shi(x)[0,1]などの定積分

\begin{alignat}{2}
&(1)  \displaystyle\int_0^1 \mathrm{shi}(x)dx=\mathrm{shi}(1)-\cosh 1+1\\
&(2)  \displaystyle\int_0^{\log 2} \mathrm{shi}(x)dx=(\log 2) \mathrm{shi}(\log 2)-\frac{1}{4}\\
&(3)  \displaystyle\int_0^1 x \,\mathrm{shi}(x)dx=\frac{1}{2}\mathrm{shi}(1)-\frac{1}{2e}\\
&(4)  \displaystyle\int_0^{\log 2} x \,\mathrm{shi}(x)dx=\frac{1}{2}(\log 2)^2 \mathrm{shi}(\log 2)-\frac{5}{8}\log 2+\frac{3}{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{shi}(x)dx&=[x\, \mathrm{shi}(x)]_0^1 -\displaystyle\int_0^1 x \cdot \frac{\sinh x}{x}dx\\
&=\mathrm{shi}(1)-\displaystyle\int_0^1 \sinh xdx\\
&=\mathrm{shi}(1) -[\cosh x]_0^1\\
&=\mathrm{shi}(1)-\cosh 1+1
\end{alignat}以上より$$\displaystyle\int_0^1 \mathrm{shi}(x)dx=\mathrm{shi}(1)-\cosh 1+1$$







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







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







\begin{alignat}{2}
(4)  \displaystyle\int_0^{\log 2} x \,\mathrm{shi}(x)dx&=\left[\frac{1}{2}x^2 \,\mathrm{shi}(x)\right]_0^{\log 2} -\displaystyle\int_0^{\log 2} \frac{1}{2}x^2 \cdot \frac{\sinh x}{x}dx\\
&=\frac{1}{2}(\log 2)^2 \mathrm{shi}(\log 2)-\frac{1}{2}\displaystyle\int_0^{\log 2} x \sinh xdx\\
&=\frac{1}{2}(\log 2)^2 \mathrm{shi}(\log 2)-\frac{1}{2} \left\{[x \cosh x]_0^{\log 2}-\displaystyle\int_0^{\log 2} \cosh xdx\right\}\\
&=\frac{1}{2}(\log 2)^2 \mathrm{shi}(\log 2)-\frac{1}{2} \left\{(\log 2) \cdot \frac{2+2^{-1}}{2}-[\sinh x]_0^{\log 2}\right\}\\
&=\frac{1}{2}(\log 2)^2 \mathrm{shi}(\log 2)-\frac{1}{2}\left(\frac{5}{4} \log 2-\frac{2-2^{-1}}{2}\right)\\
&=\frac{1}{2}(\log 2)^2 \mathrm{shi}(\log 2)-\frac{5}{8}\log 2+\frac{3}{8}
\end{alignat}以上より$$\displaystyle\int_0^{\log 2} x \,\mathrm{shi}(x)dx=\frac{1}{2}(\log 2)^2 \mathrm{shi}(\log 2)-\frac{5}{8}\log 2+\frac{3}{8}$$

コメントを残す

メールアドレスが公開されることはありません。 * が付いている欄は必須項目です