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Question Number 33883 by math khazana by abdo last updated on 26/Apr/18

$$findasimpleformoff\left(x\right)={\int }_0^{\frac{\pi }{2}}ln(1+{xsin}^{2}t)dt$$ $$with\mid x\mid <1.$$

Commented bymath khazana by abdo last updated on 01/May/18

$$wehave{f}^{{}^{\prime }}\left(x\right)={\int }_0^{\frac{\pi }{2}}\frac{{sin}^{2}t}{1+{xsin}^{2}t}dtandforx\ne 0$$ $$f^{{}^{\prime }}\left(x\right)=\frac{1}{x}{\int }_0^{\frac{\pi }{2}}\frac{1+{xsin}^{2}t-1}{1+{xsin}^{2}t}dt$$ $$=\frac{\pi }{2x}-\frac{1}{x}{\int }_0^{\frac{\pi }{2}}\frac{dt}{1+x{sin}^{2}t}but$$ $${\int }_0^{\frac{\pi }{2}}\frac{dt}{1+x{sin}^{2}t}={\int }_0^{\frac{\pi }{2}}\frac{dt}{1+x\frac{1-cos\left(2t\right)}{2}}$$ $$={\int }_0^{\frac{\pi }{2}}\frac{2dt}{2+x-xcos\left(2t\right)}{=}_{2t=u}{\int }_0^{\pi }\frac{du}{2+x-xcosu}$$ $$alsochtan\left(\frac{u}{2}\right)=tgive$$ $${\int }_0^{\frac{\pi }{2}}\frac{dt}{1+x{sin}^{2}t}={\int }_0^{+\mathrm{\infty }}\frac{1}{2+x-x\frac{1-t^2}{1+t^2}}\frac{2dt}{1+t^2}$$ $$={\int }_0^{\mathrm{\infty }}\frac{2dt}{(2+x)(1+t^2)-x(1-t^2)}$$ $$={\int }_0^{\mathrm{\infty }}\frac{2dt}{2+x+(2+x)t^2-x+{xt}^2}$$ $$={\int }_0^{\mathrm{\infty }}\frac{2dt}{2+2(1+x)t^2}={\int }_0^{\mathrm{\infty }}\frac{dt}{1+(1+x)t^2}$$ $${=}_{\sqrt{1+x}t=u}{\int }_0^{\mathrm{\infty }}\frac{1}{1+u^2}\frac{du}{\sqrt{1+x}}=\frac{\pi }{2\sqrt{1+x}}\Rightarrow$$ $$f^{{}^{\prime }}\left(x\right)=\frac{\pi }{2x}-\frac{1}{x}\frac{\pi }{2\sqrt{1+x}}=\frac{\pi }{2x}(1-\frac{1}{\sqrt{1+x}})\Rightarrow$$ $$f\left(x\right)=\frac{\pi }{2}ln\mid x\mid -\frac{\pi }{2}{\int }_{.}^x\frac{dt}{t\sqrt{1+t}}+\lambda$$ $$\int \frac{dt}{t\sqrt{1+t}}{=}_{\sqrt{1+t}=u}\int \frac{2udu}{(u^2-1)u}$$ $$=\int \frac{2du}{(u+1)(u-1)}=\int (\frac{1}{u-1}-\frac{1}{u+1})du$$ $$=ln\mid \frac{u-1}{u+1}\mid =ln\left(\frac{\sqrt{1+x}-1}{\sqrt{1+x}+1}\right)\Rightarrow$$ $$f\left(x\right)=\frac{\pi }{2}ln\mid x\mid -\frac{\pi }{2}ln\left(\frac{\sqrt{1+x}-1}{\sqrt{1+x}+1}\right)+\lambda$$ $$\lambda ={lim}_{x\to 1}(f\left(x\right)-\frac{\pi }{2}ln\mid x\mid +\frac{\pi }{2}ln\left(\frac{\sqrt{1+x}-1}{\sqrt{1+x}+1}\right))$$ $$=f\left(1\right)+\frac{\pi }{2}ln\left(\frac{\sqrt{2}-1}{\sqrt{2}+1}\right)$$ $$={\int }_0^{\frac{\pi }{2}}ln(1+{sin}^{2}t)dt+\frac{\pi }{2}ln\left(\frac{\sqrt{2}-1}{\sqrt{2}+1}\right).$$ $$$$

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