複素積分問題8

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問題

\(-1\leq \alpha\leq 1\)とする。

\(\displaystyle\int_{0}^{\infty} \displaystyle\frac{x^{\alpha}}{x^2+2x\cos\theta+1}dx\)

解答

\( f(z)=\displaystyle\frac{z^{\alpha}}{z^2+2z\cos\theta+1}\)とおき、以下の経路で積分する。

留数定理より

\(\displaystyle\int_{r}^{R} \displaystyle\frac{x^{\alpha}}{x^2+2x\cos\theta+1}dx+\displaystyle\int_{C_{2}} \displaystyle\frac{z^{\alpha}}{z^2+2z\cos\theta+1}dz\)

\(+\displaystyle\int_{R}^{r} \displaystyle\frac{x^{\alpha}e^{2\pi i\alpha}}{x^2+2x\cos\theta+1}dx+\displaystyle\int_{C_{4}} \displaystyle\frac{z^{\alpha}}{z^2+2z\cos\theta+1}dz\)

\(=2\pi i \times(留数和)\)

左辺第一項と第三項

\(R\to\infty\)、\(r\to 0\)極限で

\(\displaystyle\int_{r}^{R} \displaystyle\frac{x^{\alpha}}{x^2+2x\cos\theta+1}dx+\displaystyle\int_{R}^{r} \displaystyle\frac{x^{\alpha}e^{2\pi i\alpha}}{x^2+2x\cos\theta+1}dx\)

\(\displaystyle\int_{0}^{\infty} \displaystyle\frac{x^{\alpha}}{x^2+2x\cos\theta+1}dx-\displaystyle\int_{0}^{\infty} \displaystyle\frac{x^{\alpha}e^{2\pi i\alpha}}{x^2+2x\cos\theta+1}dx\)

\(=(1-e^{2\pi i\alpha})\displaystyle\int_{0}^{\infty} \displaystyle\frac{x^{\alpha}}{x^2+2x\cos\theta+1}dx\)

左辺第二項

\(|z|=R\to \infty\)より\(\displaystyle\frac{1}{|z^2+2z\cos\theta+1|}\leq \displaystyle\frac{1}{kR^2}\)でおさえられる。

\(\biggl|\displaystyle\int_{C_{2}} \displaystyle\frac{z^{\alpha}}{z^2+2z\cos\theta+1}dz\biggr|\)

\(\leq \displaystyle\int_{C_{2}}\displaystyle\frac{|z^{\alpha}|}{|z^2+2z\cos\theta+1|}|dz|\leq \displaystyle\frac{R^{\alpha}}{kR^2}\times 2\pi R\)

\(=\displaystyle\frac{2\pi}{kR^{1-\alpha}}\to 0\) (\(1-\alpha>0\)、\(R\to\infty\))

左辺第四項

\(|z|=r\to 0\)より\(\displaystyle\frac{1}{|z^2+2z\cos\theta+1|}\leq \displaystyle\frac{1}{kr}\)でおさえられる。

\(\biggl|\displaystyle\int_{C_{4}} \displaystyle\frac{z^{\alpha}}{z^2+2z\cos\theta+1}dz\biggr|\)

\(\leq \displaystyle\int_{C_{4}}\displaystyle\frac{|z^{\alpha}|}{|z^2+2z\cos\theta+1|}|dz|\leq \displaystyle\frac{r^{\alpha}}{kr}\times 2\pi r\)

\(=\displaystyle\frac{2\pi r^{\alpha}}{k}\to 0\) (\(\alpha>0\)、\(r\to 0\))

右辺

\(\displaystyle\frac{z^{\alpha}}{z^2+2z\cos\theta+1}=\displaystyle\frac{z^{\alpha}}{(z+e^{i\theta})(z+e^{-i\theta})}\)

より留数は

\(Res(-e^{i\theta}, f(z))=\biggl[\displaystyle\frac{z^{\alpha}}{2z+2\cos\theta}\biggr]_{-e^{i\theta}}=-\displaystyle\frac{(-e^{i\theta})^{\alpha}}{2i\sin\theta}\)

\(Res(-e^{-i\theta}, f(z))=\biggl[\displaystyle\frac{z^{\alpha}}{2z+2\cos\theta}\biggr]_{-e^{-i\theta}}=\displaystyle\frac{(-e^{-i\theta})^{\alpha}}{2i\sin\theta}\)

よって

\(右辺=2\pi i\biggl[-\displaystyle\frac{(-e^{i\theta})^{\alpha}}{2i\sin\theta}+\displaystyle\frac{(-e^{-i\theta})^{\alpha}}{2i\sin\theta}\biggr]\)

\(=-2\pi i(-1)^{\alpha}\cdot \displaystyle\frac{e^{i\alpha\theta}-e^{-i\alpha\theta}}{2i \sin\theta}=-\displaystyle\frac{2\pi i e^{\pi i\alpha}\sin\alpha\theta}{\sin\theta}\)

まとめ

留数定理に戻ると

\((1-e^{2\pi i\alpha})\displaystyle\int_{0}^{\infty} \displaystyle\frac{x^{\alpha}}{x^2+2x\cos\theta+1}dx=-\displaystyle\frac{2\pi i e^{\pi i\alpha}\sin\alpha\theta}{\sin\theta}\)

よって

\(\displaystyle\int_{0}^{\infty} \displaystyle\frac{x^{\alpha}}{x^2+2x\cos\theta+1}dx\)

\(=-\displaystyle\frac{2\pi i e^{\pi i\alpha}\sin\alpha\theta}{\sin\theta(1-e^{2\pi i\alpha})}\)

\(=\displaystyle\frac{\pi\sin\alpha\theta}{\sin\theta}\cdot\displaystyle\frac{2i}{e^{\pi i\alpha}-e^{\pi i\alpha}}\)

\(=\displaystyle\frac{\pi\sin\alpha\theta}{\sin\theta\sin\pi\alpha}\)

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