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Question Number 135382 by Bird last updated on 12/Mar/21

$$find{\int }_0^1{x}^{n}ln(1-x^4)dxwithn$$$$integrnatural$$

Answered by Dwaipayan Shikari last updated on 12/Mar/21

$$x^4=t$$$$=\frac{1}{4}{\int }_0^1{t}^{\frac{n}{4}-\frac{3}{4}}log(1-t)dt$$$$I(a,b)={\int }_0^1{t}^{a-1}{(1-t)}^{b-1}dt=\frac{\mathrm{\Gamma }\left(a\right)\mathrm{\Gamma }\left(b\right)}{\mathrm{\Gamma }(a+b)}$$$$\frac{\partial }{\partial b}I(a,b)={\int }_0^1{t}^{a-1}{(1-t)}^{b-1}log(1-t)=\frac{\mathrm{\Gamma }(a+b){\mathrm{\Gamma }}^{\prime }\left(b\right)\mathrm{\Gamma }\left(a\right)-{\mathrm{\Gamma }}^{\prime }(a+b)\mathrm{\Gamma }\left(a\right)\mathrm{\Gamma }\left(b\right))}{{\mathrm{\Gamma }}^2(a+b)}$$$$=\frac{\mathrm{\Gamma }\left(a\right)\mathrm{\Gamma }\left(b\right)}{\mathrm{\Gamma }(a+b)}(\psi \left(b\right)-\psi (a+b))$$$$b=1$$$$a=\frac{n+1}{4}$$$$\frac{1}{4}I(\frac{n+1}{4},1)=\frac{1}{4}{\int }_0^1{t}^{\frac{n}{4}-\frac{3}{4}}log(1-t)dt=\frac{\mathrm{\Gamma }\left(\frac{n+1}{4}\right)}{\mathrm{\Gamma }(n+\frac{1}{4}+1)}(-\gamma -\psi \left(\frac{n+5}{4}\right))$$$$=-\frac{4}{4n+1}(\gamma +\psi \left(\frac{n+5}{4}\right))$$

Commented by mathmax by abdo last updated on 12/Mar/21

$$\mathrm{thank}\mathrm{you}\mathrm{sir}$$

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