chi(x)[0,1]などの定積分

\begin{alignat}{2}
&(1)  \displaystyle\int_0^1 \mathrm{chi}(x)dx=\mathrm{chi}(1)-\sinh 1\\
&(2)  \displaystyle\int_0^{\log 2} \mathrm{chi}(x)dx=(\log 2) \mathrm{chi}(\log 2)-\frac{3}{4}\\
&(3)  \displaystyle\int_0^1 x \,\mathrm{chi}(x)dx=\frac{1}{2}\left\{\mathrm{chi}(1)+\frac{1}{e}-1\right\}\\
&(4)  \displaystyle\int_0^{\log 2} x \,\mathrm{chi}(x)dx=\frac{1}{2}(\log 2)^2 \mathrm{chi}(\log 2)-\frac{3}{8}\log 2+\frac{1}{8}\\
\end{alignat}








<証明>

全て、部分積分を行います。

\begin{alignat}{2}
(1)  \displaystyle\int_0^1 \mathrm{chi}(x)dx&=[x\, \mathrm{chi}(x)]_0^1 -\displaystyle\int_0^1 x \cdot \frac{\cosh x}{x}dx\\
&=\mathrm{chi}(1)-\displaystyle\int_0^1 \cosh xdx\\
&=\mathrm{chi}(1) -[\sinh x]_0^1\\
&=\mathrm{chi}(1)-\sinh 1
\end{alignat}以上より$$\displaystyle\int_0^1 \mathrm{chi}(x)dx=\mathrm{chi}(1)-\sinh 1$$







\begin{alignat}{2}
(2)  \displaystyle\int_0^{\log 2} \mathrm{chi}(x)dx&=[x\, \mathrm{chi}(x)]_0^{\log 2}-\displaystyle\int_0^{\log 2} x \cdot \frac{\cosh x}{x}dx\\
&=(\log 2)\mathrm{chi}(\log 2)-\displaystyle\int_0^{\log 2} \cosh xdx\\
&=(\log 2)\mathrm{chi}(\log 2)-[\sinh x]_0^{\log 2}\\
&=(\log 2)\mathrm{chi}(\log 2)-\frac{1}{2}\left(2-\frac{1}{2}\right)\\
&=(\log 2) \mathrm{chi}(\log 2)-\frac{3}{4}\\
\end{alignat}以上より$$\displaystyle\int_0^{\log 2} \mathrm{chi}(x)dx=(\log 2) \mathrm{chi}(\log 2)-\frac{3}{4}$$







\begin{alignat}{2}
(3)  \displaystyle\int_0^1 x \,\mathrm{chi}(x)dx&=\left[\frac{1}{2}x^2 \cdot \mathrm{chi}(x)\right]_0^1 -\displaystyle\int_0^1 \frac{1}{2}x^2 \cdot \frac{\cosh x}{x}dx\\
&=\frac{1}{2}\mathrm{chi}(1)-\frac{1}{2}\displaystyle\int_0^1 x \cosh xdx\\
&=\frac{1}{2}\mathrm{chi}(1)-\frac{1}{2}\left([x \sinh x]_0^1 -\displaystyle\int_0^1 \sinh xdx\right)\\
&=\frac{1}{2}\mathrm{chi}(1)-\frac{1}{2}\left(\sinh 1-[\cosh x]_0^1\right)\\
&=\frac{1}{2}\mathrm{chi}(1)-\frac{1}{2}(\sinh 1-\cosh 1+1)\\
&=\frac{1}{2}\mathrm{chi}(1)-\frac{1}{2}\left(\frac{e-e^{-1}}{2}-\frac{e+e^{-1}}{2}+1\right)\\
&=\frac{1}{2}\mathrm{chi}(1)-\frac{1}{2}\left(-\frac{1}{e}+1\right)\\
&=\frac{1}{2}\left\{\mathrm{chi}(1)+\frac{1}{e}-1\right\}
\end{alignat}以上より$$\displaystyle\int_0^1 x \,\mathrm{chi}(x)dx=\frac{1}{2}\left\{\mathrm{chi}(1)+\frac{1}{e}-1\right\}$$







\begin{alignat}{2}
(4)  \displaystyle\int_0^{\log 2} x \,\mathrm{chi}(x)dx&=\left[\frac{1}{2}x^2 \,\mathrm{chi}(x)\right]_0^{\log 2} -\displaystyle\int_0^{\log 2} \frac{1}{2}x^2 \cdot \frac{\cosh x}{x}dx\\
&=\frac{1}{2}(\log 2)^2 \mathrm{chi}(\log 2)-\frac{1}{2}\displaystyle\int_0^{\log 2} x \cosh xdx\\
&=\frac{1}{2}(\log 2)^2 \mathrm{chi}(\log 2)-\frac{1}{2} \left\{[x \sinh x]_0^{\log 2}-\displaystyle\int_0^{\log 2} \sinh xdx\right\}\\
&=\frac{1}{2}(\log 2)^2 \mathrm{chi}(\log 2)-\frac{1}{2} \left\{(\log 2) \cdot \frac{2-2^{-1}}{2}-[\cosh x]_0^{\log 2}\right\}\\
&=\frac{1}{2}(\log 2)^2 \mathrm{chi}(\log 2)-\frac{1}{2}\left(\frac{3}{4} \log 2-\frac{2+2^{-1}}{2}+1\right)\\
&=\frac{1}{2}(\log 2)^2 \mathrm{chi}(\log 2)-\frac{3}{8}\log 2+\frac{1}{8}
\end{alignat}以上より$$\displaystyle\int_0^{\log 2} x \,\mathrm{chi}(x)dx=\frac{1}{2}(\log 2)^2 \mathrm{chi}(\log 2)-\frac{3}{8}\log 2+\frac{1}{8}$$

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