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$\text{Batominovski's First Idenity}:$

Let $d > 1$ be an integer. Suppose that $p(X) = q(X^{d})$ and $s(X) = u(X^{d})$ for some $q(X)$, $u(X) \in \mathbb{C}[X]$ with $deg(q) > deg(u)$, and that $p(X)$ and $s(X)$ share no common roots. Assume further that $p(X)$ has no nonnegative real roots. Let $z_{1}, z_{2}, …, z_{n}$ be all the distinct roots of $p(X)$ with arguments in the interval $(0, \frac{2 \pi}{d})$, respectively, with multiplicities $m_{1}, m_{2}, …, m_{n}$ one can then show that

>$(*)$
>$$\int_{0}^{\infty} \frac{s(x)}{p(x)}dx = \frac{2 \pi i}{1 - \exp( \frac{2 \pi  i}{d})} \sum_{j}^{l} \frac{1}{(m_{j} -1)!} \! \! \! \! \, \, \lim_{z \rightarrow z_{j}} \big( \partial_{z} \big)^{m_{j}-1} \frac{(z-z_{j})^{m_{j}}s(z)}{p(z)}.$$

$\text{Proof}$

Before embarking on our journey to conquer the Conjecture, first we must make some adjustments to our integral on the RHS side of $(*)$, so notice that

$$ \int_{0}^{\infty} \frac{s(x)}{p(x)}dx =  \int_{0}^{\infty} \frac{b(x-x_{1})(x-x_{2})(x-x_{3}) \cdot \cdot \cdot (x-x_{n})}{a(z-z_{1})(z-z_{2})(z-z_{3}) \cdot \cdot \cdot (z-z_{n})}dx$$

$$ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \,  = \frac{2 \pi i}{1 - \exp( \frac{2 \pi  i}{d})} \sum_{j}^{l} \frac{1}{(m_{j} -1)!}\lim_{z \rightarrow z_{j}} \big( \partial_{z} \big)^{m_{j}-1} \frac{(z-z_{j})^{m_{j}}s(z)}{p(z)} $$

Now it's important to note that our choice of $f$ is analytic on an open set $\psi$ containing the closed upper half plane,

$$ \mathcal{H} = \Big\{ z \in \mathcal{C} | \operatorname{Im(z)} \geq 0 \Big\}.$$
Furthermore consider that $R >0$ the positively oriented contour positively oriented contour $\phi_{R} \subset \mathcal{H}$ given by

$$\big[-R, +R \big] \bigcup \big\{R \, \exp(it)| t \in [0, \pi] \big\}.$$

Now to begin on our quest we make a note that $\phi_{R} \subset \mathcal{H}$, and consider

$$\oint_{\phi_{R_{\mathcal{H}}}} \frac{s(z)}{q(z)}dz$$

Now notice that,

$$\oint_{\phi_{R_{\mathcal{H}}}} \frac{s(z)}{q(z)}dz= 2 \pi i \bigg( \sum  \mathcal{Res_{f}(P_{j}}) \cdot \operatorname{Ind_{\gamma}}(P_{j}) \bigg)$$

$$\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \,   = 2 \pi i \bigg[ \sum_{j=1}^l\,\frac{1}{\left(m_j-1\right)!}\,\lim_{z\to z_j}\,\frac{\text{d}^{m_j-1}}{\text{d}z^{m_j-1}}\,\left(\frac{\left(z-z_j\right)^{m_j}\,p(z)}{p(z)}\right) \bigg].$$

Now for every if $m_{j} = 1$ for every $j=1,2,\ldots,l$ then we get $l=\dfrac{2 \pi}{d}$ so we now have that,

$\text{#}$

$$ \oint_{\phi_{R_{\mathcal{H}}}} \frac{s(z)}{q(z)}dz = \frac{2 \pi i}{1 - \exp( \frac{2 \pi  i}{d})} \sum_{j}^{l} \frac{1}{(m_{j} -1)!}\lim_{z \rightarrow z_{j}} \big( \partial_{z} \big)^{m_{j}-1} \frac{(z-z_{j})^{m_{j}}s(z)}{p(z)}.$$

