∫[-∞..∞] 1/(x^6+1) dxの値を部分分数分解を使って求める。

$\frac{1}{x^6+1}$は偶関数なので
$$\begin{array}{rcl}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{1}{x^6+1}dx& =& 2{\int }_0^{\mathrm{\infty }}\frac{1}{x^6+1}dx\end{array}$$
積分変数を複素変数$z$に置き換えて
$$2{\int }_0^{\mathrm{\infty }}\frac{1}{x^6+1}dx=2{\int }_0^{\mathrm{\infty }}\frac{1}{z^6+1}dz$$
ここで$z^6+1=0$の解${\zeta }_k$は
$${\zeta }_k=e^{\frac{i\pi }{6}(2k+1)}(k=0,1,\cdots ,5)$$となり、${\zeta }_k$を図示すると
$z^6=-1$の解
となる。
よって$\frac{1}{z^6+1}$をヘビサイドの展開定理を使って部分分数分解すると、
$$\begin{array}{rcl}F(z):=\frac{1}{z^6+1}& =& \frac{A_0}{z-{\zeta }_0}+\frac{A_1}{z-{\zeta }_1}+\cdots ++\frac{A_5}{z-{\zeta }_5}\\ & =& \sum _{k=0}^5\frac{A_k}{z-{\zeta }_k}\end{array}$$ただし$A_k(k=0,\cdots ,5)$は
$$\begin{array}{rcl}{A}_k& =& \underset{z\to {\zeta }_k}{lim}(z-{\zeta }_k)F(z)\\ & =& \underset{z\to {\zeta }_k}{lim}\frac{z-{\zeta }_k}{z^6+1}\\ & =& \underset{z\to {\zeta }_k}{lim}\frac{1}{6z^5}& \because \text{0/0不定形なのでロピタルの定理を適用}\\ & =& \frac{1}{6({\zeta }_k{)}^5}\end{array}$$
ここで
$$\begin{array}{rcl}\frac{1}{({\zeta }_k{)}^5}& =& ({\zeta }_k{)}^{-5}\\ & =& e^{-\frac{i5\pi }{6}(2k+1)}\\ & =& e^{-i\pi (\frac{5}{3}k+\frac{5}{6})}\\ & =& e^{-i\pi (2k-\frac{1}{3}k+1-\frac{1}{6})}\\ & =& e^{-i2\pi k}{e}^{-i\pi }{e}^{-i\pi (-\frac{1}{3}k-\frac{1}{6})}\\ & =& -e^{-i\pi (-\frac{1}{3}k-\frac{1}{6})}\\ & =& -e^{i\frac{\pi }{6}(2k+1)}\\ & =& -{\zeta }_k\end{array}$$なので、
$$A_k=-\frac{{\zeta }_k}{6} (k=0,\cdots ,5)$$である。ここで${\zeta }_k$は
$$\begin{gathered} {\zeta }_0+{\zeta }_3=0 \\ {\zeta }_1+{\zeta }_4=0 \\ {\zeta }_2+{\zeta }_5=0 \end{gathered}$$という性質をもっている。これは先ほど示した図からも確かめられる。
つぎに$F(z)$の各項の中で、足すと$0$になる${\zeta }_k$のペアを含む項を足す。たとえば${\zeta }_0$と${\zeta }_3$を含む項を足すと
$$\begin{array}{rcl}\frac{\frac{-{\zeta }_0}{6}}{z-{\zeta }_0}+\frac{\frac{-{\zeta }_3}{6}}{z-{\zeta }_3}& =& -\frac{1}{6}\frac{({\zeta }_0+{\zeta }_3)z-2{\zeta }_0{\zeta }_3}{z^2-({\zeta }_0+{\zeta }_3)z+{\zeta }_0{\zeta }_3}\\ & =& \frac{1}{3}\frac{{\zeta }_0{\zeta }_3}{z^2+{\zeta }_0{\zeta }_3}\end{array}$$となる。同様の計算で${\zeta }_1$と${\zeta }_4$,${\zeta }_2$と${\zeta }_5$を含む項をそれぞれ足すと
$$\begin{array}{rcl}\frac{\frac{-{\zeta }_1}{6}}{z-{\zeta }_1}+\frac{\frac{-{\zeta }_4}{6}}{z-{\zeta }_4}& =& \frac{1}{3}\frac{{\zeta }_1{\zeta }_4}{z^2+{\zeta }_1{\zeta }_4}\\ \frac{\frac{-{\zeta }_2}{6}}{z-{\zeta }_2}+\frac{\frac{-{\zeta }_5}{6}}{z-{\zeta }_5}& =& \frac{1}{3}\frac{{\zeta }_2{\zeta }_5}{z^2+{\zeta }_2{\zeta }_5}\end{array}$$となる。よって$F(z)$は
