sin(qx^2+p^2/q)sin2px[0,∞]などの定積分

\begin{alignat}{2}
&(1) \displaystyle\int_0^{\infty} \sin\left(qx^2+\frac{p^2}{q}\right) \cos 2pxdx=\frac{1}{2}\sqrt{\frac{π}{2q}}\\
&(2) \displaystyle\int_0^{\infty} \cos\left(qx^2+\frac{p^2}{q}\right) \cos 2pxdx=\frac{1}{2}\sqrt{\frac{π}{2q}}\\
\end{alignat}



<証明>
積和の公式、及びフレネル積分を用います。

\begin{alignat}{2}
&(1) \displaystyle\int_0^{\infty} \sin\left(qx^2+\frac{p^2}{q}\right) \cos 2pxdx\\
&=\frac{1}{2}\displaystyle\int_0^{\infty} \left\{ \sin \left(qx^2+2px+\frac{p^2}{q}\right)+\sin \left(qx^2-2px+\frac{p^2}{q}\right)\right\}dx\\
&=\frac{1}{2}\displaystyle\int_0^{\infty} \left\{ \sin \left(\sqrt{q}x+\frac{p}{\sqrt{q}}\right)^2+ \sin \left(\sqrt{q}x-\frac{p}{\sqrt{q}}\right)^2\right\}dx
\end{alignat}左右の積分をそれぞれ計算します。
左の積分は \(\displaystyle \sqrt{q}x+\frac{p}{\sqrt{q}}=t\) と置きます。\((\sqrt{q}dx=dt)\)
右の積分は \(\displaystyle \sqrt{q}x-\frac{p}{\sqrt{q}}=t\) と置きます。\((\sqrt{q}dx=dt)\)
\begin{alignat}{2}
&(A) \displaystyle\int_0^{\infty} \sin \left(\sqrt{q}x+\frac{p}{\sqrt{q}}\right)^2dx=\displaystyle\int_{\frac{p}{\sqrt{q}}}^{\infty} \sin t^2\left(\frac{1}{\sqrt{q}}dt\right)=\frac{1}{\sqrt{q}}\displaystyle\int_{\frac{p}{\sqrt{q}}}^{\infty} \sin t^2 dt\\
&                          =\frac{1}{\sqrt{q}}\left(\displaystyle\int_0^{\infty} \sin t^2 dt+\displaystyle\int_{\frac{p}{\sqrt{q}}}^0 \sin t^2 dt\right)\\
&                          =\frac{1}{\sqrt{q}}\left\{\frac{1}{2}\sqrt{\frac{π}{2}}-\displaystyle\int_0^{\frac{p}{\sqrt{q}}} \sin t^2 dt\right\}\\
&                          =\frac{1}{\sqrt{q}}\left(\frac{1}{2}\sqrt{\frac{π}{2}}-\displaystyle\int_0^{\frac{p}{\sqrt{q}}} \sin x^2 dx\right)\\
\end{alignat}
\begin{alignat}{2}
&(B) \displaystyle\int_0^{\infty} \sin \left(\sqrt{q}x-\frac{p}{\sqrt{q}}\right)^2dx=\displaystyle\int_{-\frac{p}{\sqrt{q}}}^{\infty} \sin t^2\left(\frac{1}{\sqrt{q}}dt\right)=\frac{1}{\sqrt{q}}\displaystyle\int_{-\frac{p}{\sqrt{q}}}^{\infty} \sin t^2 dt\\
&                          =\frac{1}{\sqrt{q}}\left(\displaystyle\int_0^{\infty} \sin t^2 dt+\displaystyle\int_{-\frac{p}{\sqrt{q}}}^0 \sin t^2 dt\right)
\end{alignat}右の積分で \(t=-s\) と置きます。\((dt=-ds)\)
\begin{alignat}{2}
&=\frac{1}{\sqrt{q}}\left\{\frac{1}{2}\sqrt{\frac{π}{2}}+\displaystyle\int_{\frac{p}{\sqrt{q}}}^0 \sin s^2 (-ds)\right\}=\frac{1}{\sqrt{q}}\left(\frac{1}{2}\sqrt{\frac{π}{2}}+\displaystyle\int_0^{\frac{p}{\sqrt{q}}} \sin s^2 ds\right)\\
&                              =\frac{1}{\sqrt{q}}\left(\frac{1}{2}\sqrt{\frac{π}{2}}+\displaystyle\int_0^{\frac{p}{\sqrt{q}}} \sin x^2 dx\right)\\
\end{alignat}よって
\begin{alignat}{2}
&\displaystyle\int_0^{\infty} \sin\left(qx^2+\frac{p^2}{q}\right) \cos 2pxdx\\
&=\frac{1}{2}\left\{\frac{1}{\sqrt{q}}\left(\frac{1}{2}\sqrt{\frac{π}{2}}-\displaystyle\int_0^{\frac{p}{\sqrt{q}}} \sin x^2 dx\right)+\frac{1}{\sqrt{q}}\left(\frac{1}{2}\sqrt{\frac{π}{2}}+\displaystyle\int_0^{\frac{p}{\sqrt{q}}} \sin x^2 dx\right)\right\}\\
&=\frac{1}{2}\cdot \frac{1}{\sqrt{q}}\sqrt{\frac{π}{2}}=\frac{1}{2}\sqrt{\frac{π}{2q}}\\
\end{alignat}





