対数同士の積の積分
対数同士の積の積分
次の定積分を求めよ。
\[ \int_{0}^{1}\log\left(x\right)\log\left(1+x\right)dx=? \]
次の定積分を求めよ。
\[ \int_{0}^{1}\log\left(x\right)\log\left(1+x\right)dx=? \]
積分範囲が\(0\leq x\leq1\)なので\(\log\left(1+x\right)\)を展開してあとは計算すれば解けます。
また、最後で
\begin{align*} \sum_{k=1}^{\infty}\frac{\left(-1\right)^{k+1}}{k} & =\left[\sum_{k=1}^{\infty}\frac{\left(-1\right)^{k+1}x^{k}}{k}\right]_{x=1}\\ & =\left[\log\left(1+x\right)\right]_{x=1}\\ & =\log2 \end{align*} \begin{align*} \sum_{k=1}^{\infty}\frac{\left(-1\right)^{k+1}}{k^{2}} & =\sum_{k=1}^{\infty}\frac{\left(-1\right)^{2k+1}}{\left(2k\right)^{2}}+\sum_{k=1}^{\infty}\frac{\left(-1\right)^{2k-1+1}}{\left(2k-1\right)^{2}}\\ & =-\sum_{k=1}^{\infty}\frac{1}{\left(2k\right)^{2}}+\sum_{k=1}^{\infty}\frac{1}{\left(2k-1\right)^{2}}\\ & =-\sum_{k=1}^{\infty}\frac{1}{\left(2k\right)^{2}}+\sum_{k=1}^{\infty}\frac{1}{k^{2}}-\sum_{k=1}^{\infty}\frac{1}{\left(2k\right)^{2}}\\ & =-2\sum_{k=1}^{\infty}\frac{1}{\left(2k\right)^{2}}+\sum_{k=1}^{\infty}\frac{1}{k^{2}}\\ & =\left(1-\frac{1}{2}\right)\sum_{k=1}^{\infty}\frac{1}{k^{2}}\\ & =\frac{1}{2}\sum_{k=1}^{\infty}\frac{1}{k^{2}}\\ & =\frac{\zeta\left(2\right)}{2} \end{align*} を使っています。
また、最後で
\begin{align*} \sum_{k=1}^{\infty}\frac{\left(-1\right)^{k+1}}{k} & =\left[\sum_{k=1}^{\infty}\frac{\left(-1\right)^{k+1}x^{k}}{k}\right]_{x=1}\\ & =\left[\log\left(1+x\right)\right]_{x=1}\\ & =\log2 \end{align*} \begin{align*} \sum_{k=1}^{\infty}\frac{\left(-1\right)^{k+1}}{k^{2}} & =\sum_{k=1}^{\infty}\frac{\left(-1\right)^{2k+1}}{\left(2k\right)^{2}}+\sum_{k=1}^{\infty}\frac{\left(-1\right)^{2k-1+1}}{\left(2k-1\right)^{2}}\\ & =-\sum_{k=1}^{\infty}\frac{1}{\left(2k\right)^{2}}+\sum_{k=1}^{\infty}\frac{1}{\left(2k-1\right)^{2}}\\ & =-\sum_{k=1}^{\infty}\frac{1}{\left(2k\right)^{2}}+\sum_{k=1}^{\infty}\frac{1}{k^{2}}-\sum_{k=1}^{\infty}\frac{1}{\left(2k\right)^{2}}\\ & =-2\sum_{k=1}^{\infty}\frac{1}{\left(2k\right)^{2}}+\sum_{k=1}^{\infty}\frac{1}{k^{2}}\\ & =\left(1-\frac{1}{2}\right)\sum_{k=1}^{\infty}\frac{1}{k^{2}}\\ & =\frac{1}{2}\sum_{k=1}^{\infty}\frac{1}{k^{2}}\\ & =\frac{\zeta\left(2\right)}{2} \end{align*} を使っています。
\begin{align*}
\int_{0}^{1}\log\left(x\right)\log\left(x+1\right)dx & =\int_{0}^{1}\log\left(x\right)\sum_{k=1}^{\infty}\frac{\left(-1\right)^{k+1}}{k}x^{k}dx\cmt{\because-1<x\leq1\rightarrow\log\left(1+x\right)=\sum_{k=1}^{\infty}\frac{\left(-1\right)^{k+1}x^{k}}{k}}\\
& =\sum_{k=1}^{\infty}\frac{\left(-1\right)^{k+1}}{k}\int_{0}^{1}\log\left(x\right)x^{k}dx\\
& =\sum_{k=1}^{\infty}\frac{\left(-1\right)^{k+1}}{k}\left(\left[\frac{\log\left(x\right)x^{k+1}}{k+1}\right]_{0}^{1}-\int_{0}^{1}\frac{x^{k}}{k+1}dx\right)\\
& =\sum_{k=1}^{\infty}\frac{\left(-1\right)^{k}}{k}\int_{0}^{1}\frac{x^{k}}{k+1}dx\\
& =\sum_{k=1}^{\infty}\frac{\left(-1\right)^{k}}{k}\left[\frac{x^{k+1}}{\left(k+1\right)^{2}}\right]_{0}^{1}\\
& =\sum_{k=1}^{\infty}\frac{\left(-1\right)^{k}}{k\left(k+1\right)^{2}}\\
& =\sum_{k=1}^{\infty}\frac{\left(-1\right)^{k}}{\left(k+1\right)}\left(\frac{1}{k}-\frac{1}{k+1}\right)\\
& =\sum_{k=1}^{\infty}\left(-1\right)^{k}\left(\frac{1}{k\left(k+1\right)}-\frac{1}{\left(k+1\right)}\right)\\
& =\sum_{k=1}^{\infty}\left(-1\right)^{k}\left(\frac{1}{k}-\frac{1}{k+1}-\frac{1}{\left(k+1\right)^{2}}\right)\\
& =\sum_{k=1}^{\infty}\frac{\left(-1\right)^{k}}{k}+\sum_{k=1}^{\infty}\frac{\left(-1\right)^{k+1}}{k+1}+\sum_{k=1}^{\infty}\frac{\left(-1\right)^{k+1}}{\left(k+1\right)^{2}}\\
& =\sum_{k=1}^{\infty}\frac{\left(-1\right)^{k}}{k}+\sum_{k=2}^{\infty}\frac{\left(-1\right)^{k}}{k}+\sum_{k=2}^{\infty}\frac{\left(-1\right)^{k}}{k^{2}}\\
& =\sum_{k=1}^{\infty}\frac{\left(-1\right)^{k}}{k}+\sum_{k=1}^{\infty}\frac{\left(-1\right)^{k}}{k}+1+\sum_{k=1}^{\infty}\frac{\left(-1\right)^{k}}{k^{2}}+1\\
& =-2\sum_{k=1}^{\infty}\frac{\left(-1\right)^{k+1}}{k}-\sum_{k=1}^{\infty}\frac{\left(-1\right)^{k+1}}{k^{2}}+2\\
& =-2\log2-\left(\zeta\left(2\right)-\frac{2}{2^{2}}\zeta\left(2\right)\right)+2\\
& =-2\log2-\left(1-\frac{1}{2}\right)\zeta\left(2\right)+2\\
& =-2\log2-\frac{\pi^{2}}{12}+2
\end{align*}
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