(-logx-1)^μ[0,1/e]などの定積分

\begin{alignat}{2}
&(1)  \displaystyle\int_0^{\frac{1}{e}} \frac{1}{\sqrt{-\log x-1}}dx=\frac{\sqrt{π}}{e}\\
&(2)  \displaystyle\int_0^{\frac{1}{e}} \sqrt{-\log x-1}dx=\frac{\sqrt{π}}{2e}\\
&(3)  \displaystyle\int_0^{\frac{1}{e}} (-\log x-1)^μdx=\frac{1}{e}Γ(μ+1)  (μ+1 \gt 0)\\
&(4)  \displaystyle\int_0^{\frac{1}{e}} \frac{1}{\sqrt{1-\log x}}dx=e\sqrt{π}\,\mathrm{erfc}(\sqrt{2})\\
&(5)  \displaystyle\int_0^{\frac{1}{e}} \sqrt{1-\log x}dx=\frac{1}{2}e\sqrt{π}\,\mathrm{erfc}(\sqrt{2})-\frac{\sqrt{2}}{e}\\
&(6)  \displaystyle\int_0^{\frac{1}{e}} (1-\log x)^μdx=eΓ(μ+1,2)  (μ+1 \gt 0)\\
\end{alignat}









<証明>

全て始めに \(x=e^{-t}\) と置きます。\((dx=-e^{-t}dt)\)

\((1)\) から \((3)\) は途中、\(t-1=s\) と置きます。\((dt=ds)\)

\begin{alignat}{2}
(1)  \displaystyle\int_0^{\frac{1}{e}} \frac{1}{\sqrt{-\log x-1}}dx&=\displaystyle\int_{\infty}^1 \frac{1}{\sqrt{t-1}} \cdot (-e^{-t})dt=\displaystyle\int_1^{\infty} \frac{e^{-t}}{\sqrt{t-1}}dt\\
&=\displaystyle\int_0^{\infty} e^{-(s+1)}s^{-\frac{1}{2}}ds=\frac{1}{e}\displaystyle\int_0^{\infty} s^{-\frac{1}{2}}e^{-s}ds\\
&=\frac{1}{e}Γ\left(\frac{1}{2}\right)=\frac{\sqrt{π}}{e}
\end{alignat}以上より$$\displaystyle\int_0^{\frac{1}{e}} \frac{1}{\sqrt{-\log x-1}}dx=\frac{\sqrt{π}}{e}$$






\begin{alignat}{2}
(2)  \displaystyle\int_0^{\frac{1}{e}} \sqrt{-\log x-1}dx&=\displaystyle\int_{\infty}^1 \sqrt{t-1}(-e^{-t})dt=\displaystyle\int_1^{\infty} e^{-t} \sqrt{t-1}dt\\
&=\displaystyle\int_0^{\infty} e^{-(s+1)} s^{\frac{1}{2}}ds=\frac{1}{e}\displaystyle\int_0^{\infty} s^{\frac{1}{2}}e^{-s}ds\\
&=\frac{1}{e}Γ\left(\frac{3}{2}\right)=\frac{1}{e} \cdot \frac{\sqrt{π}}{2}=\frac{\sqrt{π}}{2e}
\end{alignat}以上より$$\displaystyle\int_0^{\frac{1}{e}} \frac{1}{\sqrt{-\log x-1}}dx=\frac{\sqrt{π}}{2}$$







\begin{alignat}{2}
(3)  \displaystyle\int_0^{\frac{1}{e}} (-\log x-1)^μdx&=\displaystyle\int_{\infty}^1 (t-1)^μ (-e^{-t})dt=\displaystyle\int_1^{\infty} e^{-t} (t-1)^μdt\\
&=\displaystyle\int_0^{\infty} e^{-(s+1)}s^μds=\frac{1}{e}\displaystyle\int_0^{\infty} s^μ e^{-s}ds=\frac{1}{e}Γ(μ+1)
\end{alignat}以上より$$\displaystyle\int_0^{\frac{1}{e}} (-\log x-1)^μdx=\frac{1}{e}Γ(μ+1)  (μ+1 \gt 0)$$







\((4)(5)\) では途中、\(\sqrt{1+t}=s\) と置きます。\((dt=2sds)\)

\begin{alignat}{2}
(4)  \displaystyle\int_0^{\frac{1}{e}} \frac{1}{\sqrt{1-\log x}}dx&=\displaystyle\int_{\infty}^1 \frac{1}{\sqrt{1+t}} \cdot (-e^{-t})dt=\displaystyle\int_1^{\infty} \frac{e^{-t}}{\sqrt{1+t}}dt\\
&=\displaystyle\int_{\sqrt{2}}^{\infty} \frac{e^{-(s^2-1)}}{s} \cdot 2sds=2e \displaystyle\int_{\sqrt{2}}^{\infty} e^{-s^2}ds\\
&=2e \cdot \frac{\sqrt{π}}{2}\,\mathrm{erfc}(\sqrt{2})=e\sqrt{π}\,\mathrm{erfc}(\sqrt{2})\\
\end{alignat}以上より$$\displaystyle\int_0^{\frac{1}{e}} \frac{1}{\sqrt{1-\log x}}dx=e\sqrt{π}\,\mathrm{erfc}(\sqrt{2})$$







\begin{alignat}{2}
(5)  \displaystyle\int_0^{\frac{1}{e}} \sqrt{1-\log x}dx&=\displaystyle\int_{\infty}^1 \sqrt{1+t}\cdot (-e^{-t})dt=\displaystyle\int_1^{\infty} e^{-t}\sqrt{1+t}dt\\
&=\displaystyle\int_{\sqrt{2}}^{\infty} e^{-(s^2-1)} \cdot s \cdot 2sds=2e \displaystyle\int_{\sqrt{2}}^{\infty} s^2e^{-s^2}ds\\
&=-e \displaystyle\int_{\sqrt{2}}^{\infty} s(e^{-s^2})’ds=-e \left([se^{-s^2}]_{\sqrt{2}}^{\infty}-\displaystyle\int_{\sqrt{2}}^{\infty} e^{-s^2}ds\right)\\
&=-e \left\{-\sqrt{2}e^{-2}-\frac{\sqrt{π}}{2}\,\mathrm{erfc}(\sqrt{2})\right\}\\
&=\frac{1}{2}e\sqrt{π}\,\mathrm{erfc}(\sqrt{2})-\frac{\sqrt{2}}{e}\\
\end{alignat}以上より$$\displaystyle\int_0^{\frac{1}{e}} \sqrt{1-\log x}dx=\frac{1}{2}e\sqrt{π}\,\mathrm{erfc}(\sqrt{2})-\frac{\sqrt{2}}{e}$$







$$(6)  \displaystyle\int_0^{\frac{1}{e}} (1-\log x)^μdx=\displaystyle\int_{\infty}^1 (1+t)^μ(-e^{-t})dt=\displaystyle\int_1^{\infty} e^{-t}(1+t)^μdt$$\(1+t=s\) と置きます。\((dt=ds)\)$$\displaystyle\int_2^{\infty} e^{-(s-1)}s^μds=e \displaystyle\int_2^{\infty} s^μe^{-s}ds=eΓ(μ+1,2)$$以上より$$\displaystyle\int_0^{\frac{1}{e}} (1-\log x)^μdx=eΓ(μ+1,2)  (μ+1 \gt 0)$$

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