sinc関数のn乗広義積分
sinc関数の\(n\)乗広義積分
\(n\in\mathbb{N}\)とする。
\[ \int_{0}^{\infty}sinc^{n}(x)dx=\frac{\pi}{2^{n+1}(n-1)!}\sum_{k=0}^{n}C(n,k)(-1)^{k}(n-2k)^{n-1}\sgn(n-2k) \]
\(n\in\mathbb{N}\)とする。
\[ \int_{0}^{\infty}sinc^{n}(x)dx=\frac{\pi}{2^{n+1}(n-1)!}\sum_{k=0}^{n}C(n,k)(-1)^{k}(n-2k)^{n-1}\sgn(n-2k) \]
\begin{align*}
\int_{0}^{\infty}sinc^{n}(x)dx & =\int_{0}^{\infty}\frac{\sin^{n}x}{x^{n}}dx\\
& =\int_{0}^{\infty}\int_{0}^{\infty}\cdots\cdots\int_{0}^{\infty}\int_{0}^{\infty}\sin^{n}xe^{-(a_{1}+a_{2}+\cdots\cdots+a_{n})x}da_{1}da_{2}\cdots\cdots da_{n}dx\\
& =\int_{0}^{\infty}\int_{0}^{\infty}\cdots\cdots\int_{0}^{\infty}\int_{0}^{\infty}\sin^{n}xe^{-Ax}dxda_{1}da_{2}\cdots\cdots da_{n}\qquad,\qquad A=\sum_{k=0}^{n}a_{n}\\
& =\int_{0}^{\infty}\int_{0}^{\infty}\cdots\cdots\int_{0}^{\infty}\int_{0}^{\infty}\frac{\left(e^{ix}-e^{-ix}\right)^{n}}{(2i)^{n}}e^{-Ax}dxda_{1}da_{2}\cdots\cdots da_{n}\\
& =\int_{0}^{\infty}\int_{0}^{\infty}\cdots\cdots\int_{0}^{\infty}\int_{0}^{\infty}\frac{1}{(2i)^{n}}\sum_{k=0}^{n}C(n,k)(-1)^{n-k}e^{ikx}e^{-i(n-k)x}e^{-Ax}dxda_{1}da_{2}\cdots\cdots da_{n}\\
& =\int_{0}^{\infty}\int_{0}^{\infty}\cdots\cdots\int_{0}^{\infty}\int_{0}^{\infty}\frac{1}{(2i)^{n}}\sum_{k=0}^{n}C(n,k)(-1)^{n-k}e^{-(A-i(2k-n))x}dxda_{1}da_{2}\cdots\cdots da_{n}\\
& =\int_{0}^{\infty}\int_{0}^{\infty}\cdots\cdots\int_{0}^{\infty}\int_{0}^{\infty}\frac{1}{(2i)^{n}}\sum_{k=0}^{n}C(n,k)(-1)^{n-k}\frac{1}{A-i(2k-n)}da_{1}da_{2}\cdots\cdots da_{n}\\
& =\frac{(-1)^{n}}{(2i)^{n}}\sum_{k=0}^{n}C(n,k)(-1)^{n-k}\frac{\left(-i(2k-n)\right)^{n-1}}{(n-1)!}\left\{ \log(-i(2k-n))-H_{n-1}\right\} \\
& =\frac{(-1)^{n+1}}{2^{n}i(n-1)!}\sum_{k=0}^{n}C(n,k)(-1)^{k}(2k-n)^{n-1}\left\{ \log(-i(2k-n))-H_{n-1}\right\} \\
& =\frac{(-1)^{n+1}}{2^{n}i(n-1)!}\sum_{k=0}^{n}C(n,k)(-1)^{k}(2k-n)^{n-1}\log(-i(2k-n))\\
& =\frac{(-1)^{n+1}}{2^{n+1}i(n-1)!}\sum_{k=0}^{n}\left(C(n,k)(-1)^{k}(2k-n)^{n-1}\log(-i(2k-n))+C(n,k)(-1)^{n-k}(n-2k)^{n-1}\log(-i(n-2k))\right)\qquad,\qquad k\rightarrow n-k\\
& =\frac{-1}{2^{n+1}i(n-1)!}\sum_{k=0}^{n}C(n,k)(-1)^{k}(n-2k)^{n-1}\left\{ \log\left(-i(n-2k)\right)-\log\left(-i(2k-n)\right)\right\} \\
& =\frac{-1}{2^{n+1}i(n-1)!}\sum_{k=0}^{n}C(n,k)(-1)^{k}(n-2k)^{n-1}\left\{ \log\left(-i\sgn(n-2k)\right)-\log\left(i\sgn(n-2k)\right)\right\} \\
& =\frac{-1}{2^{n+1}i(n-1)!}\sum_{k=0}^{n}C(n,k)(-1)^{k}(n-2k)^{n-1}\left(-\frac{\pi}{2}i-\frac{\pi}{2}i\right)\sgn(n-2k)\\
& =\frac{\pi}{2^{n+1}(n-1)!}\sum_{k=0}^{n}C(n,k)(-1)^{k}(n-2k)^{n-1}\sgn(n-2k)
\end{align*}
ページ情報
タイトル | sinc関数のn乗広義積分 |
URL | https://www.nomuramath.com/d6vmjrm9/ |
SNSボタン |
点列の収束と任意の部分列の収束
点列の収束と任意の部分列の収束
整列集合の基本的な性質
\[
X\left\langle \min X\right\rangle =\emptyset
\]
第3種・第4種チェビシェフ多項式の定義
\[
V_{n}(x)=\frac{\cos\left(\left(n+\frac{1}{2}\right)\cos^{\bullet}x\right)}{\cos\left(\frac{1}{2}\cos^{\bullet}x\right)}
\]
積分問題
\[
\int_{0}^{\infty}\frac{1}{1+x^{n}}dx
\]