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: 22 lip 2019, 14:16
\(\int_{0}^{\frac{\pi}{2}}{\frac{\ln{\left(\sin{x}\right)}\ln{\left(\cos{x}\right)}}{\tan{\left(x\right)}}\mbox{d}x}\\
t=-\ln{\left(\sin{x}\right)}\\
\mbox{d}t=-\frac{\cos{x}}{\sin{x}}\mbox{d}x\\
\int_{\infty}^{0}{\left(-t\right)\ln{\left(\sqrt{1-e^{-2t}}\right)}\left(-\mbox{d}t\right)}\\
\int_{0}^{\infty}{\left(-t\right)\ln{\left(\sqrt{1-e^{-2t}}\right)}\mbox{d}t}\\
-\frac{1}{2}\int_{0}^{\infty}{t\ln{\left(1-e^{-2t}\right)}\mbox{d}t}\\
\)
\(
-\frac{1}{2}\int_{0}^{\infty}{t\ln{\left(1-e^{-2t}\right)}\mbox{d}t}=\begin{vmatrix}u=\ln{\left(1-e^{-2t}\right)} & \mbox{d}v=-\frac{1}{2}t\mbox{d}t\\\mbox{d}u=\frac{2e^{-2t}}{1-e^{-2t}}\mbox{d}t & v=-\frac{1}{4}t^2\end{vmatrix}\\=
\lim_{t\to\infty}{-\frac{1}{4}t^2\ln{\left(1-e^{-2t}\right)}}-\lim_{t\to 0}{-\frac{1}{4}t^2\ln{\left(1-e^{-2t}\right)}}+\frac{1}{2}\int_{0}^{\infty}{\frac{t^2e^{-2t}}{1-e^{-2t}}\mbox{d}t}\\
=\frac{1}{2}\int_{0}^{\infty}{\sum_{n=1}^{\infty}{t^2e^{-2nt}}\mbox{d}t}\\
\)
\(
=\frac{1}{2}\sum_{n=1}^{\infty}{\int_{0}^{\infty}{t^2e^{-2nt}\mbox{d}t}}\\
\int_{0}^{\infty}{t^2e^{-2nt}\mbox{d}t}=\begin{vmatrix}u=t^2 & \mbox{d}v=e^{-2nt}\mbox{d}t \\ \mbox{d}u=2t\mbox{d}t & v=-\frac{1}{2n}e^{-2nt}\end{vmatrix}\\
\lim_{t \to \infty}{-\frac{1}{2n}t^2e^{-2nt}}-\lim_{t \to 0}{-\frac{1}{2n}t^2e^{-2nt}}+\frac{1}{n}\int_{0}^{\infty}{te^{-2nt}\mbox{d}t}\\
\int_{0}^{\infty}{t^2e^{-2nt}\mbox{d}t}=\frac{1}{n}\int_{0}^{\infty}{te^{-2nt}\mbox{d}t}\\
\int_{0}^{\infty}{te^{-2nt}\mbox{d}t}=\begin{vmatrix}u=t & \mbox{d}v =e^{-2nt} \mbox{d}t\\\mbox{d}u=\mbox{d}t & v=-\frac{1}{2n}e^{-2nt}\end{vmatrix}\\
=\lim_{t \to \infty}{-\frac{1}{2n}te^{-2nt}}-\lim_{t \to 0}{-\frac{1}{2n}te^{-2nt}}+\frac{1}{2n}\int_{0}^{\infty}{e^{-2nt}\mbox{d}t}\\
\int_{0}^{\infty}{te^{-2nt}\mbox{d}t}=\frac{1}{2n}\int_{0}^{\infty}{e^{-2nt}\mbox{d}t}\\
\int_{0}^{\infty}{te^{-2nt}\mbox{d}t}=-\frac{1}{4n^2}\left(\lim_{t \to \infty}{e^{-2nt}}-\lim_{t \to 0}{e^{-2nt}}\right)\\
\int_{0}^{\infty}{te^{-2nt}\mbox{d}t}=-\frac{1}{4n^2}\left(0-1\right)\\
\int_{0}^{\infty}{te^{-2nt}\mbox{d}t}=\frac{1}{4n^2}\\
\int_{0}^{\infty}{t^2e^{-2nt}\mbox{d}t}=\frac{1}{4n^3}\\
\)
\(
=\frac{1}{2}\sum_{n=1}^{\infty}{\frac{1}{4n^3}}\\
