Приложения на теоремата за резидуумите за пресмятане на интеграли

Твърдение 1. Нека $D\subset\mathbb{R}^2$ е отворено множество, $\{(x,y)\in\mathbb{R}^2|x^2+y^2=1\}\subset D$ и $f:D\to\mathbb{C}$ е рационална функция. Тогава $$\int_0^{2\pi}f(\cos t,\sin t)dt=\int_{\gamma}f\left(\frac{z+z^{-1}}{2},\frac{z-z^{-1}}{2i}\right)\frac{1}{iz}dz,$$ където $\gamma(t)=\exp(it)$, $t\in[0,2\pi]$.
Доказателство. Упражнение върху определението на интеграл по крива и формулите на Ойлер.

Следващите три твърдения имат спомагателна техническа роля.

Твърдение 2. Нека $\alpha,\beta\in\mathbb{R}$, $0<\beta-\alpha\leq 2\pi$, $D=\{z\in\mathbb{C}|z=|z|\exp(it), \alpha\leq t\leq \beta\}$, $\gamma_r:[\alpha,\beta]\to D$ е кривата $\gamma_r(t)=r\exp(it)$, ($r>0$) и $f:D\to\mathbb{C}$ е непрекъсната функция. Тогава
a) Ако $\lim\limits_{\substack{z\to 0}}zf(z)=0$, то $\lim\limits_{r\to 0}\int_{\gamma_{r}}f(z)dz=0$.
б) Ако $\lim\limits_{\substack{z\to \infty}}zf(z)=0$, то $\lim\limits_{r\to +\infty}\int_{\gamma_{r}}f(z)dz=0$.
Доказателство. За всяко $\varepsilon>0$ съществува $\delta>0$, такова че a) при $0<|z|<\delta$, б) при $|z|>\delta$, е вярно неравенството $|zf(z)|<\frac{\varepsilon}{\beta-\alpha}$. Тогава а) при $r<\delta$, б) при $r>\delta$ имаме $$\left|\int_{\gamma_r}f(z)\right|=\left|\int_{\gamma_r}zf(z)\frac{1}{z}\right|\leq\int_{\gamma_r}\frac{|zf(z)|}{|z|}|dz|<\frac{\varepsilon}{\beta-\alpha}\int_{\gamma_{r}}\frac{1}{|z|}|dz|=\frac{\varepsilon}{(\beta-\alpha) r}\int_{\gamma_{r}}|dz|=\frac{\varepsilon}{(\beta-\alpha)r}(\beta-\alpha) r=\varepsilon.$$

Твърдение 3. Нека $\lambda>0$, $0\leq\alpha<\beta\leq\pi$, $D=\{z\in\mathbb{C}|\alpha\leq \text{Arg }z\leq \beta\}$, $\gamma_r:[\alpha,\beta]\to D$ е кривата $\gamma_r(t)=r\exp(it)$, ($r>0$) и $f:D\to\mathbb{C}$ е непрекъсната функция, за която $\lim\limits_{\substack{z\to \infty}}f(z)=0$. Тогава $\lim\limits_{r\to +\infty}\int_{\gamma_{r}}f(z)\exp(i\lambda z)dz=0$.
Доказателство. За всяко $\varepsilon>0$ съществува $\delta>0$, такова че при $|z|>\delta$ е вярно неравенството $|f(z)|<\frac{\varepsilon\lambda}{\pi}$. Тогава при $r>\delta$ имаме
$$\left|\int_{\gamma_{r}}f(z)\exp(i\lambda z)dz\right|\leq\int_{\gamma_{r}}|f(z)||\exp(i\lambda z)||dz|<\frac{\varepsilon\lambda}{\pi}\int_{\gamma_{r}}|\exp(i\lambda z)||dz|=\frac{\varepsilon\lambda}{\pi}\int_{\alpha}^{\beta}|\exp(i\lambda \gamma_r(t))||\gamma_r'(t)|dt=$$$$=\frac{\varepsilon\lambda}{\pi}\int_{\alpha}^{\beta}|\exp(\lambda ri\exp(it))||ir\exp(it)|dt=\frac{\varepsilon\lambda}{\pi}\int_{\alpha}^{\beta}re^{-\lambda r\sin t}dt\leq\frac{\varepsilon\lambda}{\pi}\int_{0}^{\pi}re^{-\lambda r\sin t}dt=2\frac{\varepsilon\lambda}{\pi}\int_{0}^{\frac{\pi}{2}}re^{-\lambda r\sin t}dt.$$ От неравенството $\sin t\geq \frac{2}{\pi}t$, валидно в интервала $[0,\frac{\pi}{2}]$, получаваме $$\int_{0}^{\frac{\pi}{2}}re^{-\lambda r\sin t}dt\leq\int_{0}^{\frac{\pi}{2}}re^{-\lambda r\frac{2}{\pi}t}dt=\frac{r}{-\lambda r\frac{2}{\pi}}(e^{-\lambda r}-1)=\frac{\pi}{2\lambda}(1-e^{-\lambda r})<\frac{\pi}{2\lambda}.$$ Следователно $$\left|\int_{\gamma_{r}}f(z)\exp(i\lambda z)dz\right|<2\frac{\varepsilon\lambda}{\pi}\frac{\pi}{2\lambda}=\varepsilon.$$

Твърдение 4. Нека $a\in\mathbb{\mathbb{C}}$ е прост полюс на функцията $f$, $0<\beta-\alpha\leq 2\pi$, и $\gamma_r:[\alpha,\beta]\to \mathbb{C}$ е кривата зададена с $\gamma_r(t)=a+r\exp(it)$, където $r>0$. Тогава $\lim\limits_{r\to 0}\int_{\gamma_r}f(z)dz=(\beta-\alpha)i\text{Res}(f,a)$.
Доказателство. Според Твърдение 2 4) от Тема 24 съществуват $\delta>0$ и холоморфна функция $g:K(a,\delta)\to\mathbb{C}$, такива че $f(z)=\frac{a_{-1}}{z-a}+\varphi(z)$ за всяко $z\in K(a,\delta)\setminus{a}$. Тъй като $a_{-1}=\text{Res}(f,a)$, при $r<\delta$ имаме $$\int_{\gamma_{r}}f(z)dz=\int_{\gamma_{r}}\frac{\text{Res}}{z-a}+\varphi(z)dz=\text{Res}(f,a)\int_{\gamma_{r}}\frac{1}{z-a}dz+\int_{\gamma_{r}}\varphi(z)dz.$$ От друга страна $$\int_{\gamma_{r}}\frac{1}{z-a}dz=\int_{\alpha}^{\beta}\frac{1}{a+r\exp(it)-a}ir\exp(it)dt=(\beta-\alpha) i,$$ и при $r<\frac{\delta}{2}$ имаме $$\left|\int_{\gamma_{r}}\varphi(z)dz\right|\leq (\beta-\alpha) r\sup\{|\varphi(z)||z\in\overline{K(0,\delta/2)}\}.$$ Следователно $\lim\limits_{r\to 0}\int_{\gamma_r}\varphi(z)dz=0$, откъдето $\lim\limits_{r\to 0}\int_{\gamma_r}f(z)dz=(\beta-\alpha)i\text{Res}(f,a)$.

