F(x)=(n=1~∞)∑x^n/[(2n-1)×n!]
x^(-0.5)*F(x)=(n=1~∞)∑x^(n-0.5)/[(2n-1)×n!]
[x^(-0.5)*F(x)]'=(n=1~∞)∑{x^(n-0.5)/[(2n-1)×n!]}'=(n=1~∞)0.5∑{x^(n-1.5)/n!}=x^(-1.5)∑x^n/n!=x^(-1.5)*[e^x-1]
∫[x^(-0.5)*F(x)]'dx=∫x^(-1.5)*(e^x-1)dx
F(x)=x^(+0.5)*∫x^(-1.5)*(e^x-1)dx
下面积分我就不会了,望有高手解决