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Question Number 32365 by prof Abdo imad last updated on 23/Mar/18

$$letF\left(x\right)={\int }_0^{\pi }ln(1+xcos\theta )d\theta .with\mid x\mid <1$$$$findF\left(x\right).$$

Commented by prof Abdo imad last updated on 25/Mar/18

$$wehave{F}^{{}^{\prime }}\left(x\right)={\int }_0^{\pi }\frac{cos\theta }{1+xcos\theta }d\theta andforx\ne o$$$$\frac{dF}{dx}\left(x\right)=\frac{1}{x}{\int }_0^{\pi }\frac{xco\theta +1-1}{1+xcos\theta }d\theta$$$$=\frac{\pi }{x}-\frac{1}{x}{\int }_0^{\pi }\frac{d\theta }{1+xcos\theta }butch.tan\left(\frac{\theta }{2}\right)=tgive$$$${\int }_0^{\pi }\frac{d\theta }{1+xcos\theta }={\int }_0^{\mathrm{\infty }}\frac{1}{1+x\frac{1-t^2}{1+t^2}}\frac{2dt}{1+t^2}$$$$={\int }_0^{\mathrm{\infty }}\frac{2dt}{1+t^2+x-{xt}^2}={\int }_0^{\mathrm{\infty }}\frac{2dt}{1+x+(1-x)t^2}$$$$=\frac{1}{1+x}{\int }_0^{\mathrm{\infty }}\frac{2dt}{1+\frac{1-x}{1+x}{t}^2}$$$${=}_{u=\sqrt{\frac{1-x}{1+x}}t}\frac{1}{1+x}{\int }_0^{\mathrm{\infty }}\frac{2dt}{1+u^2}\frac{\sqrt{1+x}}{\sqrt{1-x}}du$$$$=\frac{2}{\sqrt{1-x^2}}\frac{\pi }{2}=\frac{\pi }{\sqrt{1-x^2}}$$$$\frac{dF}{dx}\left(x\right)=\frac{\pi }{x}-\frac{\pi }{x\sqrt{1-x^2}}\Rightarrow$$$$F\left(x\right)=\pi ln\mid x\mid -\pi \int \frac{dx}{x\sqrt{1-x^2}}+\lambda ch.x=sintgive$$$$\int \frac{dx}{x\sqrt{1-x^2}}=\int \frac{cost}{sintcost}dt=\int \frac{dt}{sint}$$$$ch.tan\left(\frac{t}{2}\right)=u\Rightarrow \int \frac{dt}{sint}=\int \frac{1}{\frac{2u}{1+u^2}}\frac{2du}{1+u^2}$$$$=\int \frac{du}{u}=ln\mid u\mid =ln\mid tan\left(\frac{t}{2}\right)\mid =ln\mid tan\left(\frac{arcsinx}{2}\right)\mid$$$$F\left(x\right)=\pi ln\mid x\mid -\pi ln\mid tan\left(\frac{arcsinx}{2}\right)\mid +\lambda$$$$\lambda =F\left(1\right)={\int }_0^{\pi }ln(1+cos\theta )d\theta$$$$$$

Commented by prof Abdo imad last updated on 25/Mar/18

$${\int }_0^{\pi }ln(1+cos\theta )d\theta ={\int }_0^{\pi }ln\left(2{cos}^2\left(\frac{\theta }{2}\right)\right)d\theta$$$$=\pi ln\left(2\right)+2{\int }_0^{\pi }ln\left(cos\left(\frac{\theta }{2}\right)\right)d\theta$$$${=}_{\frac{\theta }{2}=t}\pi ln\left(2\right)+2{\int }_0^{\frac{\pi }{2}}ln\left(cost\right)2dt$$$$=\pi ln\left(2\right)+4{\int }_0^{\frac{\pi }{2}}ln\left(cost\right)dt$$$$=\pi ln\left(2\right)+4(-\frac{\pi ln2}{2})=-\pi ln\left(2\right)finally$$$$F\left(x\right)=\pi ln\mid x\mid -\pi ln\mid tan\left(\frac{arcsinx}{2}\right)\mid -\pi ln\left(2\right).$$

Commented by prof Abdo imad last updated on 25/Mar/18

$$F\left(0\right)=0$$

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