sin(ax^2+2bx+c)[0,∞]などの定積分

\begin{alignat}{2}
&(1) \displaystyle\int_0^{\infty} \sin (ax^2+2bx)dx\\
&=\sqrt{\frac{π}{2a}}\left[\left\{\frac{1}{2}-S\left(\frac{b}{\sqrt{a}}\right)\right\}\cos \frac{b^2}{a}-\left\{\frac{1}{2}-C\left(\frac{b}{\sqrt{a}}\right)\right\} \sin \frac{b^2}{a}\right]\\
&(2) \displaystyle\int_0^{\infty} \cos (ax^2+2bx)dx\\
&=\sqrt{\frac{π}{2a}}\left[\left\{\frac{1}{2}-C\left(\frac{b}{\sqrt{a}}\right)\right\}\cos \frac{b^2}{a}+\left\{\frac{1}{2}-S\left(\frac{b}{\sqrt{a}}\right)\right\} \sin \frac{b^2}{a}\right]\\
&(3) \displaystyle\int_0^{\infty} \sin(ax^2+2bx+c)dx\\
&=\sqrt{\frac{π}{2a}}\cos \frac{b^2}{a}\left[\left\{\frac{1}{2}-C\left(\frac{b}{\sqrt{a}}\right)\right\}\sin c+\left\{\frac{1}{2}-S\left(\frac{b}{\sqrt{a}}\right)\right\} \cos c\right]\\
&+\sqrt{\frac{π}{2a}}\sin \frac{b^2}{a}\left[\left\{\frac{1}{2}-S\left(\frac{b}{\sqrt{a}}\right)\right\}\sin c-\left\{\frac{1}{2}-C\left(\frac{b}{\sqrt{a}}\right)\right\} \cos c\right]\\
&(4) \displaystyle\int_0^{\infty} \cos(ax^2+2bx+c)dx\\
&=\sqrt{\frac{π}{2a}}\cos \frac{b^2}{a}\left[\left\{\frac{1}{2}-C\left(\frac{b}{\sqrt{a}}\right)\right\}\cos c-\left\{\frac{1}{2}-S\left(\frac{b}{\sqrt{a}}\right)\right\} \sin c\right]\\
&+\sqrt{\frac{π}{2a}}\sin \frac{b^2}{a}\left[\left\{\frac{1}{2}-S\left(\frac{b}{\sqrt{a}}\right)\right\}\cos c+\left\{\frac{1}{2}-C\left(\frac{b}{\sqrt{a}}\right)\right\} \sin c\right]\\
\end{alignat}ただし、全て \(a \gt 0\)














<証明>

証明の途中、次の定積分を用います。(詳細はこちらです。)
\begin{alignat}{2}
&(A) \displaystyle\int_0^{\frac{b}{a}} \sin ax^2dx=\sqrt{\frac{π}{2a}}S\left(\frac{b}{\sqrt{a}}\right)\\
&(B) \displaystyle\int_0^{\frac{b}{a}} \cos ax^2dx=\sqrt{\frac{π}{2a}}C\left(\frac{b}{\sqrt{a}}\right)
\end{alignat}ただし、全て \(a \gt 0\)






