logx/{a^2-(logx)^2}[0,1]などの定積分

\begin{alignat}{2}
&(1) \displaystyle\int_0^1 \frac{1}{a^2+(\log x)^2}dx=\frac{1}{a}\{\mathrm{ci}(a)\sin a-\mathrm{si} (a)\cos a\}\\
&(2) \displaystyle\int_0^1 \frac{1}{a^2-(\log x)^2}dx=\frac{1}{2a}\{e^{-a}\overline{\mathrm{Ei}}(a)-e^a\mathrm{Ei}(-a)\}\\
&(3) \displaystyle\int_0^1 \frac{\log x}{a^2+(\log x)^2}dx=\mathrm{ci}(a)\cos a+\mathrm{si} (a)\sin a\\
&(4) \displaystyle\int_0^1 \frac{\log x}{a^2-(\log x)^2}dx=-\frac{1}{2}\{e^{-a}\overline{\mathrm{Ei}}(a)+e^a\mathrm{Ei}(-a)\}\\
\end{alignat}ただし、全て \(a \gt 0\)









指数積分を次のように表します。$$\mathrm{Ei}(x)=-\displaystyle\int_{-x}^{\infty} \frac{e^{-t}}{t}dt,  \overline{\mathrm{Ei}}(x)=-\displaystyle\int_x^{\infty} \frac{e^t}{t}dt$$



<証明>

\((1)\) \(\log x=-t\) と置きます。\((x=e^{-t}, dx=-e^{-t}dt)\)$$\displaystyle\int_0^1 \frac{1}{a^2+(\log x)^2}dx=\displaystyle\int_{\infty}^0 \frac{1}{a^2+t^2}\cdot (-e^{-t})dt=\displaystyle\int_0^{\infty} \frac{e^{-t}}{a^2+t^2}dt$$被積分関数をラプラス変換・逆変換します。$$=\frac{1}{a}\displaystyle\int_0^{\infty} L[e^{-t}]L^{-1}\left[\frac{a}{t^2+a^2}\right]dt=\frac{1}{a}\displaystyle\int_0^{\infty}\frac{\sin at}{t+1}dt$$\(t+1=s\) と置きます。\((dt=ds)\)
\begin{alignat}{2}
&=\frac{1}{a}\displaystyle\int_1^{\infty} \frac{\sin (as-a)}{s}ds=\frac{1}{a}\displaystyle\int_1^{\infty} \frac{\sin as \cos a-\cos as\sin a}{s}ds\\
&=\frac{1}{a}\left\{(\sin a)\left(-\displaystyle\int_1^{\infty} \frac{\cos as}{s}ds\right)-(\cos a)\left(-\displaystyle\int_1^{\infty} \frac{\sin as}{s}ds\right)\right\}\\
&=\frac{1}{a}\{\mathrm{ci}(a)\sin a-\mathrm{si} (a)\cos a\}
\end{alignat}以上より$$\displaystyle\int_0^1 \frac{1}{a^2+(\log x)^2}dx=\frac{1}{a}\{\mathrm{ci}(a)\sin a-\mathrm{si} (a)\cos a\}$$






以下 \((2)(3)(4)\) は \((1)\) と同様の流れで解きます。
\begin{alignat}{2}
&(2)  \displaystyle\int_0^1 \frac{1}{a^2-(\log x)^2}dx=\displaystyle\int_{\infty}^0 \frac{1}{a^2-t^2}\cdot (-e^{-t})dt=\displaystyle\int_0^{\infty} \frac{e^{-t}}{a^2-t^2}dt\\
&                     =-\frac{1}{a}\displaystyle\int_0^{\infty} L[e^{-t}]L^{-1}\left[\frac{a}{t^2-a^2}\right]dt=-\frac{1}{a}\displaystyle\int_0^{\infty}\frac{\sinh at}{t+1}dt\\
&                     =-\frac{1}{a}\displaystyle\int_1^{\infty} \frac{\sinh (as-a)}{s}ds=-\frac{1}{2a}\displaystyle\int_1^{\infty} \frac{e^{as-a}-e^{-as+a}}{s}ds\\
&                     =-\frac{1}{2a}\left(e^{-a}\displaystyle\int_1^{\infty} \frac{e^{as}}{s}ds-e^a\displaystyle\int_1^{\infty} \frac{e^{-as}}{s}ds\right)\\
&                     =\frac{1}{2a}\{e^{-a}\overline{\mathrm{Ei}}(a)-e^a\mathrm{Ei}(-a)\}
\end{alignat}以上より$$\displaystyle\int_0^1 \frac{1}{a^2-(\log x)^2}dx=\frac{1}{2a}\{e^{-a}\overline{\mathrm{Ei}}(a)-e^a\mathrm{Ei}(-a)\}$$







\begin{alignat}{2}
&(3) \displaystyle\int_0^1 \frac{\log x}{a^2+(\log x)^2}dx=\displaystyle\int_{\infty}^0 \frac{-t}{a^2+t^2}\cdot (-e^{-t})dt=-\displaystyle\int_0^{\infty} \frac{te^{-t}}{a^2+t^2}dt\\
&                    =-\displaystyle\int_0^{\infty} L[e^{-t}]L^{-1}\left[\frac{t}{t^2+a^2}\right]dt=-\displaystyle\int_0^{\infty}\frac{\cos at}{t+1}dt\\
&                     =\displaystyle\int_1^{\infty} \frac{\cos (as-a)}{s}ds=\displaystyle\int_1^{\infty} \frac{\cos as \cos a+\sin as\sin a}{s}ds\\
&                    =-(\cos a)\displaystyle\int_1^{\infty} \frac{\cos as}{s}ds-(\sin a)\displaystyle\int_1^{\infty} \frac{\sin as}{s}ds\\
&                    =\mathrm{ci}(a)\cos a+\mathrm{si} (a)\sin a
\end{alignat}以上より$$\displaystyle\int_0^1 \frac{\log x}{a^2+(\log x)^2}dx=\mathrm{ci}(a)\cos a+\mathrm{si} (a)\sin a$$







\begin{alignat}{2}
&(4)  \displaystyle\int_0^1 \frac{\log x}{a^2-(\log x)^2}dx=\displaystyle\int_{\infty}^0 \frac{-t}{a^2-t^2}\cdot (-e^{-t})dt=\displaystyle\int_0^{\infty} \frac{te^{-t}}{t^2-a^2}dt\\
&                     =\displaystyle\int_0^{\infty} L[e^{-t}]L^{-1}\left[\frac{t}{t^2-a^2}\right]dt=\displaystyle\int_0^{\infty}\frac{\cosh at}{t+1}dt\\
&                     =\displaystyle\int_1^{\infty} \frac{\cosh (as-a)}{s}ds=\frac{1}{2}\displaystyle\int_1^{\infty} \frac{e^{as-a}+e^{-as+a}}{s}ds\\
&                     =\frac{1}{2}\left(e^{-a}\displaystyle\int_1^{\infty} \frac{e^{as}}{s}ds+e^a\displaystyle\int_1^{\infty} \frac{e^{-as}}{s}ds\right)\\
&                     =-\frac{1}{2}\{e^{-a}\overline{\mathrm{Ei}}(a)+e^a\mathrm{Ei}(-a)\}
\end{alignat}以上より$$\displaystyle\int_0^1 \frac{\log x}{a^2-(\log x)^2}dx=-\frac{1}{2}\{e^{-a}\overline{\mathrm{Ei}}(a)+e^a\mathrm{Ei}(-a)\}$$

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