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|$$\leqq$1なので，$N$$\rightarrow$0のとき，
$$\int_{0}^{1} \frac{x^{ND}}{ \frac{1}{\omega} -x^D}dx=\int_{0}^{1} \frac{x^{ND} (\frac{\cos \theta}{R}-x^D+i\frac{\sin \theta}{R})}{ (\frac{\cos \theta}{R}-x^D)^2+(\frac{\sin \theta}{R})^2}dx$$
$$\left|\int_{0}^{1} \frac{x^{ND}}{ \frac{1}{\omega} -x^D}dx\right| \leqq\left|\int_{0}^{1} \frac{2x^{ND} }{ (\frac{\sin \theta}{R})^2}dx\right|+ \left|\int_{0}^{1} \frac{x^{ND}\frac{\sin \theta}{R} }{ (\frac{\sin \theta}{R})^2}dx\right| \rightarrow0 $$
$$\sum_{n=0}^{N-1} \frac{ \omega ^n}{1+nD}\rightarrow\int_{0}^{1} \frac{1}{1- \omega x^{D} }dx $$したがって，成り立つ．□□