\((1)\) 三角関数の変数部分を平方完成します。
\begin{alignat}{2}
&ax^2+2bx=a\left(x^2+\frac{2b}{a}x\right)=a\left(x^2+\frac{2b}{a}x+\frac{b^2}{a^2}\right)-\frac{b^2}{a}\\
&         =a\left(x+\frac{b}{a}\right)^2-\frac{b^2}{a}
\end{alignat}加法定理で三角関数を切り離します。
\begin{alignat}{2}
&  \displaystyle\int_0^{\infty} \sin (ax^2+2bx)dx\\
&=\displaystyle\int_0^{\frac{b}{a}} \sin \left\{a\left(x+\frac{b}{a}\right)^2-\frac{b^2}{a}\right\}dx\\
&=\displaystyle\int_0^{\infty} \left[\sin \left\{a\left(x+\frac{b}{a}\right)^2\right\} \cos \frac{b^2}{a}-\cos \left\{a\left(x+\frac{b}{a}\right)^2\right\} \sin \frac{b^2}{a}\right]\\
&=\cos \frac{b^2}{a}\displaystyle\int_0^{\infty} \sin \left\{a\left(x+\frac{b}{a}\right)^2\right\}dx-\sin \frac{b^2}{a} \displaystyle\int_0^{\infty }\cos \left\{a\left(x+\frac{b}{a}\right)^2\right\} dx
\end{alignat}どちらの積分も \(\displaystyle x+\frac{b}{a}=t\) と置きます。\((dx=dt)\)
\begin{alignat}{2}
&=\cos \frac{b^2}{a}\displaystyle\int_{\frac{b}{a}}^{\infty} \sin at^2 dt-\sin \frac{b^2}{a} \displaystyle\int_{\frac{b}{a}}^{\infty }\cos at^2dt\\
&=\cos \frac{b^2}{a}\left(\displaystyle\int_0^{\infty} \sin at^2 dt-\displaystyle\int_0^{\frac{b}{a}} \sin at^2 dt\right)-\sin \frac{b^2}{a} \left(\displaystyle\int_0^{\infty} \cos at^2 dt-\displaystyle\int_0^{\frac{b}{a}} \cos at^2 dt\right)\\
\end{alignat}\((A)(B)\) を代入します。
\begin{alignat}{2}
&=\cos \frac{b^2}{a}\left\{\frac{1}{2}\sqrt{\frac{π}{2a}}-\sqrt{\frac{π}{2a}}S\left(\frac{b}{\sqrt{a}}\right)\right\}-\sin \frac{b^2}{a}\left\{\frac{1}{2}\sqrt{\frac{π}{2a}}-\sqrt{\frac{π}{2a}}C\left(\frac{b}{\sqrt{a}}\right)\right\}\\
&=\sqrt{\frac{π}{2a}}\left[\left\{\frac{1}{2}-S\left(\frac{b}{\sqrt{a}}\right)\right\}\cos \frac{b^2}{a}-\left\{\frac{1}{2}-C\left(\frac{b}{\sqrt{a}}\right)\right\} \sin \frac{b^2}{a}\right]
\end{alignat}以上より$$\displaystyle\int_0^{\infty} \sin (ax^2+2bx)dx=\sqrt{\frac{π}{2a}}\left[\left\{\frac{1}{2}-S\left(\frac{b}{\sqrt{a}}\right)\right\}\cos \frac{b^2}{a}-\left\{\frac{1}{2}-C\left(\frac{b}{\sqrt{a}}\right)\right\} \sin \frac{b^2}{a}\right]$$







\((2)\) \((1)\) と同様の流れで解きます。
\begin{alignat}{2}
&  \displaystyle\int_0^{\infty} \cos (ax^2+2bx)dx\\
&=\displaystyle\int_0^{\frac{b}{a}} \cos \left\{a\left(x+\frac{b}{a}\right)^2-\frac{b^2}{a}\right\}dx\\
&=\displaystyle\int_0^{\infty} \left[\cos \left\{a\left(x+\frac{b}{a}\right)^2\right\} \cos \frac{b^2}{a}+\sin \left\{a\left(x+\frac{b}{a}\right)^2\right\} \sin \frac{b^2}{a}\right]\\
&=\cos \frac{b^2}{a}\displaystyle\int_0^{\infty} \cos \left\{a\left(x+\frac{b}{a}\right)^2\right\}dx+\sin \frac{b^2}{a} \displaystyle\int_0^{\infty }\sin \left\{a\left(x+\frac{b}{a}\right)^2\right\} dx\\
&=\cos \frac{b^2}{a}\displaystyle\int_{\frac{b}{a}}^{\infty} \cos at^2 dt+\sin \frac{b^2}{a} \displaystyle\int_{\frac{b}{a}}^{\infty }\sin at^2dt\\
&=\cos \frac{b^2}{a}\left(\displaystyle\int_0^{\infty} \cos at^2 dt-\displaystyle\int_0^{\frac{b}{a}} \cos at^2 dt\right)+\sin \frac{b^2}{a} \left(\displaystyle\int_0^{\infty} \sin at^2 dt-\displaystyle\int_0^{\frac{b}{a}} \sin at^2 dt\right)\\
&=\cos \frac{b^2}{a}\left\{\frac{1}{2}\sqrt{\frac{π}{2a}}-\sqrt{\frac{π}{2a}}C\left(\frac{b}{\sqrt{a}}\right)\right\}+\sin \frac{b^2}{a}\left\{\frac{1}{2}\sqrt{\frac{π}{2a}}-\sqrt{\frac{π}{2a}}S\left(\frac{b}{\sqrt{a}}\right)\right\}\\
&=\sqrt{\frac{π}{2a}}\left[\left\{\frac{1}{2}-C\left(\frac{b}{\sqrt{a}}\right)\right\}\cos \frac{b^2}{a}+\left\{\frac{1}{2}-S\left(\frac{b}{\sqrt{a}}\right)\right\} \sin \frac{b^2}{a}\right]
\end{alignat}以上より$$\displaystyle\int_0^{\infty} \cos (ax^2+2bx)dx=\sqrt{\frac{π}{2a}}\left[\left\{\frac{1}{2}-C\left(\frac{b}{\sqrt{a}}\right)\right\}\cos \frac{b^2}{a}+\left\{\frac{1}{2}-S\left(\frac{b}{\sqrt{a}}\right)\right\} \sin \frac{b^2}{a}\right]$$








