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 Author: akotronis [ Sat Jul 09, 2016 5:49 am ] Post subject: A trigonometric - logarithmic integral Evaluate $$\displaystyle\int_{0}^{\pi/2}4\,\cos^2x\,\ln^2(\cos x)\,dx$$. NOTE

 Author: Grigorios Kostakos [ Sat Jul 09, 2016 5:50 am ] Post subject: Re: A trigonometric - logarithmic integral \begin{align*}\displaystyle\int_{0}^{\frac{\pi}{2}}{4\cos^2x\,\log^2(\cos x)\,dx}&\mathop{=\!=\!=\!=\!=\!=\!=\!=\!=\!=}\limits^{\begin{subarray}{c}{t\,=\,\cos^2 x}\\{-\frac{1}{\sqrt{t}\sqrt{1-t}}\,dt\,=\,dx}\\\end{subarray}}\,4\int_{0}^{1}{\frac{t}{\sqrt{t}\sqrt{1-t}}\log^2\bigl({\sqrt{t}\,}\bigr)\,dt}\\&=\int_{0}^{1}{\frac{\sqrt{t}}{\sqrt{1-t}}\log^2{t}\,dt}\,.\end{align*}We have that \begin{align*} I(m)&=\int_{0}^{1}{\frac{t^{m-\frac{1}{2}}}{\sqrt{1-t}}\,dt}=\int_{0}^{1}{t^{m+\frac{1}{2}-1}({1-t})^{\frac{1}{2}-1}\,dt}\\&={\rm{B}}\bigl({m+\tfrac{1}{2},\tfrac{1}{2}}\bigr)=\frac{\sqrt{\pi}\,\Gamma\bigl({m+\tfrac{1}{2}}\bigr)}{\Gamma({m+1})}\quad\Rightarrow\\ \frac{d^2}{dm^2}I(m)&=\int_{0}^{1}{\frac{t^{m-\frac{1}{2}}}{\sqrt{1-t}}\log^2{t}\,dt}\\ &=\frac{\sqrt{\pi}\,\Gamma\bigl({m+\tfrac{1}{2}}\bigr)}{\Gamma({m+1})}\Bigl({\psi'\bigl({m+\tfrac{1}{2}}\bigr)-\psi'({m+1})+\Bigr({\psi\bigl({m+\tfrac{1}{2}}\bigr)-\psi({m+1})}\Bigr)^2}\Bigr)\quad\stackrel{m=1}{\Longrightarrow}\\ \int_{0}^{1}{\frac{\sqrt{t}}{\sqrt{1-t}}\log^2{t}\,dt}&=\frac{\sqrt{\pi}\,\Gamma\bigl({\tfrac{3}{2}}\bigr)}{\Gamma({2})}\Bigl({\psi'\bigl({\tfrac{3}{2}}\bigr)-\psi'({2})+\Bigr({\psi\bigl({\tfrac{3}{2}}\bigr)-\psi({2})}\Bigr)^2}\Bigr)\\ &=\frac{\sqrt{\pi}\,\sqrt{\pi}}{2}\Bigl({\frac{\pi^2}{2}-4-\frac{\pi^2}{6}+1+\bigl({2-2\log2-\gamma-1+\gamma}\bigr)^2}\Bigr)\\ &=\pi\Bigl({2\log^22-2\log2+\frac{\pi^2}{6}-1}\Bigr)\,.\end{align*}

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