フレネル積分の級数表示
フレネル積分の級数表示
フレネル積分を級数表示すると次のようになる。
\[ S\left(x\right)=\sum_{k=0}^{\infty}\left(-1\right)^{k}\frac{x^{2\alpha k+\alpha+1}}{\left(2k+1\right)!\left(2\alpha k+\alpha+1\right)} \]
\begin{align*} C\left(x\right) & =\sum_{k=0}^{\infty}\left(-1\right)^{k}\frac{x^{2\alpha k+1}}{\left(2k\right)!\left(2\alpha k+1\right)} \end{align*}
フレネル積分を級数表示すると次のようになる。
(1)
\begin{align*} S\left(x\right) & =\int_{0}^{x}\sin\left(x^{2}\right)dx\\ & =\sum_{k=0}^{\infty}\left(-1\right)^{k}\frac{x^{4k+3}}{\left(2k+1\right)!\left(4k+3\right)} \end{align*}(2)
\begin{align*} C\left(x\right) & =\int_{0}^{x}\cos\left(x^{2}\right)dx\\ & =\sum_{k=0}^{\infty}\left(-1\right)^{k}\frac{x^{4k+1}}{\left(2k\right)!\left(4k+1\right)} \end{align*}(3)
\(\alpha\in\mathbb{C}\setminus\mathbb{Z}^{-}\)とする。\[ S\left(x\right)=\sum_{k=0}^{\infty}\left(-1\right)^{k}\frac{x^{2\alpha k+\alpha+1}}{\left(2k+1\right)!\left(2\alpha k+\alpha+1\right)} \]
(4)
\(\alpha\in\mathbb{C}\setminus\left\{ -\frac{1}{2},-\frac{1}{4},-\frac{1}{8},\cdots\right\} \)とする。\begin{align*} C\left(x\right) & =\sum_{k=0}^{\infty}\left(-1\right)^{k}\frac{x^{2\alpha k+1}}{\left(2k\right)!\left(2\alpha k+1\right)} \end{align*}
(1)
\begin{align*} S\left(x\right) & =\int_{0}^{x}\sin\left(x^{2}\right)dx\\ & =\int_{0}^{x}\sum_{k=0}^{\infty}\left(-1\right)^{k}\frac{\left(x^{2}\right)^{2k+1}}{\left(2k+1\right)!}dx\\ & =\sum_{k=0}^{\infty}\frac{\left(-1\right)^{k}}{\left(2k+1\right)!}\int_{0}^{x}x^{4k+2}dx\\ & =\sum_{k=0}^{\infty}\frac{\left(-1\right)^{k}}{\left(2k+1\right)!}\cdot\frac{1}{\left(4k+3\right)}\left[x^{4k+3}\right]_{0}^{x}\\ & =\sum_{k=0}^{\infty}\left(-1\right)^{k}\frac{x^{4k+3}}{\left(2k+1\right)!\left(4k+3\right)} \end{align*}(2)
\begin{align*} C\left(x\right) & =\int_{0}^{x}\cos\left(x^{2}\right)dx\\ & =\int_{0}^{x}\sum_{k=0}^{\infty}\left(-1\right)^{k}\frac{\left(x^{2}\right)^{2k}}{\left(2k\right)!}dx\\ & =\sum_{k=0}^{\infty}\frac{\left(-1\right)^{k}}{\left(2k\right)!}\int_{0}^{x}x^{4k}dx\\ & =\sum_{k=0}^{\infty}\frac{\left(-1\right)^{k}}{\left(2k\right)!}\cdot\frac{1}{\left(4k+1\right)}\left[x^{4k+1}\right]_{0}^{x}\\ & =\sum_{k=0}^{\infty}\left(-1\right)^{k}\frac{x^{4k+1}}{\left(2k\right)!\left(4k+1\right)} \end{align*}(3)
\begin{align*} S\left(x\right) & =\int_{0}^{x}\sin\left(x^{\alpha}\right)dx\\ & =\int_{0}^{x}\sum_{k=0}^{\infty}\left(-1\right)^{k}\frac{\left(x^{\alpha}\right)^{2k+1}}{\left(2k+1\right)!}dx\\ & =\sum_{k=0}^{\infty}\frac{\left(-1\right)^{k}}{\left(2k+1\right)!}\int_{0}^{x}x^{2\alpha k+\alpha}dx\\ & =\sum_{k=0}^{\infty}\frac{\left(-1\right)^{k}}{\left(2k+1\right)!}\cdot\frac{1}{\left(2\alpha k+\alpha+1\right)}\left[x^{2\alpha k+\alpha+1}\right]_{0}^{x}\\ & =\sum_{k=0}^{\infty}\left(-1\right)^{k}\frac{x^{2\alpha k+\alpha+1}}{\left(2k+1\right)!\left(2\alpha k+\alpha+1\right)} \end{align*}(4)
\begin{align*} C\left(x\right) & =\int_{0}^{x}\cos\left(x^{\alpha}\right)dx\\ & =\int_{0}^{x}\sum_{k=0}^{\infty}\left(-1\right)^{k}\frac{\left(x^{\alpha}\right)^{2k}}{\left(2k\right)!}dx\\ & =\sum_{k=0}^{\infty}\frac{\left(-1\right)^{k}}{\left(2k\right)!}\int_{0}^{x}x^{2\alpha k}dx\\ & =\sum_{k=0}^{\infty}\frac{\left(-1\right)^{k}}{\left(2k\right)!}\cdot\frac{1}{\left(2\alpha k+1\right)}\left[x^{2\alpha k+1}\right]_{0}^{x}\\ & =\sum_{k=0}^{\infty}\left(-1\right)^{k}\frac{x^{2\alpha k+1}}{\left(2k\right)!\left(2\alpha k+1\right)} \end{align*}ページ情報
| タイトル | フレネル積分の級数表示 |
| URL | https://www.nomuramath.com/r0vwv7ye/ |
| SNSボタン |
ガウス積分の応用
\[
\int_{0}^{\infty}x^{\gamma}e^{-\alpha x^{b}}dx=\frac{\alpha^{-\frac{1+\gamma}{b}}}{1+\gamma}\Gamma\left(\frac{1+\gamma}{b}+1\right)
\]
フレネル積分の一般化
\[
\int_{0}^{\infty}\sin\left(x^{a}\right)dx=\Gamma\left(\frac{1}{a}+1\right)\sin\frac{\pi}{2a}
\]
ガウス積分の一般化
\[
\int_{0}^{\infty}e^{-\alpha x^{b}}dx=\frac{1}{\alpha^{\frac{1}{b}}}\Gamma\left(\frac{1}{b}+1\right)
\]
フレネル積分の値
\[
\int_{0}^{\infty}\sin\left(x^{2}\right)dx=\frac{\sqrt{2\pi}}{4}
\]