∫ 0 1 P n ( 1 − 2 x 2 ) sin ( x y ) d x = 1 2 π ( J n + 1 2 ( 1 2 y ) ) 2 \int_{0}^{1}P_{n}\left(1-2x^{2}\right)\sin\left(xy\right)\mathrm{d}x=\tfrac{1}{2}\pi\left(J_{n+\frac{1}{2}}\left(\tfrac{1}{2}y\right)\right)^{2}