Ei(-ax)e^{-μx}[0,∞]などの定積分

\begin{alignat}{2}
&(1)  \displaystyle\int_0^{\infty} \mathrm{Ei}(-ax)dx=-\frac{1}{a}\\
&(2)  \displaystyle\int_0^{\infty} \mathrm{Ei}(-ax)e^{-μx}dx=-\frac{1}{μ}\log \left(1+\frac{μ}{a}\right)\\
&(3)  \displaystyle\int_0^{\infty} \mathrm{Ei}(ax)e^{-μx}dx=-\frac{1}{μ}\log \left(\frac{μ}{a}-1\right)  (μ \gt a)\\
\end{alignat}ただし、全て \(μ,a \gt 0\)











<証明>

全て \(t=axy\) と置きます。\((dt=axdy)\)

\begin{alignat}{2}
(1)  \displaystyle\int_0^{\infty} \mathrm{Ei}(-ax)dx&=\displaystyle\int_0^{\infty} \left(-\displaystyle\int_{ax}^{\infty} \frac{e^{-t}}{t}dt\right)dx\\
&=\displaystyle\int_0^{\infty} \left(-\displaystyle\int_1^{\infty} \frac{e^{-axy}}{axy} \cdot axdy\right)dx\\
&=\displaystyle\int_0^{\infty} \left(-\displaystyle\int_1^{\infty} \frac{e^{-axy}}{y}dy\right)dx\\
&=-\displaystyle\int_1^{\infty} \frac{1}{y}\left(\displaystyle\int_0^{\infty} e^{-ayx}dx\right)dy\\
&=-\displaystyle\int_1^{\infty} \frac{1}{y}\left[-\frac{e^{-ayx}}{ay}\right]_0^{\infty} dy\\
&=-\displaystyle\int_1^{\infty} \frac{1}{y} \cdot \frac{1}{ay}dy=-\frac{1}{a} \displaystyle\int_1^{\infty} \frac{1}{y^2}dy\\
&=-\frac{1}{a}\left[-\frac{1}{y}\right]_1^{\infty}=-\frac{1}{a}\\
\end{alignat}以上より$$\displaystyle\int_0^{\infty} \mathrm{Ei}(-ax)dx=-\frac{1}{a}$$







\begin{alignat}{2}
(2)  \displaystyle\int_0^{\infty} \mathrm{Ei}(-ax)e^{-μx}dx&=\displaystyle\int_0^{\infty} \left(-\displaystyle\int_{ax}^{\infty} \frac{e^{-t}}{t}dt\right)e^{-μx}dx\\
&=\displaystyle\int_0^{\infty} \left(-\displaystyle\int_1^{\infty} \frac{e^{-axy}}{axy} \cdot axdy\right)e^{-μx}dx\\
&=\displaystyle\int_0^{\infty} \left(-\displaystyle\int_1^{\infty} \frac{e^{-axy}}{y}dy\right)e^{-μx}dx\\
&=-\displaystyle\int_1^{\infty} \frac{1}{y}\left\{\displaystyle\int_0^{\infty} e^{-(ay+μ)x}dx\right\}dy\\
&=-\displaystyle\int_1^{\infty} \frac{1}{y}\left[-\frac{e^{-(ay+μ)x}}{ay+μ}\right]_0^{\infty} dy\\
&=-\displaystyle\int_1^{\infty} \frac{1}{y} \cdot \frac{1}{ay+μ}dy=-\frac{1}{μ}\displaystyle\int_1^{\infty} \left(\frac{1}{y}-\frac{1}{y+\frac{μ}{a}}\right)dy\\
&-\frac{1}{μ}\left[\log |y| -\log \left|y+\frac{μ}{a}\right|\right]_1^{\infty}=-\frac{1}{μ}\left[\log \left|\frac{y}{y+\frac{μ}{a}}\right|\right]_1^{\infty}\\
&=-\frac{1}{μ}\left(-\log \frac{1}{1+\frac{μ}{a}}\right)=-\frac{1}{μ}\log \left(1+\frac{μ}{a}\right)
\end{alignat}以上より$$\displaystyle\int_0^{\infty} \mathrm{Ei}(-ax)e^{-μx}dx=-\frac{1}{μ}\log \left(1+\frac{μ}{a}\right)$$







\begin{alignat}{2}
(3)  \displaystyle\int_0^{\infty} \mathrm{Ei}(ax)e^{-μx}dx&=\displaystyle\int_0^{\infty} \left(-\displaystyle\int_{-ax}^{\infty} \frac{e^{-t}}{t}dt\right)e^{-μx}dx\\
&=\displaystyle\int_0^{\infty} \left(-\displaystyle\int_{-1}^{\infty} \frac{e^{-axy}}{axy} \cdot axdy\right)e^{-μx}dx\\
&=\displaystyle\int_0^{\infty} \left(-\displaystyle\int_{-1}^{\infty} \frac{e^{-axy}}{y}dy\right)e^{-μx}dx\\
&=-\displaystyle\int_{-1}^{\infty} \frac{1}{y}\left\{\displaystyle\int_0^{\infty} e^{-(ay+μ)x}dx\right\}dy\\
&=-\displaystyle\int_{-1}^{\infty} \frac{1}{y}\left[-\frac{e^{-(ay+μ)x}}{ay+μ}\right]_0^{\infty} dy\\
&=-\displaystyle\int_{-1}^{\infty} \frac{1}{y} \cdot \frac{1}{ay+μ}dy=-\frac{1}{μ}\displaystyle\int_{-1}^{\infty} \left(\frac{1}{y}-\frac{1}{y+\frac{μ}{a}}\right)dy\\
&-\frac{1}{μ}\left[\log |y| -\log \left|y+\frac{μ}{a}\right|\right]_{-1}^{\infty}=-\frac{1}{μ}\left[\log \left|\frac{y}{y+\frac{μ}{a}}\right|\right]_{-1}^{\infty}\\
&=-\frac{1}{μ}\left(-\log \frac{1}{\frac{μ}{a}-1}\right)=-\frac{1}{μ}\log \left(\frac{μ}{a}-1\right)
\end{alignat}以上より$$\displaystyle\int_0^{\infty} \mathrm{Ei}(ax)e^{-μx}dx=-\frac{1}{μ}\log \left(\frac{μ}{a}-1\right)$$

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