無限級数その３

［証明］
$$\sum _{n=0}^{N-1}\frac{{\omega }^n}{1+nD}={\int }_0^1\sum _{i=0}^{N-1}{\omega }^n{x}^{nD}dx={\int }_0^1\frac{1-(\omega x^D{)}^N}{1-\omega x^D}dx$$
$${\int }_0^1\frac{(\omega x^D{)}^N}{1-\omega x^D}dx={\omega }^{N-1}{\int }_0^1\frac{x^{ND}}{\frac{1}{\omega }-x^D}dx$$
$\omega =R(\cos \theta +i\sin \theta )$とおく．
ここで，$\left|\omega \right|$$\leqq$1なので，$N$$\to$0のとき，
$${\int }_0^1\frac{x^{ND}}{\frac{1}{\omega }-x^D}dx={\int }_0^1\frac{x^{ND}(\frac{\cos \theta }{R}-x^D+i\frac{\sin \theta }{R})}{(\frac{\cos \theta }{R}-x^D{)}^2+(\frac{\sin \theta }{R}{)}^2}dx$$
$$\left|{\int }_0^1\frac{x^{ND}}{\frac{1}{\omega }-x^D}dx\right|\leqq \left|{\int }_0^1\frac{2x^{ND}}{(\frac{\sin \theta }{R}{)}^2}dx\right|+\left|{\int }_0^1\frac{x^{ND}\frac{\sin \theta }{R}}{(\frac{\sin \theta }{R}{)}^2}dx\right|\to 0$$
$$\sum _{n=0}^{N-1}\frac{{\omega }^n}{1+nD}\to {\int }_0^1\frac{1}{1-\omega x^D}dx$$したがって，成り立つ．□□