無限級数その２

［証明］
$\omega=R( \cos \theta +i \sin \theta ),0 \lt R \lt 1$とおく．
$$ \sum_{n=1}^{N}\frac{ \omega^n}{n}=\sum_{n=1}^{N} \omega^n\int_{0}^{1} x^{n-1} dx=\omega\int_{0}^{1} \frac{1- (\omega x)^{N}}{1-\omega x}dx$$
ここで，$N$$\rightarrow$0のとき，
$$\left|\omega\int_{0}^{1} \frac{(\omega x)^{N}}{1-\omega x}dx\right| \leqq\left|\int_{0}^{1} \frac{x^{N}}{ \frac{1}{\omega} -x}dx\right|$$
$$\int_{0}^{1} \frac{x^{N}}{ \frac{1}{\omega} -x}dx=\int_{0}^{1} \frac{x^{N} (\frac{\cos \theta}{R}-x+i\frac{\sin \theta}{R})}{ (\frac{\cos \theta}{R}-x)^2+(\frac{\sin \theta}{R})^2}dx$$
$$\left|\int_{0}^{1} \frac{x^{N}}{ \frac{1}{\omega} -x}dx\right| \leqq\left|\int_{0}^{1} \frac{2x^{N} }{ (\frac{\sin \theta}{R})^2}dx\right|+ \left|\int_{0}^{1} \frac{x^{N}\frac{\sin \theta}{R} }{ (\frac{\sin \theta}{R})^2}dx\right| \rightarrow0 $$
$$ \sum_{n=1}^{N}\frac{\omega^n}{n}\rightarrow\omega\int_{0}^{1}\frac{1} {1-\omega x}dx$$とくに，$R=1$，$\theta＝ \frac{2 \pi }{p} $のとき，
$$\int_{0}^{1} \frac{1}{ \frac{1}{\omega} -x}dx=\int_{0}^{1} \frac{ \cos \theta-x+i{\sin \theta}}{ ({\cos \theta}-x)^2+({\sin \theta})^2}dx $$
$$\int_{0}^{1} \frac{1}{ \frac{1}{\omega} -x}dx =\int_{0}^{1} \frac{ \cos \theta-x}{ ({\cos \theta}-x)^2+({\sin \theta})^2}dx +i\int_{0}^{1} \frac{ {\sin \theta}}{ ({\cos \theta}-x)^2+({\sin \theta})^2}dx $$
実数部分は，
$$\int_{0}^{1} \frac{ \cos \theta-x}{ ({\cos \theta}-x)^2+({\sin \theta})^2} dx=- \frac{\log \left| { ({\cos \theta}-1)^2+({\sin \theta})^2} \right| }{2}=- \log({2 \sin \frac{ \theta }{2} }) $$
虚数部分は，$x- \cos \theta =t \sin \theta $と置換して，
$$\frac{- \cos \theta }{ \sin \theta}= \tan(\theta -\frac{\pi}{2})， \frac{1- \cos \theta }{ \sin \theta}= \tan\frac{\theta}{2}$$
$$\int_{0}^{1} \frac{ 1}{ ({\cos \theta}-x)^2+({\sin \theta})^2}dx= \int_{\tan(\theta -\frac{\pi}{2})}^{\tan\frac{\theta}{2}} \frac{1}{1+t^2}dt= \frac{ \pi }{2}-\frac{\theta}{2} $$
$$ \sum_{n=1}^{∞}\frac{\omega^n}{n}=\omega\int_{0}^{1}\frac{1} {1-\omega x}dx=- \log({2 \sin \frac{ \pi }{p} })+i(\frac{ \pi }{2}-\frac{\pi}{p} )$$
よって，成り立つ．□□