let-f-x-0-pi-2-ln-1-xcost-dt-with-x-lt-1-1-developp-f-at-integr-serie-2-find-a-explicit-form-of-f-x-3-find-the-values-of-integrals-0-pi-2-ln-1-cost-dt-and-0-pi-2-ln-1-cost-dt-4-

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Question Number 59247 by maxmathsup by imad last updated on 06/May/19

$$letf\left(x\right)={\int }_0^{\frac{\pi }{2}}ln(1-xcost)dtwith\mid x\mid <1$$$$1)developpfatintegrserie$$$$2)findaexplicitformoff\left(x\right)$$$$3)findthevaluesofintegrals{\int }_0^{\frac{\pi }{2}}ln(1-cost)dtand{\int }_0^{\frac{\pi }{2}}ln(1+cost)dt$$$$4)calculate{U}_n={\int }_0^{\frac{\pi }{2}}ln(1-\frac{2}{n}cost)dtwithnintegrandn⩾2$$$$studytheconvergenceof{U}_{n}and\mathrm{\Sigma }{U}_n$$

Commented by Mr X pcx last updated on 09/May/19

$$2)wehave{f}^{{}^{\prime }}\left(x\right)={\int }_0^{\frac{\pi }{2}}\frac{-cost}{1-xcost}dt$$$$forx\ne 0f^{{}^{\prime }}\left(x\right)=\frac{1}{x}{\int }_0^{\frac{\pi }{2}}\frac{1-xcost-1}{1-xcost}dt$$$$=\frac{\pi }{2x}-\frac{1}{x}{\int }_0^{\frac{\pi }{2}}\frac{dt}{1-xcost}but$$$$chang.tan\left(\frac{t}{2}\right)=ugive$$$${\int }_0^{\frac{\pi }{2}}\frac{dt}{1-xcost}={\int }_0^1\frac{1}{1-x\frac{1-u^2}{1+u^2}}\frac{2du}{1+u^2}$$$$={\int }_0^1\frac{2du}{1+u^2-x+{xu}^2}={\int }_0^1\frac{2du}{(1+x)u^2+1-x}$$$$=\frac{2}{1+x}{\int }_0^1\frac{du}{u^2+\frac{1-x}{1+x}}$$$${=}_{u=\sqrt{\frac{1-x}{1+x}}\alpha }\frac{2}{1+x}{\int }_0^{\sqrt{\frac{1+x}{1-x}}}\frac{1}{\frac{1-x}{1+x}(1+{\alpha }^2)}\sqrt{\frac{1-x}{1+x}}d\alpha$$$$=\frac{2}{1-x}\frac{\sqrt{1-x}}{\sqrt{1+x}}{\int }_0^{\sqrt{\frac{1+x}{1-x}}}\frac{d\alpha }{1+{\alpha }^2}$$$$=\frac{2}{\sqrt{1-x^2}}{\left[arctan\left(\alpha \right)\right]}_0^{\sqrt{\frac{1+x}{1-x}}}$$$$=\frac{2}{\sqrt{1-x^2}}arctan\left(\sqrt{\frac{1+x}{1-x}}\right)\Rightarrow$$$$f^{{}^{\prime }}\left(x\right)=\frac{\pi }{2x}-\frac{2}{x\sqrt{1-x^2}}arctan\left(\sqrt{\frac{1+x}{1-x}}\right)\Rightarrow$$$$f\left(x\right)=\frac{\pi }{2}ln\mid x\mid -\int \frac{2}{x\sqrt{1-x^2}}arctan\left(\sqrt{\frac{1+x}{1-x}}\right)dx+c$$$$\int \frac{2}{x\sqrt{1-x^2}}arctan\left(\sqrt{\frac{1+x}{1-x}}\right)dx$$$${=}_{x=cost}\int \frac{2}{cost.sint}arctan\left(\sqrt{\frac{2{cos}^2\left(\frac{t}{2}\right)}{2{sin}^2\left(\frac{t}{2}\right)}}\right)(-sint)dt$$$$=-2\int \frac{1}{sint}arctan\left(\frac{1}{tan\left(\frac{t}{2}\right)}\right)dt$$$$=-2\int \frac{1}{sint}(\frac{\pi }{2}-\frac{t}{2})dt$$$$=-\pi \int \frac{dt}{sint}+\int \frac{t}{sint}dt\dots .becontinued\dots$$$$$$

Commented by maxmathsup by imad last updated on 09/May/19

$$erroratline16wehave\int \frac{2}{x\sqrt{1-x^2}}arctan\left(\sqrt{\frac{1+x}{1-x}}\right)dx$$$$=-2\int \frac{1}{cost}(\frac{\pi }{2}-\frac{t}{2})dt+c=-\pi \int \frac{dt}{cost}+\int \frac{t}{cost}dt+c\Rightarrow$$$$f\left(x\right)=\frac{\pi }{2}ln\mid x\mid +\pi \int \frac{dt}{cost}-\int \frac{t}{cost}dt+c\dots .$$

Answered by Mr X pcx last updated on 09/May/19

$$wehave{ln}^{{}^{\prime }}(1-u)=\frac{-1}{1-u}=-\sum _{n=0}^{\mathrm{\infty }}{u}^n$$$$with\mid u\mid <1\Rightarrow ln(1-u)=-\sum _{n=0}^{\mathrm{\infty }}\frac{u^{n+1}}{n+1}$$$$=-\sum _{n=1}^{\mathrm{\infty }}\frac{u^n}{n}wehave\mid xcost\mid <1\Rightarrow$$$$ln(1-xcost)=-\sum _{n=1}^{\mathrm{\infty }}\frac{x^n{cos}^{n}t}{n}\Rightarrow$$$${\int }_0^{\frac{\pi }{2}}ln(1-xcost)dt=-\sum _{n=1}^{\mathrm{\infty }}\frac{x^n}{n}{\int }_0^{\frac{\pi }{2}}{cos}^n\left(t\right)$$$$=-\sum _{n=1}^{\mathrm{\infty }}\frac{A_n}{n}{x}^n{withA}_n={\int }_0^{\frac{\pi }{2}}{cos}^{n}tdt$$$$\left(integralofwallis\right).$$

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