Твърдение 5. Нека $P$ и $Q$ са полиноми с реални коефициенти, за които $\deg Q\geq \deg P+2$. Нека $f=\frac{P}{Q}$ и $M=\{z\in\mathbb{C}|\Im z>0, Q(z)=0\}$. Тогава
а) Ако $Q(x)\neq 0$ за всяко $x\in\mathbb{R}$, то $$\int_{-\infty}^{+\infty}f(x)dx=2\pi i\sum_{a\in M}\text{Res}(f,a),$$
б) Ако $N=\{x\in\mathbb{R}|Q(x)=0\}\neq\emptyset$ и за всяко $x\in N$ имаме $Q'(x)\neq 0$, (т. е. всички реални нули на $Q$ са прости), то $$v.p.\int_{-\infty}^{+\infty}f(x)dx=2\pi i\sum_{a\in M}\text{Res}(f,a)+\pi i\sum_{b\in N}\text{Res}(f,b),$$
Доказателство.
а) От условието $\deg Q\geq \deg P+2$ и $Q(z)\neq 0$ за всяко $z\in\mathbb{R}$ виждаме, че интегралът $\int_{-\infty}^{+\infty}f(x)dx$ е сходящ и $$\int_{-\infty}^{+\infty}f(x)dx=\lim_{R\to+\infty}\int_{-R}^{R}f(x)dx.$$ Нека $R>0$ е такова, че $M\subset K(0,R)$ и $\Gamma_R$ е положително ориентираната граница на компакта $K_R=\{z\in\mathbb{C}||z|\leq R,\Im z\geq 0\}$.
Тогава от една страна от теоремата за разидуумите (Твърдение 3 от Тема 28) $$\int_{\Gamma_R}f(z)dz=2\pi i\sum_{a\in M}\text{Res}(f,a),$$ а от друга страна $$\int_{\Gamma_R}f(z)dz=\int_{-R}^{R}f(x)dx+\int_{\gamma_R}f(z)dz,$$ където $\gamma_R(t)=R\exp(i t)$, $t\in[0,\pi]$. Следователно $$\label{keyeq}
\int_{-R}^{R}f(x)dx+\int_{\gamma_R}f(z)dz=2\pi i\sum_{a\in M}\text{Res}(f,a).
$$
Тъй като $\deg Q\geq \deg P+2$ виждаме, че $zf(z)\to 0$ при $z\to\infty$. Следователно $\int_{\gamma_{R}}f(z)dz\to 0$ при $R\to +\infty$ (Твърдение 2, б)) и след граничен преход в последното равенство, при $R\to+\infty$, получаваме $$\int_{-\infty}^{+\infty}f(x)dx=2\pi i\sum_{a\in M}\text{Res}(f,a).$$
б) Нека $N=\{x_1,\ldots,x_n\}$. Тогава по определение $$v.p.\int_{-\infty}^{+\infty}f(x)dx=\lim\limits_{\substack{R\to +\infty \\\varepsilon_1\to 0\\…\\\varepsilon_n\to 0}}\left(\int_{-R}^{x_1-\varepsilon_1}f(x)dx+\sum_{j=1}^{n-1}\int_{x_j+\varepsilon_j}^{x_{j+1}-\varepsilon_{j+1}}f(x)dx+\int_{x_n+\varepsilon_n}^Rf(x)dx\right).$$ Нека $R>0$ е такова, че $M\cup N\subset K(0,R)$, а $\varepsilon_j>0$, $j\in\{1,\ldots,n\}$ са такива, че $M\subset\text{int }K_{R,\varepsilon_1,\ldots,\varepsilon_n}$, където $$K_{R,\varepsilon_1,\ldots,\varepsilon_n}=\{z\in\mathbb{C}||z|\leq R,|z-x_j|\geq \varepsilon_j,j\in\{1,\ldots,n\}\},$$ и $\Gamma_{R,\varepsilon_1,\ldots,\varepsilon_n}$ е положително ориентираната граница на компакта $K_{R,\varepsilon_1,\ldots,\varepsilon_n}$. Тогава от една страна $$\int_{\Gamma_{R,\varepsilon_1,\ldots,\varepsilon_n}}f(z)=2\pi i\sum_{a\in M}\text{Res}(f,a)$$ (Твърдение 3 от Тема 28), а от друга страна $$\int_{\Gamma_{R,\varepsilon_1,\ldots,\varepsilon_n}}f(z)=\int_{-R}^{x_1-\varepsilon_1}f(x)dx+\sum_{j=1}^{n-1}\int_{x_j+\varepsilon_j}^{x_{j+1}-\varepsilon_{j+1}}f(x)dx+\int_{x_n+\varepsilon_n}^Rf(x)dx-\sum_{j=1}^{n-1}\int_{\gamma_{\epsilon_j}}f(z)dz+\int_{\gamma_{R}}f(z)dz,$$ където $\gamma_{\varepsilon_j}(t)=x_j+\varepsilon_j\exp(i t)$, $j\in\{1,\ldots,n\}$ и $\gamma_R=R\exp(it)$, $t\in[0,\pi]$. Следователно \begin{equation}\label{maineq}
\int_{-R}^{x_1-\varepsilon_1}f(x)dx+\sum_{j=1}^{n-1}\int_{x_j+\varepsilon_j}^{x_{j+1}-\varepsilon_{j+1}}f(x)dx+\int_{x_n+\varepsilon_n}^Rf(x)dx-\sum_{j=1}^{n}\int_{\gamma_{\epsilon_j}}f(z)dz+\int_{\gamma_{R}}f(z)dz=2\pi i\sum_{a\in M}\text{Res}(f,a).
\end{equation} Тъй като елементите на $N$ са прости полюси на $f$, от Твърдение 4 получаваме, че $$\lim\limits_{\varepsilon_j\to 0}\int_{\gamma_{\epsilon_j}}f(z)dz=\pi i\text{Res}(f,x_j), \quad j\in\{1,\ldots,n\}.$$ Също така, от условието $\deg Q\geq \deg P+2$ виждаме, че $zf(z)\to 0$ при $z\to\infty$. Следователно $$\int_{\gamma_{R}}f(z)dz\to 0$$ при $R\to +\infty$ (Твърдение 2, б)). След граничен преход в (1) при $R\to+\infty$, $\varepsilon_j\to 0$, $j\in\{1,\ldots,n\}$ получаваме
$$v.p.\int_{-\infty}^{+\infty}f(x)dx=2\pi i\sum_{a\in M}\text{Res}(f,a)+\pi i\sum_{b\in N}\text{Res}(f,b),$$

Твърдение 6. Нека $P$ и $Q$ са полиноми с реални коефициенти, за които $\deg Q\geq \deg P+1$, $m>0$, и $f(x)=\frac{P(x)}{Q(x)}\exp(i m x)$. Тогава
а) Ако $Q(x)\neq 0$ за всяко $x\in \mathbb{R}$ и $M=\{z\in\mathbb{C}|\Im z>0, Q(z)=0\}$, то $$\int_{-\infty}^{+\infty}f(x)dx=2\pi i\sum_{a\in M}\text{Res}(f,a),$$
б) Ако $M=\{z\in\mathbb{C}|\Im z>0, Q(z)=0\}\neq\emptyset$, $N=\{x\in\mathbb{R}|Q(x)=0\}\neq\emptyset$ и за всяко $x\in N$ имаме $Q'(x)\neq 0$, (т. е. всички реални нули на $Q$ са прости), то $$v.p.\int_{-\infty}^{+\infty}f(x)dx=2\pi i\sum_{a\in M}\text{Res}(f,a)+\pi i\sum_{b\in N}\text{Res}(f,b),$$
Доказателство. Условието $\deg Q\geq \deg P+1$ осигурява сходимостта на несобствения интеграл в а) и съществуването на главната стойност (v.p.) на несобствения интеграл в б). Оттук нататък доказателството е дословно повторение на доказателството на Твърдение 5, като вместо Твърдение 2, прилагаме Твърдение 3.

