# 请问一下，这个积分怎么求呀

\displaystyle \int_{-\infty}^{+\infty}\frac{x^4dx}{(2+3x^2)^4}

\displaystyle f(z)=\frac{z^4}{(2+3z^2)^4}

\displaystyle z=\sqrt{\frac{2}{3}}i

\displaystyle \sqrt{\frac{2}{3}}i=a,\ \ z-a=t\Rightarrow z=a+t\\{}\\ f(z)=\frac{z^4}{(2+3z^2)^4}=\frac{z^4}{3^4(z-a)^4(z+a)^4}\\{}\\ =\frac{(t+a)^4}{3^4t^4(t+2a)^4}\\{}\\ =\frac{1}{3^4t^4}\cdot\frac{a^4+4a^3t+6a^2t^2+4at^3+t^4}{16a^4+32a^3t+24a^2t^2+8at^3+t^4}\\{}\\ =\frac{1}{3^4t^4}\left(\frac{1}{16}+\frac{t}{8a}+\frac{t^2}{32a^2}-\frac{t^3}{32a^3}+\cdots\right)\\{}\\ \Rightarrow\mathop{\mathrm{Res}}\limits_{z=a}f(z)=-\frac{1}{3^4\cdot32a^3}\\{}\\ \Rightarrow\mathop{\mathrm{Res}}\limits_{z=\sqrt{\frac{2}{3}}i}f(z)=-\frac{1}{3^4\cdot32\left(\sqrt{\frac{2}{3}}i\right)^3}\\{}\\ =\frac{-i}{32\sqrt{2^3\cdot3^5}}=\frac{-i}{576\sqrt{6}}\\{}\\ \Rightarrow\int_{-\infty}^{+\infty}\frac{x^4}{(2+3x^2)^4}dx=2\pi i\cdot\frac{-i}{576\sqrt{6}}=\frac{\pi}{288\sqrt{6}}

3个月前 评论
Mars_step （楼主） 3个月前

\displaystyle f(z)=\frac{z^4}{(2+3z^2)^4}

\displaystyle z=\sqrt{\frac{2}{3}}i

\displaystyle \sqrt{\frac{2}{3}}i=a,\ \ z-a=t\Rightarrow z=a+t\\{}\\ f(z)=\frac{z^4}{(2+3z^2)^4}=\frac{z^4}{3^4(z-a)^4(z+a)^4}\\{}\\ =\frac{(t+a)^4}{3^4t^4(t+2a)^4}\\{}\\ =\frac{1}{3^4t^4}\cdot\frac{a^4+4a^3t+6a^2t^2+4at^3+t^4}{16a^4+32a^3t+24a^2t^2+8at^3+t^4}\\{}\\ =\frac{1}{3^4t^4}\left(\frac{1}{16}+\frac{t}{8a}+\frac{t^2}{32a^2}-\frac{t^3}{32a^3}+\cdots\right)\\{}\\ \Rightarrow\mathop{\mathrm{Res}}\limits_{z=a}f(z)=-\frac{1}{3^4\cdot32a^3}\\{}\\ \Rightarrow\mathop{\mathrm{Res}}\limits_{z=\sqrt{\frac{2}{3}}i}f(z)=-\frac{1}{3^4\cdot32\left(\sqrt{\frac{2}{3}}i\right)^3}\\{}\\ =\frac{-i}{32\sqrt{2^3\cdot3^5}}=\frac{-i}{576\sqrt{6}}\\{}\\ \Rightarrow\int_{-\infty}^{+\infty}\frac{x^4}{(2+3x^2)^4}dx=2\pi i\cdot\frac{-i}{576\sqrt{6}}=\frac{\pi}{288\sqrt{6}}

3个月前 评论
Mars_step （楼主） 3个月前