複素線積分
{\displaystyle {\begin{aligned}&-{\frac {4}{3}}i\left[\oint _{C_{1}}{\frac {\frac {z}{(z+{\sqrt {3}}i)(z-{\sqrt {3}}i)\left(z+{\frac {i}{\sqrt {3}}}\right)}}{z-{\frac {i}{\sqrt {3}}}}}\,dz+\oint _{C_{2}}{\frac {\frac {z}{(z+{\sqrt {3}}i)(z-{\sqrt {3}}i)\left(z-{\frac {i}{\sqrt {3}}}\right)}}{z+{\frac {i}{\sqrt {3}}}}}\,dz\right]\\&=-{\frac {4}{3}}i\left[2\pi i\left({\frac {z}{(z+{\sqrt {3}}i)(z-{\sqrt {3}}i)(z+{\frac {i}{\sqrt {3}}})}}\right){\Bigg |}_{z={\frac {i}{\sqrt {3}}}}+2\pi i\left({\frac {z}{(z+{\sqrt {3}}i)(z-{\sqrt {3}}i)(z-{\frac {i}{\sqrt {3}}})}}\right){\Bigg |}_{z=-{\frac {i}{\sqrt {3}}}}\right]\\&={\frac {8\pi }{3}}\left[{\frac {\frac {i}{\sqrt {3}}}{({\frac {i}{\sqrt {3}}}+{\sqrt {3}}i)({\frac {i}{\sqrt {3}}}-{\sqrt {3}}i)({\frac {i}{\sqrt {3}}}+{\frac {i}{\sqrt {3}}})}}+{\frac {-{\frac {i}{\sqrt {3}}}}{(-{\frac {i}{\sqrt {3}}}+{\sqrt {3}}i)(-{\frac {i}{\sqrt {3}}}-{\sqrt {3}}i)(-{\frac {i}{\sqrt {3}}}-{\frac {i}{\sqrt {3}}})}}\right]\\&={\frac {8\pi }{3}}\left[{\frac {\frac {i}{\sqrt {3}}}{({\frac {4}{\sqrt {3}}}i)(-{\frac {2}{i{\sqrt {3}}}})({\frac {2}{{\sqrt {3}}i}})}}+{\frac {-{\frac {i}{\sqrt {3}}}}{({\frac {2}{\sqrt {3}}}i)(-{\frac {4}{\sqrt {3}}}i)(-{\frac {2}{\sqrt {3}}}i)}}\right]\\&={\frac {8\pi }{3}}\left[{\frac {\frac {i}{\sqrt {3}}}{i({\frac {4}{\sqrt {3}}})({\frac {2}{\sqrt {3}}})({\frac {2}{\sqrt {3}}})}}+{\frac {-{\frac {i}{\sqrt {3}}}}{-i({\frac {2}{\sqrt {3}}})({\frac {4}{\sqrt {3}}})({\frac {2}{\sqrt {3}}})}}\right]\\&={\frac {8\pi }{3}}\left[{\frac {\frac {1}{\sqrt {3}}}{({\frac {4}{\sqrt {3}}})({\frac {2}{\sqrt {3}}})({\frac {2}{\sqrt {3}}})}}+{\frac {\frac {1}{\sqrt {3}}}{({\frac {2}{\sqrt {3}}})({\frac {4}{\sqrt {3}}})({\frac {2}{\sqrt {3}}})}}\right]\\&={\frac {8\pi }{3}}\left[{\frac {\frac {1}{\sqrt {3}}}{\frac {16}{3{\sqrt {3}}}}}+{\frac {\frac {1}{\sqrt {3}}}{\frac {16}{3{\sqrt {3}}}}}\right]\\&={\frac {8\pi }{3}}\left[{\frac {3}{16}}+{\frac {3}{16}}\right]=\pi .\end{aligned}}}
ゆえに
∫ C z z 2 + 6 z + 8 d z = 2 ∫ 0 ∞ x x 2 + 6 x + 8 d x {\displaystyle \int _{C}{{\sqrt {z}} \over z^{2}+6z+8}\,dz=2\int _{0}^{\infty }{{\sqrt {x}} \over x^{2}+6x+8}\,dx}
留数定理を使うか、もしくはコーシーの積分公式を使う(まず被積分関数を部分分数分解して、2個の単純な円の周りの積分に書き直してから和をとる)かして、
π i ( i 2 − i ) = ∫ 0 ∞ x x 2 + 6 x + 8 d x = π ( 1 − 1 2 ) . ◻ {\displaystyle \pi i\left({i \over {\sqrt {2}}}-i\right)=\int _{0}^{\infty }{{\sqrt {x}} \over x^{2}+6x+8}\,dx=\pi \left(1-{1 \over {\sqrt {2}}}\right).\quad \square }
を得る。
{\displaystyle {\begin{aligned}\int _{R}^{\varepsilon }{{\sqrt {z}} \over z^{2}+6z+8}\,dz&=\int _{R}^{\varepsilon }{e^{{1 \over 2}\mathrm {Log} (z)} \over z^{2}+6z+8}\,dz\\&=\int _{R}^{\varepsilon }{e^{{1 \over 2}(\log {|z|}+i\arg {z})} \over z^{2}+6z+8}\,dz\\&=\int _{R}^{\varepsilon }{e^{{1 \over 2}\log {|z|}}e^{(1/2)(2\pi i)} \over z^{2}+6z+8}\,dz\\&=\int _{R}^{\varepsilon }{e^{{1 \over 2}\log {|z|}}e^{\pi i} \over z^{2}+6z+8}\,dz\\&=\int _{R}^{\varepsilon }{-{\sqrt {z}} \over z^{2}+6z+8}\,dz\\&=\int _{\varepsilon }^{R}{{\sqrt {z}} \over z^{2}+6z+8}\,dz.\end{aligned}}}