Твърдение 7. Нека $P$ и $Q$ са полиноми с реални коефициенти, за които $\deg Q\geq \deg P+2$. Нека $m\in\mathbb{N}\cup\{0\}$, $f_m(x)=\frac{P(x)}{Q(x)}(\ln x)^m$ и $g_m(z)=\frac{P(z)}{Q(z)}(\log z)^{m+1}$, където $\log z=\ln|z|+i\arg z$, $0<\arg z<2\pi$. Тогава
а) Ако $Q(x)\neq 0$ за всяко $x>0$ и $M=\{z\in\mathbb{C}|Q(z)=0\}$, то $$\int_{0}^{+\infty}f_0(x)dx=-\sum_{a\in M}\text{Res}(g_0,a),$$ и $$\int_{0}^{+\infty}f_m(x)dx=-\frac{1}{m+1}\sum_{a\in M}\text{Res}(g_m,a)-\frac{1}{m+1}\sum_{k=0}^{m-1}\binom{m+1}{k}(2\pi i)^{m-k}\int_{0}^{+\infty}f_k(x)dx.$$
б) Ако $M=\{z\in\mathbb{C}|\Im z>0, Q(z)=0\}\neq\emptyset$, $N=\{x>0|Q(x)=0\}\neq\emptyset$ и за всяко $x\in N$ имаме $Q'(x)\neq 0$, (т. е. всички реални нули на $Q$ са прости), $g_{m,0}(z)=\frac{P(z)}{Q(z)}(\log_0z)^{m+1}$, $g_{m,1}(z)=\frac{P(z)}{Q(z)}(\log_1z)^{m+1}$, то $$v.p.\int_{0}^{+\infty}f_0(x)dx=-\sum_{a\in M}\text{Res}(g_0,a)-\frac{1}{2}\sum_{b\in N}\text{Res}(g_{0,0},b)-\frac{1}{2}\sum_{b\in N}\text{Res}(g_{0,1},b),$$ и при $m\geq 1$
$$v.p.\int_{0}^{+\infty}f_m(x)dx=-\frac{1}{m+1}\sum_{a\in M}\text{Res}(g_m,a)-\frac{1}{2(m+1)}\sum_{b\in N}\text{Res}(g_{m,0},b)-\frac{1}{2(m+1)}\sum_{b\in N}\text{Res}(g_{m,1},b)-$$$$-\frac{1}{m+1}\sum_{k=0}^{m-1}\binom{m+1}{k}(2\pi i)^{m-k}v.p.\int_{0}^{+\infty}f_k(x)dx.$$
Доказателство. Условието $\deg Q\geq \deg P+2$ осигурява сходимостта на несобствените интеграли в а) и съществуването на главните стойности на интегралите в б).
а) Нека $R>0,\varepsilon>0,\varepsilon_1>0,\varepsilon_2>0$ и $\Gamma_{R,\varepsilon,\varepsilon_1,\varepsilon_2}$ е положително ориентираната граница на компакта $$K_{R,\varepsilon,\varepsilon_1,\varepsilon_2}=\{z\in\mathbb{C}\setminus{0}|z=|z|\exp(it), \varepsilon\leq|z|\leq R,\varepsilon_1\leq t\leq 2\pi-\varepsilon_2\},$$ като $R,\varepsilon,\varepsilon_1,\varepsilon_2$ са такива, че $M\subset\text{int}K_{R,\varepsilon,\varepsilon_1,\varepsilon_2}$. Тогава от една страна, от теоремата за резидуумите (Твърдение 3 от Тема 28) имаме
$$\int_{\Gamma_{R,\varepsilon,\varepsilon_1,\varepsilon_2}}g_m(z)dz=2\pi i\sum_{a\in M}\text{Res}(g_m,a),$$ а от друга страна $$\int_{\Gamma_{R,\varepsilon,\varepsilon_1,\varepsilon_2}}g_m(z)dz=\int_{\gamma_{R,\varepsilon,\varepsilon_1}}g_m(z)dz+\int_{\gamma_{R,\varepsilon_1,\varepsilon_2}}g_m(z)dz-\int_{\gamma_{R,\varepsilon,\varepsilon_2}}g_m(z)dz-\int_{K_{\varepsilon,\varepsilon_1,\varepsilon_2}}g_m(z)dz,$$ където $$\gamma_{R,\varepsilon,\varepsilon_1}(t)=t\exp(i\varepsilon_1),\quad t\in[\varepsilon,R],$$
$$\gamma_{R,\varepsilon_1,\varepsilon_2}(t)=R\exp(it),\quad t\in[\varepsilon_1,2\pi-\varepsilon_2],$$
$$\gamma_{R,\varepsilon,\varepsilon_2}(t)=t\exp(i(2\pi-\varepsilon_2)),\quad t\in[\varepsilon,R],$$
$$\gamma_{R,\varepsilon_1,\varepsilon_2}(t)=\varepsilon\exp(it),\quad t\in[\varepsilon_1,(2\pi-\varepsilon_2)].$$
Тогава
$$\int_{\Gamma_{R,\varepsilon,\varepsilon_1,\varepsilon_2}}g_m(z)dz=\int_{\varepsilon}^Rg_m(\gamma_{R,\varepsilon,\varepsilon_1}(t))\gamma_{R,\varepsilon,\varepsilon_1}'(t)dt+\int_{\varepsilon_1}^{2\pi-\varepsilon_2}g_m(\gamma_{R,\varepsilon_1,\varepsilon_2}(t))\gamma_{R,\varepsilon_1,\varepsilon_2}'(t)dt-$$$$-\int_{\varepsilon}^Rg_m(\gamma_{R,\varepsilon,\varepsilon_2}(t))\gamma_{R,\varepsilon,\varepsilon_2}'(t)dt-\int_{\varepsilon_1}^{2\pi-\varepsilon_2}g_m(\gamma_{\varepsilon,\varepsilon_1,\varepsilon_2}(t))\gamma_{\varepsilon,\varepsilon_1,\varepsilon_2}'(t)dt.$$