\((3)(4)\) 加法定理で三角関数を切り離し \((1)(2)\) の結果を代入します。
\begin{alignat}{2}
&(3) \displaystyle\int_0^{\infty} \sin (ax^2+2bx+c)dx\\
&=\displaystyle\int_0^{\infty} \{\sin(ax^2+2bx)\cos c+\cos(ax^2+2bx)\sin c\}dx\\
&=\sqrt{\frac{π}{2a}}\left[\left\{\frac{1}{2}-S\left(\frac{b}{\sqrt{a}}\right)\right\}\cos \frac{b^2}{a}-\left\{\frac{1}{2}-C\left(\frac{b}{\sqrt{a}}\right)\right\} \sin \frac{b^2}{a}\right]\cos c+\sqrt{\frac{π}{2a}}\left[\left\{\frac{1}{2}-C\left(\frac{b}{\sqrt{a}}\right)\right\}\cos \frac{b^2}{a}+\left\{\frac{1}{2}-S\left(\frac{b}{\sqrt{a}}\right)\right\} \sin \frac{b^2}{a}\right] \sin c\\
&=\sqrt{\frac{π}{2a}}\cos \frac{b^2}{a}\left[\left\{\frac{1}{2}-C\left(\frac{b}{\sqrt{a}}\right)\right\}\sin c+\left\{\frac{1}{2}-S\left(\frac{b}{\sqrt{a}}\right)\right\} \cos c\right]+\sqrt{\frac{π}{2a}}\sin \frac{b^2}{a}\left[\left\{\frac{1}{2}-S\left(\frac{b}{\sqrt{a}}\right)\right\}\sin c-\left\{\frac{1}{2}-C\left(\frac{b}{\sqrt{a}}\right)\right\} \cos c\right]\\
\end{alignat}以上より$$\displaystyle\int_0^{\infty} \sin (ax^2+2bx+c)dx=\sqrt{\frac{π}{2a}}\cos \frac{b^2}{a}\left[\left\{\frac{1}{2}-C\left(\frac{b}{\sqrt{a}}\right)\right\}\sin c+\left\{\frac{1}{2}-S\left(\frac{b}{\sqrt{a}}\right)\right\} \cos c\right]+\sqrt{\frac{π}{2a}}\sin \frac{b^2}{a}\left[\left\{\frac{1}{2}-S\left(\frac{b}{\sqrt{a}}\right)\right\}\sin c-\left\{\frac{1}{2}-C\left(\frac{b}{\sqrt{a}}\right)\right\} \cos c\right]$$







\begin{alignat}{2}
&(4) \displaystyle\int_0^{\infty} \cos (ax^2+2bx+c)dx\\
&=\displaystyle\int_0^{\infty} \{\cos(ax^2+2bx)\cos c-\sin(ax^2+2bx)\sin c\}dx\\
&=\sqrt{\frac{π}{2a}}\left[\left\{\frac{1}{2}-C\left(\frac{b}{\sqrt{a}}\right)\right\}\cos \frac{b^2}{a}+\left\{\frac{1}{2}-S\left(\frac{b}{\sqrt{a}}\right)\right\} \sin \frac{b^2}{a}\right]\cos c-\sqrt{\frac{π}{2a}}\left[\left\{\frac{1}{2}-S\left(\frac{b}{\sqrt{a}}\right)\right\}\cos \frac{b^2}{a}-\left\{\frac{1}{2}-C\left(\frac{b}{\sqrt{a}}\right)\right\} \sin \frac{b^2}{a}\right] \sin c\\
&=\sqrt{\frac{π}{2a}}\cos \frac{b^2}{a}\left[\left\{\frac{1}{2}-C\left(\frac{b}{\sqrt{a}}\right)\right\}\cos c-\left\{\frac{1}{2}-S\left(\frac{b}{\sqrt{a}}\right)\right\} \sin c\right]+\sqrt{\frac{π}{2a}}\sin \frac{b^2}{a}\left[\left\{\frac{1}{2}-S\left(\frac{b}{\sqrt{a}}\right)\right\}\cos c+\left\{\frac{1}{2}-C\left(\frac{b}{\sqrt{a}}\right)\right\} \sin c\right]\\
\end{alignat}以上より$$\displaystyle\int_0^{\infty} \cos (ax^2+2bx+c)dx=\sqrt{\frac{π}{2a}}\cos \frac{b^2}{a}\left[\left\{\frac{1}{2}-C\left(\frac{b}{\sqrt{a}}\right)\right\}\cos c-\left\{\frac{1}{2}-S\left(\frac{b}{\sqrt{a}}\right)\right\} \sin c\right]+\sqrt{\frac{π}{2a}}\sin \frac{b^2}{a}\left[\left\{\frac{1}{2}-S\left(\frac{b}{\sqrt{a}}\right)\right\}\cos c+\left\{\frac{1}{2}-C\left(\frac{b}{\sqrt{a}}\right)\right\} \sin c\right]$$

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