{1}{16}{\int }_0^1{t}^{-\frac{3}{4}}\sqrt{1-t}\ln tdt=\frac{1}{16}\cdot \frac{\partial }{\partial \alpha }{\mid }_{\alpha =\frac{1}{4}}{\int }_0^1{t}^{\alpha -1}\sqrt{1-t}dt$$$$=\frac{1}{16}\cdot \frac{\partial }{\partial \alpha }{\mid }_{\alpha =\frac{1}{4}}\beta (\alpha ,\frac{3}{2})=\frac{1}{16}\cdot \frac{\partial }{\partial \alpha }\frac{\mathrm{\Gamma }\left(\alpha \right)\mathrm{\Gamma }\left(\frac{1}{2}\right)\frac{1}{2}}{\mathrm{\Gamma }(\alpha +\frac{3}{2})}{}_{\alpha =\frac{1}{4}}$$$$=\frac{\sqrt{\pi }}{32}\cdot {(\frac{{\mathrm{\Gamma }}^{\prime }\left(\alpha \right)}{\mathrm{\Gamma }(\alpha +\frac{3}{2})}-\frac{\mathrm{\Gamma }\left(\alpha \right)\psi (\alpha +\frac{3}{2})}{\mathrm{\Gamma }(\alpha +\frac{3}{2})})}_{\alpha =\frac{1}{4}}=\frac{\sqrt{\pi }}{32}\left(\frac{\mathrm{\Gamma }\left(\frac{1}{4}\right)\psi \left(\frac{1}{4}\right)-\mathrm{\Gamma }\left(\frac{1}{4}\right)\psi \left(\frac{7}{4}\right)}{\mathrm{\Gamma }\left(\frac{7}{4}\right)}\right)$$$$=\frac{\sqrt{\pi }}{32}\cdot \frac{\mathrm{\Gamma }\left(\frac{1}{4}\right)}{\frac{3}{4}\mathrm{\Gamma }\left(\frac{3}{4}\right)}(\psi \left(\frac{1}{4}\right)-\psi \left(\frac{3}{4}\right)-\frac{4}{3})$$$$=\frac{\sqrt{\pi }}{24}\cdot \frac{{\mathrm{\Gamma }}^2\left(\frac{1}{4}\right)}{\pi \sqrt{2}}(-\pi \cot \left(\frac{\pi }{4}\right)-\frac{4}{3})=-\frac{\sqrt{2\pi }}{48\pi }\left(\frac{3\pi +4}{3}\right){\mathrm{\Gamma }}^2\left(\frac{1}{4}\right)$$

Commented by Ar Brandon last updated on 04/Jul/22

Commented by Ar Brandon last updated on 04/Jul/22

Commented by Tawa11 last updated on 06/Jul/22

$$\mathrm{Great}\mathrm{sir}$$

Answered by Mathspace last updated on 04/Jul/22

$$\mathrm{{\rm Y}}={\int }_0^1\sqrt{1-x^4}lnxdxchangement$$$$x^4=tgive$$$$\mathrm{{\rm Y}}={\int }_0^1\sqrt{1-t}ln\left(t^{\frac{1}{4}}\right).\frac{1}{4}{t}^{\frac{1}{4}-1}dt$$$$=\frac{1}{16}{\int }_0^1{t}^{\frac{1}{4}-1}{(1-t)}^{\frac{1}{2}}lntdt$$$$letf\left(a\right)={\int }_0^1{t}^{a-1}{(1-t)}^{\frac{1}{2}}dt\Rightarrow$$$$f^{{}^{\prime }}\left(a\right)={\int }_0^1{t}^{a-1}{(1-t)}^{\frac{1}{2}}lnt\Rightarrow$$$${\int }_0^1{t}^{\frac{1}{4}-1}{(1-t)}^{\frac{1}{2}}lntdt=f^{{}^{\prime }}\left(\frac{1}{4}\right)$$$$wehavef\left(a\right)=B(a,\frac{3}{2})$$$$=\frac{\mathrm{\Gamma }\left(a\right).\mathrm{\Gamma }\left(\frac{3}{2}\right)}{\mathrm{\Gamma }(a+\frac{3}{2})}$$$$\mathrm{\Gamma }\left(\frac{3}{2}\right)=\mathrm{\Gamma }(\frac{1}{2}+1)=\frac{1}{2}\mathrm{\Gamma }\left(\frac{1}{2}\right)=\frac{\sqrt{\pi }}{2}$$$$\mathrm{\Gamma }(a+\frac{3}{2})=\mathrm{\Gamma }(a+\frac{1}{2}+1)$$$$=(a+\frac{1}{2})\mathrm{\Gamma }(a+\frac{1}{2})$$$$f\left(a\right)=\frac{\sqrt{\pi }}{2}\times \frac{\mathrm{\Gamma }\left(a\right)}{\mathrm{\Gamma }(a+\frac{3}{2})}\Rightarrow$$$$f^{{}^{\prime }}\left(a\right)=\frac{\sqrt{\pi }}{2}.\frac{{\mathrm{\Gamma }}^{{}^{\prime }}\left(a\right)\mathrm{\Gamma }(a+\frac{3}{2})-\mathrm{\Gamma }\left(a\right){\mathrm{\Gamma }}^{{}^{\prime }}(a+\frac{3}{2})}{{\mathrm{\Gamma }}^2(a+\frac{3}{2})}$$$$\Rightarrow f^{{}^{\prime }}\left(\frac{1}{4}\right)=\frac{\sqrt{\pi }}{2}.\frac{{\mathrm{\Gamma }}^{{}^{\prime }}\left(\frac{1}{4}\right)\mathrm{\Gamma }(\frac{1}{4}+\frac{3}{2})-\mathrm{\Gamma }\left(\frac{1}{4}\right){\mathrm{\Gamma }}^{{}^{\prime }}(\frac{1}{4}+\frac{3}{2})}{{\mathrm{\Gamma }}^2(\frac{1}{4}+\frac{3}{2})}$$$$=\frac{\sqrt{\pi }}{2}.\frac{{\mathrm{\Gamma }}^{{}^{\prime }}\left(\frac{1}{4}\right)\mathrm{\Gamma }\left(\frac{7}{4}\right)-\mathrm{\Gamma }\left(\frac{1}{4}\right){\mathrm{\Gamma }}^{{}^{\prime }}\left(\frac{7}{4}\right)}{{\mathrm{\Gamma }}^2\left(\frac{7}{4}\right)}$$$$and\mathrm{{\rm Y}}=\frac{1}{16}{f}^{{}^{\prime }}\left(\frac{1}{4}\right)$$

Commented by Tawa11 last updated on 06/Jul/22

$$\mathrm{Great}\mathrm{sir}$$

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