Тъй като $$g_m(\gamma_{R,\varepsilon,\varepsilon_1}(t))=\frac{P(t\exp(i\varepsilon_1))}{Q(t\exp(i\varepsilon_1))}[\log(t\exp(i\varepsilon_1))]^{m+1}=\frac{P(t\exp(i\varepsilon_1))}{Q(t\exp(i\varepsilon_1))}(\ln t+i\varepsilon_1)^{m+1}$$ и $$g_m(\gamma_{R,\varepsilon,\varepsilon_2}(t))=\frac{P(t\exp(i(2\pi -\varepsilon_2)))}{Q(t\exp(i(2\pi -\varepsilon_2)))}[\log(t\exp(i(2\pi -\varepsilon_2)))]^{m+1}=\frac{P(t\exp(i(2\pi -\varepsilon_2)))}{Q(t\exp(i(2\pi -\varepsilon_2))}(\ln t+i(2\pi -\varepsilon_2))^{m+1},$$ по известни теореми от реалния анализ за граничен преход под интеграла получаваме $$\lim\limits_{\substack{\varepsilon_1\to 0\\\varepsilon_2\to 0}}\int_{\varepsilon}^Rg_m(\gamma_{R,\varepsilon,\varepsilon_1}(t))\gamma_{R,\varepsilon,\varepsilon_1}'(t)dt=\int_{\varepsilon}^R\frac{P(t)}{Q(t)}(\ln t)^{m+1}dt=\int_{\varepsilon}^Rg_m(t)dt,$$$$\lim\limits_{\substack{\varepsilon_1\to 0\\\varepsilon_2\to 0}}\int_{\varepsilon_1}^{2\pi-\varepsilon_2}g_m(\gamma_{R,\varepsilon_1,\varepsilon_2}(t))\gamma_{R,\varepsilon_1,\varepsilon_2}'(t)dt=\int_{0}^{2\pi}\frac{P(R\exp(it))}{Q(R\exp(it))}(\log(R\exp(it)))^{m+1}Ri\exp(it)dt,$$$$\lim\limits_{\substack{\varepsilon_1\to 0\\\varepsilon_2\to 0}}\int_{\varepsilon}^{R}g_m(\gamma_{R,\varepsilon,\varepsilon_2}(t))\gamma_{R,\varepsilon,\varepsilon_2}'(t)dt=\int_{\varepsilon}^R\frac{P(t)}{Q(t)}(\ln t+2\pi i)^{m+1}dt,$$$$\lim\limits_{\substack{\varepsilon_1\to 0\\\varepsilon_2\to 0}}\int_{\varepsilon_1}^{2\pi-\varepsilon_2}g_m(\gamma_{\varepsilon,\varepsilon_1,\varepsilon_2}(t))\gamma_{\varepsilon,\varepsilon_1,\varepsilon_2}'(t)dt=\int_{0}^{2\pi}\frac{P(\varepsilon\exp(it))}{Q(\varepsilon\exp(it))}(\log(\varepsilon\exp(it)))^{m+1}\varepsilon i\exp(it)dt.$$
Следователно
\begin{equation}\label{intaflim}
\lim\limits_{\substack{\varepsilon_1\to 0\\\varepsilon_2\to 0}}\int_{\Gamma_{R,\varepsilon,\varepsilon_1,\varepsilon_2}}g_m(z)dz=\int_{\varepsilon}^Rg_m(t)dt-\int_{\varepsilon}^{R}\frac{P(t)}{Q(t)}(\ln t+2\pi i)^{m+1}dt+\int_{\gamma_{R}}g_m(z)dz-\int_{\gamma_{\epsilon}}g_m(z)dz,
\end{equation}
където $\gamma_{R}(t)=R\exp(it), \gamma_{\varepsilon}(t)=\varepsilon\exp(it), t\in[0,2\pi]$.
Тъй като $$(\ln t+2\pi i)^{m+1}=\sum_{k=0}^{m+1}\binom{m+1}{k}(\ln t)^k(2\pi i)^{m+1-k}=(\ln t)^{m+1}+\sum_{k=0}^{m}\binom{m+1}{k}(\ln t)^k(2\pi i)^{m+1-k},$$
виждаме, че $$\int_{\varepsilon}^{R}\frac{P(t)}{Q(t)}(\ln t+2\pi i)^{m+1}dt=\int_{\varepsilon}^{R}\frac{P(t)}{Q(t)}(\ln t)^{m+1}dt+\sum_{k=0}^{m}\binom{m+1}{k}(2\pi i)^{m+1-k}\int_{\varepsilon}^{R}\frac{P(t)}{Q(t)}(\ln t)^kdt=$$\begin{equation}\label{powm}
=\int_{\varepsilon}^{R}g_m(t)dt+\sum_{k=0}^{m}\binom{m+1}{k}(2\pi i)^{m+1-k}\int_{\varepsilon}^{R}f_k(t)dt
\end{equation}
Замествайки (3) в (2) получаваме $$\lim\limits_{\substack{\varepsilon_1\to 0\\\varepsilon_2\to 0}}\int_{\Gamma_{R,\varepsilon,\varepsilon_1,\varepsilon_2}}g_m(z)dz=-\sum_{k=0}^{m}\binom{m+1}{k}(2\pi i)^{m+1-k}\int_{\varepsilon}^{R}f_k(t)dt+\int_{\gamma_{R}}g_m(z)dz-\int_{\gamma_{\epsilon}}g_m(z)dz,$$
От друга страна $$\lim\limits_{\substack{\varepsilon_1\to 0\\\varepsilon_2\to 0}}\int_{\Gamma_{R,\varepsilon,\varepsilon_1,\varepsilon_2}}g_m(z)dz=2\pi i\sum_{a\in M}\text{Res}(g_m,a),$$ т. е.