=\frac{1}{8}\sum_{n=1}^{\infty}{\frac{1}{n^3}}
\)
Ciekawy jestem czy gdzieś nie popełniłem błędu
t=-\ln{\left(\sin{x}\right)}\\
\mbox{d}t=-\frac{\cos{x}}{\sin{x}}\mbox{d}x\\
\int_{\infty}^{0}{\left(-t\right)\ln{\left(\sqrt{1-e^{-2t}}\right)}\left(-\mbox{d}t\right)}\\
\int_{0}^{\infty}{\left(-t\right)\ln{\left(\sqrt{1-e^{-2t}}\right)}\mbox{d}t}\\
-\frac{1}{2}\int_{0}^{\infty}{t\ln{\left(1-e^{-2t}\right)}\mbox{d}t}\\
\)
\(
-\frac{1}{2}\int_{0}^{\infty}{t\ln{\left(1-e^{-2t}\right)}\mbox{d}t}=\begin{vmatrix}u=\ln{\left(1-e^{-2t}\right)} & \mbox{d}v=-\frac{1}{2}t\mbox{d}t\\\mbox{d}u=\frac{2e^{-2t}}{1-e^{-2t}}\mbox{d}t & v=-\frac{1}{4}t^2\end{vmatrix}\\=
\lim_{t\to\infty}{-\frac{1}{4}t^2\ln{\left(1-e^{-2t}\right)}}-\lim_{t\to 0}{-\frac{1}{4}t^2\ln{\left(1-e^{-2t}\right)}}+\frac{1}{2}\int_{0}^{\infty}{\frac{t^2e^{-2t}}{1-e^{-2t}}\mbox{d}t}\\
=\frac{1}{2}\int_{0}^{\infty}{\sum_{n=1}^{\infty}{t^2e^{-2nt}}\mbox{d}t}\\
\)
\(
=\frac{1}{2}\sum_{n=1}^{\infty}{\int_{0}^{\infty}{t^2e^{-2nt}\mbox{d}t}}\\
\int_{0}^{\infty}{t^2e^{-2nt}\mbox{d}t}=\begin{vmatrix}u=t^2 & \mbox{d}v=e^{-2nt}\mbox{d}t \\ \mbox{d}u=2t\mbox{d}t & v=-\frac{1}{2n}e^{-2nt}\end{vmatrix}\\
\lim_{t \to \infty}{-\frac{1}{2n}t^2e^{-2nt}}-\lim_{t \to 0}{-\frac{1}{2n}t^2e^{-2nt}}+\frac{1}{n}\int_{0}^{\infty}{te^{-2nt}\mbox{d}t}\\
\int_{0}^{\infty}{t^2e^{-2nt}\mbox{d}t}=\frac{1}{n}\int_{0}^{\infty}{te^{-2nt}\mbox{d}t}\\
\int_{0}^{\infty}{te^{-2nt}\mbox{d}t}=\begin{vmatrix}u=t & \mbox{d}v =e^{-2nt} \mbox{d}t\\\mbox{d}u=\mbox{d}t & v=-\frac{1}{2n}e^{-2nt}\end{vmatrix}\\
=\lim_{t \to \infty}{-\frac{1}{2n}te^{-2nt}}-\lim_{t \to 0}{-\frac{1}{2n}te^{-2nt}}+\frac{1}{2n}\int_{0}^{\infty}{e^{-2nt}\mbox{d}t}\\
\int_{0}^{\infty}{te^{-2nt}\mbox{d}t}=\frac{1}{2n}\int_{0}^{\infty}{e^{-2nt}\mbox{d}t}\\
\int_{0}^{\infty}{te^{-2nt}\mbox{d}t}=-\frac{1}{4n^2}\left(\lim_{t \to \infty}{e^{-2nt}}-\lim_{t \to 0}{e^{-2nt}}\right)\\
\int_{0}^{\infty}{te^{-2nt}\mbox{d}t}=-\frac{1}{4n^2}\left(0-1\right)\\
\int_{0}^{\infty}{te^{-2nt}\mbox{d}t}=\frac{1}{4n^2}\\
\int_{0}^{\infty}{t^2e^{-2nt}\mbox{d}t}=\frac{1}{4n^3}\\
\)
\(
=\frac{1}{2}\sum_{n=1}^{\infty}{\frac{1}{4n^3}}\\
=\frac{1}{8}\sum_{n=1}^{\infty}{\frac{1}{n^3}}
\)
Ciekawy jestem czy gdzieś nie popełniłem błędu