\begin{alignat}{2}
&(2) \displaystyle\int_0^{\infty} \cos\left(qx^2+\frac{p^2}{q}\right) \cos 2pxdx\\
&=\frac{1}{2}\displaystyle\int_0^{\infty} \left\{ \cos \left(qx^2+2px+\frac{p^2}{q}\right)+\cos \left(qx^2-2px+\frac{p^2}{q}\right)\right\}dx\\
&=\frac{1}{2}\displaystyle\int_0^{\infty} \left\{ \cos \left(\sqrt{q}x+\frac{p}{\sqrt{q}}\right)^2+ \cos \left(\sqrt{q}x-\frac{p}{\sqrt{q}}\right)^2\right\}dx
\end{alignat}左右の積分をそれぞれ計算します。
左の積分は \(\displaystyle \sqrt{q}x+\frac{p}{\sqrt{q}}=t\) と置きます。\((\sqrt{q}dx=dt)\)
右の積分は \(\displaystyle \sqrt{q}x-\frac{p}{\sqrt{q}}=t\) と置きます。\((\sqrt{q}dx=dt)\)
\begin{alignat}{2}
&(A) \displaystyle\int_0^{\infty} \cos \left(\sqrt{q}x+\frac{p}{\sqrt{q}}\right)^2dx=\displaystyle\int_{\frac{p}{\sqrt{q}}}^{\infty} \cos t^2\left(\frac{1}{\sqrt{q}}dt\right)=\frac{1}{\sqrt{q}}\displaystyle\int_{\frac{p}{\sqrt{q}}}^{\infty} \cos t^2 dt\\
&                          =\frac{1}{\sqrt{q}}\left(\displaystyle\int_0^{\infty} \cos t^2 dt+\displaystyle\int_{\frac{p}{\sqrt{q}}}^0 \cos t^2 dt\right)\\
&                          =\frac{1}{\sqrt{q}}\left\{\frac{1}{2}\sqrt{\frac{π}{2}}-\displaystyle\int_0^{\frac{p}{\sqrt{q}}} \cos t^2 dt\right\}\\
&                          =\frac{1}{\sqrt{q}}\left(\frac{1}{2}\sqrt{\frac{π}{2}}-\displaystyle\int_0^{\frac{p}{\sqrt{q}}} \cos x^2 dx\right)\\
\end{alignat}
\begin{alignat}{2}
&(B) \displaystyle\int_0^{\infty} \cos \left(\sqrt{q}x-\frac{p}{\sqrt{q}}\right)^2dx=\displaystyle\int_{-\frac{p}{\sqrt{q}}}^{\infty} \cos t^2\left(\frac{1}{\sqrt{q}}dt\right)=\frac{1}{\sqrt{q}}\displaystyle\int_{-\frac{p}{\sqrt{q}}}^{\infty} \cos t^2 dt\\
&                          =\frac{1}{\sqrt{q}}\left(\displaystyle\int_0^{\infty} \cos t^2 dt+\displaystyle\int_{-\frac{p}{\sqrt{q}}}^0 \cos t^2 dt\right)
\end{alignat}右の積分で \(t=-s\) と置きます。\((dt=-ds)\)
\begin{alignat}{2}
&=\frac{1}{\sqrt{q}}\left\{\frac{1}{2}\sqrt{\frac{π}{2}}+\displaystyle\int_{\frac{p}{\sqrt{q}}}^0 \cos s^2 (-ds)\right\}=\frac{1}{\sqrt{q}}\left(\frac{1}{2}\sqrt{\frac{π}{2}}+\displaystyle\int_0^{\frac{p}{\sqrt{q}}} \cos s^2 ds\right)\\
&                              =\frac{1}{\sqrt{q}}\left(\frac{1}{2}\sqrt{\frac{π}{2}}+\displaystyle\int_0^{\frac{p}{\sqrt{q}}} \cos x^2 dx\right)\\
\end{alignat}よって
\begin{alignat}{2}
&\displaystyle\int_0^{\infty} \cos\left(qx^2+\frac{p^2}{q}\right) \cos 2pxdx\\
&=\frac{1}{2}\left\{\frac{1}{\sqrt{q}}\left(\frac{1}{2}\sqrt{\frac{π}{2}}-\displaystyle\int_0^{\frac{p}{\sqrt{q}}} \cos x^2 dx\right)+\frac{1}{\sqrt{q}}\left(\frac{1}{2}\sqrt{\frac{π}{2}}+\displaystyle\int_0^{\frac{p}{\sqrt{q}}} \cos x^2 dx\right)\right\}\\
&=\frac{1}{2}\cdot \frac{1}{\sqrt{q}}\sqrt{\frac{π}{2}}=\frac{1}{2}\sqrt{\frac{π}{2q}}\\
\end{alignat}

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