$$2\pi i\sum_{a\in M}\text{Res}(g_m,a)=-\sum_{k=0}^{m}\binom{m+1}{k}(2\pi i)^{m+1-k}\int_{\varepsilon}^{R}f_k(t)dt+\int_{\gamma_{R}}g_m(z)dz-\int_{\gamma_{\epsilon}}g_m(z)dz,$$
откъдето при $m=0$ имаме $$\int_{\varepsilon}^Rf_0(t)dt=-\sum_{a\in M}\text{Res}(g_0,a)+\frac{1}{2\pi i}\int_{\gamma_{R}}g_0(z)dz-\frac{1}{2\pi i}\int_{\gamma_{\epsilon}}g_0(z)dz,$$
а при $m\geq 1$ имаме $$2\pi i\binom{m+1}{m}\int_{\varepsilon}^Rf_m(z)dz=-2\pi i\sum_{a\in M}\text{Res}(g_m,a)-\sum_{k=0}^{m-1}\binom{m+1}{k}(2\pi i)^{m+1-k}\int_{\varepsilon}^{R}f_k(t)dt+$$$$+\int_{\gamma_{R}}g_m(z)dz-\int_{\gamma_{\epsilon}}g_m(z)dz,$$
т. е. $$\int_{\varepsilon}^Rf_m(z)dz=-\frac{1}{m+1}\sum_{a\in M}\text{Res}(g_m,a)-\frac{1}{m+1}\sum_{k=0}^{m-1}\binom{m+1}{k}(2\pi i)^{m-k}\int_{\varepsilon}^{R}f_k(t)dt+$$$$+\frac{1}{2\pi i (m+1)}\int_{\gamma_{R}}g_m(z)dz-\frac{1}{2\pi i(m+1)}\int_{\gamma_{\epsilon}}g_m(z)dz,$$
След граничен преход при $\varepsilon\to 0$ и $R\to+\infty$, прилагайки Твърдение 2, получаваме съответно $$\int_{0}^{+\infty}f_0(z)dz=-\sum_{a\in M}\text{Res}(g_0,a)$$ и
$$\int_{0}^{+\infty}f_m(x)dx=-\frac{1}{m}\sum_{a\in M}\text{Res}(g_m,a)-\frac{1}{m}\sum_{k=0}^{m-1}\binom{m+1}{k}(2\pi i)^{m-k}\int_{0}^{+\infty}f_k(x)dx.$$
б) Нека $R>0,\varepsilon>0,\delta_1>0,\delta_2>0,\varepsilon_1>0,\ldots,\varepsilon_n>0$ и $\Gamma_{R,\varepsilon,\varepsilon_1,\ldots,\varepsilon_n,,\delta_1,\delta_2}$ е положително ориентираната граница на компакта $$K_{R,\varepsilon,\varepsilon_1,\ldots,\varepsilon_n,\delta_1,\delta_2}=\{z\in\mathbb{C}\setminus{0}|z=|z|\exp(it), \varepsilon\leq|z|\leq R,\delta_1\leq t\leq 2\pi-\delta_2,|z-x_1|\geq\varepsilon_1,\ldots,|z-x_n|\geq\varepsilon_n\},$$ като $R,\varepsilon,\varepsilon_1,\ldots,\varepsilon_n,,\delta_1,\delta_2$ са такива, че $M\cup N\subset \text{int }K_{R,\varepsilon,\varepsilon_1,\ldots,\varepsilon_n,\delta_1,\delta_2}$. Тогава от теоремата за резидуумите (Твърдение 3 от Тема 28) имаме
\begin{equation}\label{resthapl}
\int_{\Gamma_{R,\varepsilon,\varepsilon_1,\ldots,\varepsilon_n,\delta_1,\delta_2}}g_m(z)dz=2\pi i\sum_{a\in M}\text{Res}(g_m,a),
\end{equation} а от друга страна $$\int_{\Gamma_{R,\varepsilon,\varepsilon_1,\ldots,\varepsilon_n,\delta_1,\delta_2}}g_m(z)dz=\int_{\gamma_{\varepsilon,\varepsilon_1,\delta_1}}g_m(z)dz+\sum_{j=1}^{n-1}\int_{\gamma_{\varepsilon_j,\varepsilon_{j+1},\delta_1}}g_m(z)dz+\int_{\gamma_{\varepsilon_n,R,\delta_1}}g_m(z)dz-\sum_{j=1}^n\int_{\gamma_{\varepsilon_j,\delta_1}}g_m(z)dz+$$$$+\int_{\gamma_{R,\delta_1,\delta_2}}g_m(z)dz-\int_{\gamma_{\varepsilon_n,R,\delta_2}}g_m(z)dz-\sum_{j=1}^{n-1}\int_{\gamma_{\varepsilon_j,\varepsilon_{j+1},\delta_2}}g_m(z)dz-\int_{\gamma_{\varepsilon,\varepsilon_1,\delta_2}}g_m(z)dz-\sum_{j=1}^{n}\int_{\gamma_{\varepsilon_j,\delta_2}}g_m(z)dz-\int_{\gamma_{\varepsilon,\delta_1,\delta_2}}g_m(z)dz,$$
където $$\gamma_{\varepsilon,\varepsilon_1,\delta_1}(t)=t\exp(i\delta_1),\quad t\in[\varepsilon,x_1-\varepsilon_1],$$$$\gamma_{\varepsilon_j,\varepsilon_{j+1},\delta_1}(t)=t\exp(i\delta_1),\quad t\in[x_j+\varepsilon_j, x_{j+1}-\varepsilon_{j+1}], \quad j\in\{1,\ldots,n\}$$$$\gamma_{\varepsilon_n,R,\delta_1}(t)=t\exp(i\delta_1),\quad t\in[x_n+\varepsilon_n,R],$$$$\gamma_{\varepsilon_j,\delta_1}(t)=[x_j+\varepsilon_j\exp\left(it\right)]\exp(i\delta_1),\quad t\in\left[0,\pi\right],$$$$\gamma_{R,\delta_1,\delta_2}(t)=R\exp(it),\quad t\in[\delta_1,2\pi-\delta_2],$$$$\gamma_{\varepsilon_n,R,\delta_2}(t)=t\exp(i(2\pi-\delta_2)),\quad t\in[x_n+\varepsilon_n,R],$$$$\gamma_{\varepsilon_j,\varepsilon_{j+1},\delta_2}(t)=t\exp(i(2\pi-\delta_2)),\quad t\in[x_j+\varepsilon_j, x_{j+1}-\varepsilon_{j+1}],$$$$\gamma_{\varepsilon,\varepsilon_1,\delta_2}(t)=t\exp(i(2\pi-\delta_2)),\quad t\in[\varepsilon, x_1-\varepsilon_1],$$$$\gamma_{\varepsilon_j,\delta_2}(t)=[x_j+\varepsilon_j\exp(it)]\exp(i(2\pi-\delta_2)),\quad t\in[-\pi,0], \quad j\in\{1,\ldots,n\},$$$$\gamma_{\varepsilon,\delta_1,\delta_2}(t)=\varepsilon\exp(it),\quad t\in[\delta_1,2\pi-\delta_2].$$
Както в a) се убеждаваме, че $$\lim\limits_{\substack{\delta_1\to 0\\\delta_2\to 0}}\int_{\Gamma_{R,\varepsilon,\varepsilon_1,\ldots,\varepsilon_n,\delta_1,\delta_2}}g_m(z)dz=\int_{\varepsilon}^{x_1-\varepsilon_1}g_m(t)dt+\sum_{j=1}^{n-1}\int_{x_j+\varepsilon_j}^{x_{j+1}-\varepsilon_{j+1}}g_m(t)dt+\int_{x_n+\varepsilon_n}^{R}g_m(t)dt-$$$$-\int_{\varepsilon}^{x_1-\varepsilon_1}\frac{P(t)}{Q(t)}(\ln t+2\pi i)^{m+1}dt-\sum_{j=1}^{n-1}\int_{x_j+\varepsilon_j}^{x_{j+1}-\varepsilon_{j+1}}\frac{P(t)}{Q(t)}(\ln t+2\pi i)^{m+1}dt-\int_{x_n+\varepsilon_n}^{R}\frac{P(t)}{Q(t)}(\ln t+2\pi i)^{m+1}dt-$$$$-\sum_{j=1}^n\int_{\gamma_{0,\varepsilon_j}}g_m(z)dz-\sum_{j=1}^{n}\int_{\gamma_{1,\varepsilon_j}}g_m(z)dz+\int_{\gamma_{R}}g_m(z)dz-\int_{\gamma_{\varepsilon}}g_m(z)dz,$$ където $$\gamma_{0,\varepsilon_j}(t)=x_j+\varepsilon_j\exp(it), t\in [0,\pi],$$$$\gamma_{1,\varepsilon_j}(t)=x_j+\varepsilon_j\exp(it), t\in[\pi,2\pi],$$$$\gamma_{R}(t)=R\exp(it),t\in[0,2\pi],$$$$\gamma_{\varepsilon}(t)=\varepsilon\exp(it),t\in[0,2\pi].$$
Прилагайки (3), (4) и замествайки в съответните интервали, от последната формула получаваме $$2\pi i\sum_{a\in M}\text{Res}(g_m,a)=-\sum_{k=0}^{m}\binom{m+1}{k}(2\pi i)^{m+1-k}\left[\int_{\varepsilon}^{x_1-\varepsilon_1}f_k(t)dt+\sum_{j=1}^{n-1}\int_{x_j+\varepsilon_j}^{x_{j+1}-\varepsilon_{j+1}}f_k(t)dt+\int_{x_n+\varepsilon_n}^{R}f_k(t)dt\right]-$$
\begin{equation}\label{form1}
-\sum_{j=1}^n\int_{\gamma_{0,\varepsilon_j}}g_m(z)dz-\sum_{j=1}^{n}\int_{\gamma_{1,\varepsilon_j}}g_m(z)dz+\int_{\gamma_{R}}g_m(z)dz-\int_{\gamma_{\varepsilon}}g_m(z)dz.