$$\begin{array}{rcl}F(z)& =& \frac{1}{3}\frac{{\zeta }_0{\zeta }_3}{z^2+{\zeta }_0{\zeta }_3}+\frac{1}{3}\frac{{\zeta }_1{\zeta }_4}{z^2+{\zeta }_1{\zeta }_4}+\frac{1}{3}\frac{{\zeta }_2{\zeta }_5}{z^2+{\zeta }_2{\zeta }_5}\\ & =& \frac{1}{3}(\frac{{\zeta }_0{\zeta }_3}{z^2+{\zeta }_0{\zeta }_3}+\frac{{\zeta }_1{\zeta }_4}{z^2+{\zeta }_1{\zeta }_4}+\frac{{\zeta }_2{\zeta }_5}{z^2+{\zeta }_2{\zeta }_5})\end{array}$$となる。よって求める積分$I$は
$$\begin{array}{rcl}I& =& 2{\int }_0^{\mathrm{\infty }}F(z)dz\\ & =& \frac{2}{3}{\int }_0^{\mathrm{\infty }}\frac{{\zeta }_0{\zeta }_3}{z^2+{\zeta }_0{\zeta }_3}+\frac{{\zeta }_1{\zeta }_4}{z^2+{\zeta }_1{\zeta }_4}+\frac{{\zeta }_2{\zeta }_5}{z^2+{\zeta }_2{\zeta }_5}dz\\ & =& \frac{2}{3}{\int }_0^{\mathrm{\infty }}\frac{1}{{\left(\frac{z}{\sqrt{{\zeta }_0{\zeta }_3}}\right)}^2+1}+\frac{1}{{\left(\frac{z}{\sqrt{{\zeta }_1{\zeta }_4}}\right)}^2+1}+\frac{1}{{\left(\frac{z}{\sqrt{{\zeta }_1{\zeta }_4}}\right)}^2+1}dz\end{array}$$ここで$\frac{z}{\sqrt{{\zeta }_n{\zeta }_m}}=v$と置換して$dz=\sqrt{{\zeta }_n{\zeta }_m}dv$なので
$$\begin{array}{rcl}& =& \frac{2}{3}{\int }_0^{\mathrm{\infty }}\frac{\sqrt{{\zeta }_0{\zeta }_3}}{v^2+1}+\frac{\sqrt{{\zeta }_1{\zeta }_4}}{v^2+1}+\frac{\sqrt{{\zeta }_2{\zeta }_5}}{v^2+1}dv\\ & =& \frac{2}{3}{[\sqrt{{\zeta }_0{\zeta }_3}\arctan \left(v\right)+\sqrt{{\zeta }_1{\zeta }_4}\arctan \left(v\right)+\sqrt{{\zeta }_2{\zeta }_5}\arctan \left(v\right)]}_0^{\mathrm{\infty }}\\ & =& \frac{2}{3}(\frac{\pi }{2}\sqrt{{\zeta }_0{\zeta }_3}+\frac{\pi }{2}\sqrt{{\zeta }_1{\zeta }_4}+\frac{\pi }{2}\sqrt{{\zeta }_2{\zeta }_5})\end{array}$$
ここで足すと$0$になる${\zeta }_k$どうしの積は
$$\begin{gathered} {\zeta }_0{\zeta }_3=e^{i\frac{\pi }{6}+i\frac{7\pi }{6}}=e^{i\frac{8\pi }{6}}=e^{-i\frac{2\pi }{3}} \\ {\zeta }_1{\zeta }_4=i(-i)=1 \\ {\zeta }_2{\zeta }_5=e^{i\frac{5\pi }{6}+i\frac{-\pi }{6}}=e^{i\frac{4\pi }{6}}=e^{i\frac{2\pi }{3}} \end{gathered}$$なので
$$\begin{gathered} =\frac{2}{3}(\frac{\pi }{2}\sqrt{e^{-i\frac{2\pi }{3}}}+\frac{\pi }{2}\sqrt{1}+\frac{\pi }{2}\sqrt{e^{i\frac{2\pi }{3}}}) \\ =\frac{\pi }{3}(e^{-i\frac{\pi }{3}}+1+e^{i\frac{\pi }{3}}) \\ =\frac{\pi }{3}(\frac{1}{2}+1+\frac{1}{2}) \\ =\frac{2\pi }{3} \end{gathered}$$となり命題が示された。