sin(ax)/(x+b)[0,∞]などの定積分

\begin{alignat}{2}
&(1) \displaystyle\int_0^{\infty} \frac{\sin (ax)}{x+b}dx=\sin(ab)\mathrm{ci
} (ab) – \cos (ab) \mathrm{si} (ab)\\
&(2) \displaystyle\int_0^{\infty} \frac{\cos (ax)}{x+b}dx=-\sin(ab)\mathrm{si
} (ab) – \cos (ab) \mathrm{ci} (ab)\\
&(3) \displaystyle\int_0^{\infty} \frac{\sin (ax)}{b-x}dx=\sin(ab)\mathrm{ci
} (ab) – \cos (ab) \{\mathrm{si} (ab)+π\}\\
&(4) \displaystyle\int_0^{\infty} \frac{\cos (ax)}{b-x}dx=\cos (ab)\mathrm{ci
} (ab) + \sin (ab) \{\mathrm{si} (ab)+π\}\\
\end{alignat}ただし、全て \(a,b \gt 0\)








<証明>

予め、次の計算をしておきます。

\((A)(B)\) では \(t=bs, dt=bds\) と置いています。
\((C)(D)\) では \(t=-bs, dt=-bds\) と置いています。
\begin{alignat}{2}
&(A) \displaystyle\int_b^{\infty} \frac{\sin (at)}{t}dt=\displaystyle\int_1^{\infty} \frac{\sin (abs)}{s}ds=-\mathrm{si}(ab)\\
&(B) \displaystyle\int_b^{\infty} \frac{\cos (at)}{t}dt=\displaystyle\int_1^{\infty} \frac{\cos (abs)}{s}ds=-\mathrm{ci}(ab)\\
&\\
&(C) \displaystyle\int_{-b}^{\infty} \frac{\sin (at)}{t}dt=\displaystyle\int_1^{-\infty} \frac{\sin (abs)}{s}ds=-\displaystyle\int_{-\infty}^1 \frac{\sin (abs)}{s}ds\\
&=\displaystyle\int_{-\infty}^{\infty} \frac{\sin (abs)}{s}ds-\displaystyle\int_1^{\infty} \frac{\sin (abs)}{s}ds=π+\mathrm{si}(ab)\\
&(D) \displaystyle\int_{-b}^{\infty} \frac{\cos (at)}{t}dt=\displaystyle\int_1^{-\infty} \frac{\cos (abs)}{s}ds=-\displaystyle\int_{-\infty}^1 \frac{\cos (abs)}{s}ds\\
&=-\left\{\displaystyle\int_{-\infty}^{\infty} \frac{\cos (abs)}{s}ds-\displaystyle\int_1^{\infty} \frac{\cos (abs)}{s}ds\right\}=\displaystyle\int_1^{\infty} \frac{\cos (abs)}{s}ds=-\mathrm{ci}(ab)
\end{alignat}





\((1)\) \(x+b=t\) と置きます。\((dx=dt)\)
\begin{alignat}{2}
&\displaystyle\int_0^{\infty} \frac{\sin (ax)}{x+b}dx=\displaystyle\int_b^{\infty} \frac{\sin (at-ab)}{t}dt=\displaystyle\int_b^{\infty} \frac{\sin (at) \cos (ab)- \cos (at) \sin (ab)}{t}dt\\
&             =\cos (ab) \displaystyle\int_b^{\infty} \frac{\sin (at)}{t}dt-\sin (ab)\displaystyle\int_b^{\infty} \frac{\cos (at)}{t}dt\\
&             =-\cos (ab) \mathrm{si}(ab)+\sin (ab) \mathrm{ci}(ab)\\
\end{alignat}以上より$$\displaystyle\int_0^{\infty} \frac{\sin (ax)}{x+b}dx=\sin(ab)\mathrm{ci
} (ab) – \cos (ab) \mathrm{si} (ab)$$







\((2)\) \(x+b=t\) と置きます。\((dx=dt)\)
\begin{alignat}{2}
&\displaystyle\int_0^{\infty} \frac{\cos (ax)}{x+b}dx=\displaystyle\int_b^{\infty} \frac{\cos (at-ab)}{t}dt=\displaystyle\int_b^{\infty} \frac{\cos (at) \cos (ab)+ \sin (at) \sin (ab)}{t}dt\\
&             =\cos (ab) \displaystyle\int_b^{\infty} \frac{\cos (at)}{t}dt+\sin (ab)\displaystyle\int_b^{\infty} \frac{\sin (at)}{t}dt\\
&             =-\cos (ab) \mathrm{ci}(ab)-\sin (ab) \mathrm{si}(ab)\\
\end{alignat}以上より$$\displaystyle\int_0^{\infty} \frac{\cos (ax)}{x+b}dx=-\sin(ab)\mathrm{si
} (ab) – \cos (ab) \mathrm{ci} (ab)$$







\((3)\) \(x-b=t\) と置きます。\((dx=dt)\)
\begin{alignat}{2}
&\displaystyle\int_0^{\infty} \frac{\sin (ax)}{b-x}dx=\displaystyle\int_{-b}^{\infty} \frac{\sin (at+ab)}{-t}dt=-\displaystyle\int_{-b}^{\infty} \frac{\sin (at) \cos (ab)+ \cos (at) \sin (ab)}{t}dt\\
&             =-\cos (ab) \displaystyle\int_{-b}^{\infty} \frac{\sin (at)}{t}dt- \sin (ab)\displaystyle\int_{-b}^{\infty} \frac{\cos (at)}{t}dt\\
&             =-\cos (ab) \{π+\mathrm{si}(ab)\}+ \sin (ab) \mathrm{ci}(ab)\\
\end{alignat}以上より$$\displaystyle\int_0^{\infty} \frac{\sin (ax)}{b-x}dx=\sin(ab)\mathrm{ci
} (ab) – \cos (ab) \{\mathrm{si} (ab)+π\}$$







\((4)\) \(x-b=t\) と置きます。\((dx=dt)\)
\begin{alignat}{2}
&\displaystyle\int_0^{\infty} \frac{\cos (ax)}{b-x}dx=\displaystyle\int_{-b}^{\infty} \frac{\cos (at+ab)}{-t}dt=-\displaystyle\int_{-b}^{\infty} \frac{\cos (at) \cos (ab)- \sin (at) \sin (ab)}{t}dt\\
&             =-\cos (ab) \displaystyle\int_{-b}^{\infty} \frac{\cos (at)}{t}dt+ \sin (ab)\displaystyle\int_{-b}^{\infty} \frac{\sin (at)}{t}dt\\
&             =\cos (ab) \mathrm{ci}(ab)+ \sin (ab)\{π+\mathrm{si}(ab)\} \\
\end{alignat}以上より$$\displaystyle\int_0^{\infty} \frac{\cos (ax)}{b-x}dx=\cos (ab)\mathrm{ci
} (ab) + \sin (ab) \{\mathrm{si} (ab)+π\}$$

“sin(ax)/(x+b)[0,∞]などの定積分” への2件の返信

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