\end{equation}
По определение $$v.p.\int_{0}^{+\infty}f_m(t)dt=\lim\limits_{\substack{\varepsilon\to 0 \\\varepsilon_1\to 0 \\ \ldots \\\varepsilon_n\to 0 \\ R\to +\infty}}\left[\int_{\varepsilon}^{x_1-\varepsilon_1}f_m(t)dt+\sum_{j=1}^{n-1}\int_{x_j+\varepsilon_j}^{x_{j+1}-\varepsilon_{j+1}}f_m(t)dt+\int_{x_n+\varepsilon_n}^Rf_m(z)dz\right].$$
От Твърдениe 2 имаме, че $\lim\limits_{R\to\infty}\int_{\gamma_R}g_m(z)dz=0$ и $\lim\limits_{\varepsilon\to 0}\int_{\gamma_{\varepsilon}}g_m(z)dz=0$. Също така $\log z=\log_0 z$, при $z\in\{z\in\mathbb{C}|\Re z>0,\Im z\geq 0\}$, $\log z=\log_1 z$, при $z\in\{z\in\mathbb{C}|\Re z>0, \Im z\leq 0\}$ и клоновете $\log_0$, и $\log_1$ са холоморфни в дясната полуравнина. Следователно за всяко $j\in\{1,\ldots,n\}$, от Твърдение 4 получаваме $$\int_{\gamma_{0,\varepsilon_j}}g_m(z)dz=\int_{\gamma_{0,\varepsilon_j}}g_{m,0}(z)dz=\pi i\text{Res}(g_{m,0},x_j)$$ и $$\int_{\gamma_{1,\varepsilon_j}}g_m(z)dz=\int_{\gamma_{1,\varepsilon_j}}g_{m,1}(z)dz=\pi i\text{Res}(g_{m,1},x_j).$$ Следователно при $\varepsilon\to 0$, $\varepsilon_1\to 0,\varepsilon_n\to 0$, $R\to+\infty$, от (5) получаваме $$2\pi i\sum_{a\in M}\text{Res}(g_m,a)=-\sum_{k=0}^{m}\binom{m+1}{k}(2\pi i)^{m+1-k}v.p.\int_{0}^{+\infty}f_k(t)dt-\pi i\sum_{j=1}^n\text{Res}(g_{m,0},x_j)-\pi i\sum_{j=1}^n\text{Res}(g_{m,1},x_j).$$ Оттук при $m=0$ имаме $$v.p.\int_{0}^{+\infty}f_0(x)dx=-\sum_{a\in M}\text{Res}(g_0,a)-\frac{1}{2}\sum_{b\in N}\text{Res}(g_{0,0},b)-\frac{1}{2}\sum_{b\in N}\text{Res}(g_{0,1},b),$$ и при $m\geq 1$ $$v.p.\int_{0}^{+\infty}f_m(x)dx=-\frac{1}{m+1}\sum_{a\in M}\text{Res}(g_m,a)-\frac{1}{2(m+1)}\sum_{b\in N}\text{Res}(g_{m,0},b)-\frac{1}{2(m+1)}\sum_{b\in N}\text{Res}(g_{m,1},b)-$$$$-\frac{1}{m+1}\sum_{k=0}^{m-1}\binom{m+1}{k}(2\pi i)^{m-k}v.p.\int_{0}^{+\infty}f_k(x)dx.$$

Твърдение 8. Нека $P$ и $Q$ са полиноми с реални коефициенти, за които $\deg Q\geq \deg P+1$. Нека $m\in\mathbb{N}\cup\{0\}$, $0<\alpha<1$, $f_m(x)=\frac{P(x)}{Q(x)}x^{-\alpha}\ln^m x$ и $g_m(z)=\frac{P(z)}{Q(z)}\exp(-\alpha\log z)(\log z)^m$, където $\log z=\ln|z|+i\arg z$, $0<\arg z<2\pi$. Тогава
а) Ако $Q(x)\neq 0$ за всяко $x>0$ и $M=\{z\in\mathbb{C}|Q(z)=0\}$, то $$\int_{0}^{+\infty}f_0(x)dx=\frac{\pi\exp(\pi i\alpha)}{\sin\pi\alpha}\sum_{a\in M}\text{Res}(g_0,a),$$ и $$\int_{0}^{+\infty}f_m(x)dx=\frac{\pi\exp(\pi i\alpha)}{\sin\pi\alpha}\sum_{a\in M}\text{Res}(g_m,a)+\frac{\exp(-\pi i(\alpha+1/2))}{2\sin\pi\alpha}\sum_{k=0}^{m-1}\binom{m}{k}(2\pi i)^{m-k}\int_{0}^{+\infty}f_k(x)dx.$$
б) Ако $M=\{z\in\mathbb{C}|\Im z>0, Q(z)=0\}$, $N=\{x>0|Q(x)=0\}\neq\emptyset$ и за всяко $x\in N$ имаме $Q'(x)\neq 0$, (т. е. всички реални нули на $Q$ са прости), $g_{m,0}(z)=\frac{P(z)}{Q(z)}\exp(-\alpha\log_0 z)(\log_0z)^{m}$, $g_{m,1}(z)=\frac{P(z)}{Q(z)}\exp(-\alpha\log_1 z)(\log_1z)^{m}$, то $$v.p.\int_{0}^{+\infty}f_0(x)dx=\frac{\pi\exp(\pi i\alpha)}{\sin\pi\alpha}\left(\sum_{a\in M}\text{Res}(g_0,a)+\frac{1}{2}\sum_{b\in N}\text{Res}(g_{0,0},b)+\frac{1}{2}\sum_{b\in N}\text{Res}(g_{0,1},b)\right),$$ и
$$v.p.\int_{0}^{+\infty}f_m(x)dx=\frac{\pi\exp(\pi i\alpha)}{\sin\pi\alpha}\left(\sum_{a\in M}\text{Res}(g_m,a)+\frac{1}{2}\sum_{b\in N}\text{Res}(g_{m,0},b)+\frac{1}{2}\sum_{b\in N}\text{Res}(g_{m,1},b)\right)+$$$$+\frac{\exp(-\pi i(\alpha+1/2))}{2\sin\pi\alpha}\sum_{k=0}^{m-1}\binom{m}{k}(2\pi i)^{m-k}v.p.\int_{0}^{+\infty}f_k(x)dx.$$
Доказателство. a) Отчитайки, че $\exp(-\alpha(\ln t+2\pi i))=t^{-\alpha}\exp(-2\pi i\alpha)$ и следвайки дословно доказателството на Твърдение 7 a), в съотвествие с означенията, получаваме \begin{equation}\label{resa0}
2\pi i\sum_{a\in M}\text{Res}(g_m,a)=\int_{\varepsilon}^{R}f_m(t)dt-\exp(-2\pi i\alpha)\int_{\varepsilon}^{R}\frac{P(t)}{Q(t)}t^{-\alpha}(\ln t+2\pi i)^{m}dt+\int_{\gamma_{R}}g_m(z)dz-\int_{\gamma_{\varepsilon}}g_m(z)dz.
