Hot Integral - 2

Let $p(x)$ be the characteristic function of the interval $[0,1]$ . Its Fourier transform is $(\mathcal{F}p)(x) \; = \; \int_0^1 e^{-itx}\,dt \; = \; \frac{1 - e^{-ix}}{ix} \; = \; \frac{2e^{-\frac12ix}\sin\frac12x}{x}$ so that $\frac{\sin x}{x} \; = \; e^{ix}(\mathcal{F}p)(2x) \;.$ Thus $\left(\frac{\sin x}{x}\right)^n \; = \; e^{inx} \big[(\mathcal{F}p)(2x)\big]^n \; = \; e^{inx} \big[\mathcal{F}p^{(n)}\big](2x) \;,$ where $p^{(n)}$ is the $n$ -fold convolution of $p$ with itself. Thus the integral we want is $\begin{array}{rcl} \displaystyle \int_0^\infty \left(\frac{\sin x}{x}\right)^n \cos mx \,dx & = & \displaystyle \tfrac12\int_{\mathbb{R}} \left(\frac{\sin x}{x}\right)^n e^{-imx}\,dx \\ & = & \displaystyle \tfrac12 \int_{\mathbb{R}} e^{inx} \big[\mathcal{F}p^{(n)}\big](2x) e^{-imx}\,dx \\ & = & \displaystyle \tfrac14 \int_{\mathbb{R}} \big[\mathcal{F}p^{(n)}\big](x)\, e^{\frac12i(n-m)x}\,dx \\ & = & \tfrac12\pi p^{(n)}\big(\tfrac{n-m}{2}\big) \;. \end{array}$ for $0 \le m \le n$ . The value of $p^{(n)}$ is most easily found using the Laplace transform, since $\begin{array}{rcl} \big[\mathcal{L}p^{(n)}\big](z) & = & \displaystyle \big[(\mathcal{L}p)(z)\big]^n \; =\; \left(\frac{1 - e^{-z}}{z}\right)^n \\ & = & \displaystyle \sum_{k=0}^n {n \choose k}(-1)^k z^{-n} e^{-kz} \end{array}$ Standard Laplace transform considerations give us that $\begin{array}{rcl} p^{(n)}(u) & = & \displaystyle \sum_{k=0}^n {n \choose k}(-1)^k \frac{(u-k)^{n-1}}{(n-1)!} \chi_{[k,\infty)}(u) \\ & =& \displaystyle \sum_{k=0}^{\lfloor u \rfloor } {n \choose k}(-1)^k \frac{(u-k)^{n-1}}{(n-1)!} \end{array}$ and hence $\begin{array}{rcl} \displaystyle \int_0^\infty \left(\frac{\sin x}{x}\right)^n \cos mx \,dx & = & \displaystyle \tfrac12\pi \sum_{k=0}^{\lfloor \frac12(n-m)\rfloor} {n \choose k}(-1)^k \frac{(\frac{n-m}{2}-k)^{n-1}}{(n-1)!} \\ & = &\displaystyle \frac{\pi}{2^n (n-1)!} \sum_{k=0}^{\lfloor \frac12(n-m)\rfloor} {n \choose k}(-1)^k (n - m - 2k)^{n-1} \end{array}$ for $0 \le m \le n$ , as required. Note that, since $p^{(n)}(u) = 0$ for $u < 0$ , this integral is equal to $0$ for $m > n$ .