오늘의 계산 64 - 난잡한 적분

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Calculation 1. We have $${\int }_0^{\mathrm{\infty }}\frac{x^8-4x^6+9x^4-5x^2+1}{x^{12}-10x^{10}+37x^8-42x^6+26x^4-8x^2+1}\mathrm{d}x=\frac{\pi }{2}.$$ This is Problem 11148 of American Mathematical Monthly, Vol.112, April 2005.

Proof. For each $f(z)\in \mathbb{C}[z]$, we define $f^{\ast }(z)=\overline{f(\overline{z})}$. This is the same as taking conjugate to the coefficients appearing in $f(z)$. Let $$\begin{array}{rl}P(z)& =z^8-4z^6+9z^4-5z^2+1,\\ Q(z)& =z^{12}-10z^{10}+37z^8-42z^6+26z^4-8z^2+1.\end{array}$$ The key observation is that all the zeros of the polynomial $$q(z)=z^3-(2+i)z^2-(1-i)z+1$$ lies in the upper half-plane $\mathbb{H}=\{z\in \mathbb{C}:\mathrm{Im}(z)>0\}$, and $$Q(z)=q(z)q^{\ast }(z)q(-z)q^{\ast }(-z)$$ is a factorization of $Q(z)$ into coprime factors. Now let $p(z)$ be the unique polynomial such that $\mathrm{deg}p<\mathrm{deg}q$ and $$P(z)\equiv p(z)q^{\ast }(z)q(-z)q^{\ast }(-z)(\mathrm{mod}q(z)).$$ Such $p(z)$ exists because $q(z)$ and $q^{\ast }(z)q(-z)q^{\ast }(-z)$ are coprime. Also, such $p(z)$ can be computed using the extended Euclidean algorithm. After a tedious computation, it follows that $$p(z)=\frac{-i{z}^2-(1-2i)z+1}{4}.$$ Then by the symmetry and the uniqueness of partial fraction decomposition, it follows that $$\frac{P(z)}{Q(z)}=\frac{p(z)}{q(z)}+\frac{p^{\ast }(z)}{q^{\ast }(z)}+\frac{p(-z)}{q(-z)}+\frac{p^{\ast }(-z)}{q^{\ast }(-z)}.$$ Moreover, using the fact that all the zeros of $q(z)$ lies in $\mathbb{H}$, we can invoke Cauchy's integral theorem to write $$\begin{array}{rl}\mathrm{PV}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{x^n}{q(x)}\mathrm{d}x& =\underset{R\to \mathrm{\infty }}{lim}{\int }_{-R}^R\frac{x^n}{q(x)}\mathrm{d}x\\ & =\underset{R\to \mathrm{\infty }}{lim}{\int }_{-\pi }^0\frac{i(R{e}^{i\theta }{)}^{n+1}}{q(R{e}^{i\theta })}\mathrm{d}\theta \\ & =\{\begin{array}{ll}0,& n<\mathrm{deg}q-1,\\ \pi i,& n=\mathrm{deg}q-1.\end{array}\end{array}$$ Therefore we conclude that $${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{P(x)}{Q(x)}\mathrm{d}x=4\mathrm{Re}\left[\mathrm{PV}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{p(x)}{q(x)}\mathrm{d}x\right]=4\pi i([z^2]p(z))=\pi .$$ The answer is half of the above integral, hence the claim follows. $◻$

 

아래는 또 다른 예제입니다.

 

Calculation 2. $${\int }_0^{\mathrm{\infty }}\frac{x^{14}-15x^{12}+82x^{10}-190x^8+184x^6-60x^4+16x^2}{x^{16}-20x^{14}+156x^{12}-616x^{10}+1388x^8-1792x^6+1152x^4-224x^2+16}\mathrm{d}x=\frac{\pi }{2}.$$

Proof. Apply the same argument as above with the choices $$q(z)=z^4+(3-i)z^3-(1+3i)z^2-(6+2i)z-2$$ and $$p(z)=\frac{-i{z}^3-3i{z}^2-2iz}{4}.$$ We assure the reader that all the zeros of $q(z)$ lie in the upper half-plane, and the denominator of the integral admits the same form of factorization into coprime factors as before. $◻$