\end{equation} Оттук при $m=0$ получаваме $$2\pi i\sum_{a\in M}\text{Res}(g_0,a)=(1-\exp(-2\pi i\alpha))\int_0^{+\infty}f_0(t)dt+\int_{\gamma_{R}}g_0(z)dz-\int_{\gamma_{\varepsilon}}g_0(z)dz,$$
и тъй като $\lim\limits_{R\to\infty}\int_{\gamma_R}g_m(z)dz=0$ и $\lim\limits_{\varepsilon\to 0}\int_{\gamma_{\varepsilon}}g_m(z)dz=0$ (Твърдение 2), след граничен преход при $\varepsilon\to 0$ и $R\to+\infty$, имаме $$\int_0^{+\infty}f_0(t)dt=\frac{2\pi i}{1-\exp(-2\pi i\alpha)}\sum_{a\in M}\text{Res}(g_0,a).$$
Отчитайки, че \begin{equation}\label{exptosin}
\frac{2\pi i}{1-\exp(-2\pi i\alpha )}=\frac{2\pi i\exp(\pi i\alpha)}{\exp(\pi i\alpha)-\exp(-\pi i\alpha)}=\frac{\pi\exp(\pi i\alpha)}{\sin\pi\alpha},
\end{equation} получаваме
$$\int_0^{+\infty}f_0(t)dt=\frac{\pi\exp(\pi i\alpha)}{\sin\pi \alpha}\sum_{a\in M}\text{Res}(g_0,a).$$
При $m>0$ използваме, че $$
(\ln t+2\pi i)^{m}=\sum_{k=0}^{m}\binom{m}{k}(\ln t)^k(2\pi i)^{m-k}=(\ln t)^m+\sum_{k=0}^{m-1}\binom{m}{k}(\ln t)^k(2\pi i)^{m-k},$$
при което
\begin{equation}\label{binomlog}
\frac{P(t)}{Q(t)}t^{-\alpha}(\ln t+2\pi i)^m=f_m(t)+\sum_{k=0}^{m-1}\binom{m}{k}\frac{P(t)}{Q(t)}t^{-\alpha}(\ln t)^k(2\pi i)^{m-k}=f_m(t)+\sum_{k=0}^{m-1}\binom{m}{k}(2\pi i)^{m-k}f_k(t)
\end{equation} и след заместване в (6) получаваме $$2\pi i\sum_{a\in M}\text{Res}(g_m,a)=(1-\exp(-2\pi i\alpha))\int_{\varepsilon}^{R}f_m(t)dt-\exp(-2\pi i\alpha)\sum_{k=0}^{m-1}\binom{m}{k}(2\pi i)^{m-k}\int_{\varepsilon}^{R}f_k(t)dt.$$ След граничен преход при $\varepsilon\to 0$ и $R\to+\infty$ получаваме $$\int_{0}^{+\infty}f_m(x)dx=\frac{2\pi i}{1-\exp(-2\pi i\alpha)}\sum_{a\in M}\text{Res}(g_m,a)+\frac{\exp(-2\pi i\alpha)}{1-\exp(-2\pi i\alpha)}\sum_{j=0}^{m-1}\binom{m}{j}(2\pi i)^{m-j}\int_{0}^{+\infty}f_j(x)dx.$$ Тъй като \begin{equation}\label{exptosin2}
\frac{\exp(-2\pi i\alpha)}{1-\exp(-2\pi i\alpha)}=\frac{\exp(-\pi i\alpha)}{\exp(\pi i\alpha)-\exp(-\pi i\alpha)}=\frac{\exp(-\pi i\alpha)}{2i\sin\pi\alpha}=\frac{\exp(-\pi i/2)\exp(-\pi i\alpha)}{2\sin\pi\alpha}=\frac{\exp(-\pi i(\alpha+1/2))}{2\sin\pi\alpha},
\end{equation} вземайки предвид (7) и (9) получаваме $$\int_{0}^{+\infty}f_m(x)dx=\frac{\pi\exp(\pi i\alpha)}{\sin\pi\alpha}\sum_{a\in M}\text{Res}(g_m,a)+\frac{\exp(-\pi i(\alpha+1/2))}{2\sin\pi\alpha}\sum_{k=0}^{m-1}\binom{m}{k}(2\pi i)^{m-k}\int_{0}^{+\infty}f_k(x)dx.$$
б) Следвайки дословно доказателството на на Твърдение 7 б), в съотвествие с означенията, получаваме
$$2\pi i\sum_{a\in M}\text{Res}(g_m,a)=\int_{\varepsilon}^{x_1-\varepsilon_1}f_m(t)dt+\sum_{j=1}^{n-1}\int_{x_j+\varepsilon_j}^{x_{j+1}-\varepsilon_{j+1}}f_m(t)dt+\int_{\varepsilon_n}^{R}f_m(t)dt-$$$$-\exp(-2\pi i\alpha)\left[\int_{\varepsilon}^{x_1-\varepsilon_1}\frac{P(t)}{Q(t)}t^{-\alpha}(\ln t+2\pi i)^{m}dt+\sum_{j=1}^{n-1}\int_{x_j+\varepsilon_j}^{x_{j+1}-\varepsilon_{j+1}}\frac{P(t)}{Q(t)}t^{-\alpha}(\ln t+2\pi i)^{m}dt\right.+$$$$+\left.\int_{x_n+\varepsilon_n}^{R}\frac{P(t)}{Q(t)}t^{-\alpha}(\ln t+2\pi i)^{m}dt\right]-$$
\begin{equation}\label{mainform2}
-\sum_{j=1}^n\int_{\gamma_{0,\varepsilon_j}}g_m(z)dz-\sum_{j=1}^{n}\int_{\gamma_{1,\varepsilon_j}}g_m(z)dz+\int_{\gamma_{R}}g_m(z)dz-\int_{\gamma_{\varepsilon}}g_m(z)dz,
\end{equation}
където $$\gamma_{0,\varepsilon_j}(t)=x_j+\varepsilon_j\exp(it), t\in [0,\pi],$$ $$\gamma_{1,\varepsilon_j}(t)=x_j+\varepsilon_j\exp(it), t\in[\pi,2\pi],$$$$\gamma_{R}(t)=R\exp(it),t\in[0,2\pi],$$$$\gamma_{\varepsilon}(t)=\varepsilon\exp(it),t\in[0,2\pi].$$
По определение $$v.p.\int_{0}^{+\infty}f_m(t)dt=\lim\limits_{\substack{\varepsilon\to 0 \\\varepsilon_1\to 0 \\ \ldots \\\varepsilon_n\to 0 \ R\to +\infty}}\left[\int_{\varepsilon}^{x_1-\varepsilon_1}f_m(t)dt+\sum_{j=1}^{n-1}\int_{x_j+\varepsilon_j}^{x_{j+1}-\varepsilon_{j+1}}f_m(t)dt+\int_{x_n+\varepsilon_n}^Rf_m(z)dz\right].$$
