sinc関数のn乗広義積分

\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*}