To prove $\text{#}$ notice that if $m_{j} = 1$ for every $j =1,2,..l$ we have that

$$ \frac{s(x)}{p(x)}=\sum_{j=1}^{\frac{2 \pi}{d}}\,\frac{d(s'(x_{j})) \, \big(x^{d}-x_{j}^{d})}{p'\left(z_j\right)\,\left(z^d-z_j^d\right)} =  \sum_{j=1}^{\frac{2 \pi}{d}}\,\frac{u'\left(z_j^d\right)\,\left(x^d-z_j^d\right)}{q'\left(z_j^d\right)\,\left(x^d-z_j^d\right)}.$$

Now unfortunately what we done at this point isn't enough to realize a full proof of the conjecture we still need to make a road-trip around $\phi_{R}.$ We can begin making this journey by noticing through Cauchy's Theorem that,

$$\oint_{\phi_{R_{\mathcal{H}}}} \frac{s(z)}{p(z)}dz = \oint_{\phi_{R^1_{\mathcal{H}}}}\frac{s(z)}{p(z)}dz  + \oint_{\phi_{R^{2}_{\mathcal{H}}}}\frac{s(z)}{p(z)}dz.$$

Now it's worthwhile to claim that,

$$\oint_{\gamma_{R^{1}_{\mathcal{H}}}}\frac{s(z)}{p(z)}dz \rightarrow \int_{-\infty}^{c} \frac{p(x)}{q(x)}dx + \int_{c}^{\infty} \frac{p(x)}{q(x)}dx  \, \, \text{as} \, R \rightarrow \infty.$$

Before one start's making any qualms make note using the tringle inequality for our choice of $p(z)$, since $|z| > R$ we have that,

$$|p(z)| \leq |a_{m}z^{m}| + \cdot \cdot \cdot |a_{1}z| + |a_{o}| \leq |a_{m}|R^{m} + \cdot \cdot \cdot + |a_{1}|R + |a_{o}|.$$

On a similar note,

$$|s(z)| \leq |b_{m}x^{m}| + \cdot \cdot \cdot |b_{1}x| + |b_{o}| \leq |b_{m}|R^{m} + \cdot \cdot \cdot + |b_{1}|R + |b_{o}|.$$

After taking care of $p(z)$, and $s(z)$ respectively we have the estimate,

$$|f(z)| = \frac{|f(z)|}{|q(z)|} \leq \frac{|b_{m}|R^{m} + \cdot \cdot \cdot + |b_{1}|R + |b_{o}|}{|a_{n}|R^{n}/2} \leq \frac{M}{R^{n-m}}.$$

Now we have the following,

$$\bigg| \oint_{\phi_{R^{2}_{\mathcal{H}}}} \frac{p(z)}{q(z)}dz \bigg| \leq \frac{M(2 \pi R)}{R^{n-m}} = \frac{2M \pi}{R^{n-m-1}}.$$

It's to safe to say that,

$$\lim_{R \rightarrow \infty} \bigg| \oint_{\phi_{R^{2}_{\mathcal{H}}}} \frac{p(z)}{q(z)} \bigg| \rightarrow 0.$$

Putting everything together it's easy to note that,

$$\lim_{R \rightarrow \infty} \oint_{\gamma_{R_{\mathcal{H}}}} \frac{p(z)}{q(z)}dz =  \lim_{R \rightarrow \infty}\oint_{\gamma_{R^{1}_{\mathcal{H}}}} \frac{p(z)}{q(z)}dz + \oint_{\gamma_{R^{2}_{\mathcal{H}}}} \frac{p(z)}{q(z)}dz  $$

$$\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, = \frac{2 \pi i}{1 - \exp( \frac{2 \pi  i}{d})} \sum_{j}^{l} \frac{1}{(m_{j} -1)!}\lim_{z \rightarrow z_{j}} \big( \partial_{z} \big)^{m_{j}-1} \frac{(z-z_{j})^{m_{j}}s(z)}{p(z)}.$$

Thu, 11 Jul 2019 18:51 GMT