Beyond Dirichlet

For any integer $m \ge 0$ define $f_m(y) \; = \; \int_0^\infty e^{-xy} \sin^{2m} x\,dx \qquad \qquad y > 0 \;.$ Integrating by parts twice, we deduce that $f_m(y) \; = \; \frac{2m(2m-1)}{y^2}f_{m-1}(y) - \frac{4m^2}{y^2}f_{m}(y) \; = \; \frac{2m(2m-1)}{y^2 + 4m^2}f_{m-1}(y)$ for all $m \ge 1$ . Since $f_0(y) = y^{-1}$ , it follows that $f_{m}(y) \; = \; \frac{(2m)!}{y \prod_{j=1}^m (y^2 + 4j^2)}$ For integers $m \ge n \ge 1$ we have $\begin{array}{rcl} \displaystyle I_{m,n} & = & \displaystyle \int_0^\infty \frac{\sin^{2m}x}{x^{2n}}\,dx \; =\; \frac{1}{(2n-1)!}\int_0^\infty \sin^{2m}x\left(\int_0^\infty y^{2n-1}e^{-xy}\,dy\right)\,dx \\ & =& \displaystyle \frac{1}{(2n-1)!}\int_0^\infty y^{2n-1}\left(\int_0^\infty e^{-xy}\sin^{2m}x\,dx\right)\,dy \; = \; \frac{1}{(2n-1)!}\int_0^\infty y^{2n-1} f_m(y)\,dy \end{array}$ (the reversal of the order of integration is justified by Tonelli's and Fubini's Theorems). Thus $I_{m,n} \; = \; \frac{(2m)!}{(2n-1)!} \int_0^\infty \frac{y^{2(n-1)}}{\prod_{j=1}^m (y^2 + 4j^2)}\,dy \qquad \qquad m \ge n \ge 1 \;.$ The integrand can be expanded by partial fractions (in $y^2$ ), writing $\frac{y^{2(n-1)}}{\prod_{j=1}^m (y^2 + 4j^2)} \; = \; \sum_{j=1}^m \frac{A^{(m,n)}_j}{y^2 + 4j^2}$ which makes $I_{m,n} \; = \; \frac{\pi (2m)!}{4(2n-1)!} \sum_{j=1}^m \tfrac{1}{j}A^{(m,n)}_j$ Applying the Cover-Up Rule to determine the coefficients $A^{(m,n)}_j$ , we deduce that $A^{(m,n)}_j \; = \; (-1)^{n-j}\frac{2\,j^{2n}}{4^{m-n}(m-j)!(m+j)!}$ and hence $I_{m,n} \; =\; \frac{2\pi}{4^{1+m-n}(2n-1)!}\sum_{j=1}^m (-1)^{n-j} j^{2n-1}{2m \choose m+j} \qquad m \ge n \ge 1 \;.$ For this problem, $\begin{array}{rcl} \displaystyle \int_0^\infty \frac{\sin^{2m+2}x}{x^4}\,dx \; = \; I_{m+1,2} & = & \displaystyle \frac{2\pi}{4^{m} 3!}\sum_{j=1}^{m+1} (-1)^j j^3 {2m+2 \choose m+1+j} \\ & = & \displaystyle \frac{2\pi}{4^{m} 3!} \times \frac{(m+1)(m+2)}{2(2m+1)(2m-1)}{2m+2 \choose m+2} \\ & = & \displaystyle \frac{\pi(m+1)}{3(2m-1)4^{m}}{2m \choose m} \; = \; \frac{\pi(m+1)^2}{3(2m-1)4^{m}}C_{m} \end{array}$ for any integer $m\ge 1$ . This makes the answer $\boxed{1341}$ .