■シンク関数の積分実験(その6)
【2】∫(0,∞)Πsin(kx)/(kx)dx=?
∫(0,∞)sinc(x)=π/2
∫(0,∞)sinc(x)sinc(2x)dx=π/4
∫(0,∞)sinc(x)sinc(2x)sinc(4x)dx=π/8
∫(0,∞)sinc(x)sinc(2x)sinc(4x)sinc(8x)dx=π/16
∫(0,∞)sinc(x)=π/2
∫(0,∞)sinc(x)sinc(3x)dx=π/6
∫(0,∞)sinc(x)sinc(3x)sinc(9x)dx=π/18
∫(0,∞)sinc(x)sinc(3x)sinc(9x)sinc(27x)dx=π/54
置換積分により
∫(0,∞)Πsin(x*n^i)/(x*n^i)dx=1/n^k∫(0,∞)sin(x)/(x)dx
すなわち,本質的に同じ積分となる.
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