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。

(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*}

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フレネル積分の級数表示
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