0259132人目の素数さん
2019/02/01(金) 09:51:13.08ID:nDpUgAzk部分分数分解で
1/{(y-1^2)(y-3^2)・・・・(y-(2k-1)^2)}
= (-1)^k {1/[(2k-1)!!]^2 + (1/4)^(k-1)・yΣ[j=1,k] (-1)^j /((2j-1)・(k-j)!(k+j-1)![y-(2j-1)^2]) }
1/{x(x^2-π^2)(x^2-(3π)^2)・・・・[x^2-((2k-1)π)^2]}
= (-1/π^2)^k {1/(((2k-1)!!)^2・x) + (1/4)^(k-1)・xΣ[j=1,k] (-1)^j /{(2j-1)・(k-j)!(k+j-1)![x^2-((2j-1)π)^2]}
∫[0,∞] sin(x)・x/(x^2 - (Lπ)^2) dx = (π/2)(-1)^L,
特に∫[0,∞] sin(x)/x dx = π/2,