级数公式&椭圆级数变换公式(二)之推导

展开全文∑(n=1…∞)(-q)n^2(0<q<1)=-1/2+[(-1)N/√(-πlnq)]∫(0,∞)e(x^2)/lnq[cos(2N+1)x/cosx]dx(N→∞)=-1/2+[(-1)N/√(-πlnq)]∫(0,∞)∑(k=
展开全文
∑(n=1…∞)(-q)n^2(0<q<1)

=-1/2+[(-1)N/√(-πlnq)]∫(0,∞)e(x^2)/lnq[cos(2N+1)x/cosx]dx(N→∞)

=-1/2+[(-1)N/√(-πlnq)]∫(0,∞)∑(k=0…∞)[(x2/lnq)k/k!][cos(2N+1)x/cosx]dx(N→∞)

=-1/2+[1/√(-πlnq)]∑(k=0…∞)[(1/lnq)k/k!]∑(n=0…∞)∫(nπ,nπ+π)[x2k(-1)Ncos(2N+1)x/cosx]dx(N→∞)

=-1/2+[1/√(-πlnq)]∑(k=0…∞)[(1/lnq)k/k!]∑(n=0…∞)π2k+1(n+1/2)2k

=-1/2+[π/√(-πlnq)]∑(n=1…∞)e(1/lnq)(nπ-π/2)^2

令 q=e-πx(x>0)

α(x)=∑(n=1…∞)(-1)ne-πxn^2

β(x)=∑(n=1…∞)e-πx(n-1/2)^2

得2β(1/x)=[1+2α(x)]√x

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