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A trigonometric - logarithmic integral
https://www.mathimatikoi.org/forum/viewtopic.php?f=3&t=801
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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
Proposed by Paolo Perfetti

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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