От Твърдениe 2 имаме, че $\lim\limits_{R\to\infty}\int_{\gamma_R}g_m(z)dz=0$ и $\lim\limits_{\varepsilon\to 0}\int_{\gamma_{\varepsilon}}g_m(z)dz=0$. Също така $\log z=\log_0 z$, при $z\in\{z\in\mathbb{C}|\Re z>0,\Im z> 0\}$, $\log z=\log_1 z$, при $z\in\{z\in\mathbb{C}|\Re z>0, \Im z< 0\}$ и клоновете $\log_0$, и $\log_1$ са холоморфни в дясната полуравнина. Следователно за всяко $j\in\{1,\ldots,n\}$, от Твърдение 4 получаваме $$\int_{\gamma_{0,\varepsilon_j}}g_m(z)dz=\int_{\gamma_{0,\varepsilon_j}}g_{m,0}(z)dz=\pi i\text{Res}(g_{m,0},x_j)$$ и $$\int_{\gamma_{1,\varepsilon_j}}g_m(z)dz=\int_{\gamma_{1,\varepsilon_j}}g_{m,0}(z)dz=\pi i\text{Res}(g_{m,1},x_j).$$
Следователно при $m=0$, след преобразуване и граничен преход в (10), при $\varepsilon\to 0$, $\varepsilon_1\to 0,\ldots,\varepsilon_n\to 0$, $R\to +\infty$ получаваме $$2\pi i\sum_{a\in M}\text{Res}(g_0,a)=(1-\exp(-2\pi i\alpha))v.p.\int_{0}^{+\infty}f_0(t)dt-\sum_{j=1}^n\pi i\text{Res}(g_{0,0},x_j)-\sum_{j=1}^{n}\pi i\text{Res}(g_{0,1},x_j),$$
откъдето \begin{equation}\label{result1b}
v.p.\int_{0}^{+\infty}f_0(x)dx=\frac{2\pi i}{1-\exp(-2\pi i\alpha )}\left(\sum_{a\in M}\text{Res}(g_0,a)+\frac{1}{2}\sum_{b\in N}\text{Res}(g_{0,0},b)+\frac{1}{2}\sum_{b\in N}\text{Res}(g_{0,1},b)\right).
\end{equation}
При $m>0$, от (8), след заместване в (10) получаваме $$2\pi i\sum_{a\in M}\text{Res}(g_m,a)=(1-\exp(-2\pi i\alpha))\left[\int_{\varepsilon}^{x_1-\varepsilon_1}f_m(t)dt+\sum_{j=1}^{n-1}\int_{x_j+\varepsilon_j}^{x_{j+1}-\varepsilon_{j+1}}f_m(t)dt+\int_{x_n+\varepsilon_n}^{R}f_m(t)dt\right]-$$$$-\exp(-2\pi i\alpha)\sum_{k=0}^{m-1}\binom{m}{k}(2\pi i)^{m-k}\left[\int_{\varepsilon}^{x_1-\varepsilon_1}f_k(t)dt+\sum_{j=1}^{n-1}\int_{x_j+\varepsilon_j}^{x_{j+1}-\varepsilon_{j+1}}f_k(t)dt+\int_{x_n+\varepsilon_n}^{R}f_k(t)dt\right]-$$
\begin{equation}\label{form2}
-\sum_{j=1}^n\int_{\gamma_{0,\varepsilon_j}}g_m(z)dz-\sum_{j=1}^{n}\int_{\gamma_{1,\varepsilon_j}}g_m(z)dz+\int_{\gamma_{R}}g_m(z)dz-\int_{\gamma_{\varepsilon}}g_m(z)dz.
\end{equation}
Следователно при граничен преход в (12), при $\varepsilon\to 0$, $\varepsilon_1\to 0,\ldots,\varepsilon_n\to 0$, $R\to +\infty$ получаваме $$2\pi i\sum_{a\in M}\text{Res}(g_m,a)=(1-\exp(-2\pi i\alpha))v.p.\int_{0}^{+\infty}f_m(t)dt-\exp(-2\pi i\alpha)\sum_{k=0}^{m-1}\binom{m}{k}(2\pi i)^{m-k}v. p.\int_{0}^{+\infty}f_k(t)dt-$$$$-\sum_{j=1}^n\pi i\text{Res}(g_{m,0},x_j)-\sum_{j=1}^{n}\pi i\text{Res}(g_{m,1},x_j),$$
откъдето $$v.p.\int_{0}^{+\infty}f_m(x)dx=\frac{2\pi i}{1-\exp(-2\pi i\alpha )}\left(\sum_{a\in M}\text{Res}(g_m,a)+\frac{1}{2}\sum_{b\in N}\text{Res}(g_{m,0},b)+\frac{1}{2}\sum_{b\in N}\text{Res}g_{m,1},b)\right)+$$\begin{equation}\label{result2b}
+\frac{\exp(-2\pi i\alpha)}{1-\exp(-2\pi i\alpha)}\sum_{k=0}^{m-1}\binom{m}{k}(2\pi i)^{m-k}v.p.\int_{0}^{+\infty}f_k(x)dx.
\end{equation}
Отчитайки (7) и (9), замествайки съответно в (11) и (13) получаваме верността на твърдението.

Упражнение 1. Пресметнете интегралите
1) $\int_0^{\pi}\sin^{2n}xdx$, $n\in\mathbb{N}$, 2) $\int_0^{+\infty}\frac{1}{(1+x^2)^n}dx$, $n\in\mathbb{N}$, 3) $v.p.\int_{-\infty}^{\infty}\frac{1}{x^3-x^2+4x-4}dx$, 4) $\int_0^{+\infty}\frac{1}{1+x^n}dx$, $n\in\mathbb{N}\setminus\{1\}$,
5) $v.p.\int_0^{+\infty}\frac{1}{1-x^n}dx$, $n\in\mathbb{N}\setminus\{1\}$, 6) $\int_0^{\infty}\frac{x\sin mx}{x^2+a^2}$, $m>0, a\geq 0$, 7) $v.p.\int_{-\infty}^{\infty}\frac{x\cos mx}{(x-1)(x^2+1)^2}dx$, $m>0$,
8) $\int_0^{\infty}\frac{1}{x^{\alpha}(1+x)}dx$, $0<\alpha<1$, 9) $v.p.\int_0^{\infty}\frac{1}{x^{\alpha}(1-x)}dx$, $0<\alpha<1$, 10) $v.p.\int_0^{+\infty}\frac{\ln^3 x}{x^2-a^2}dx, a>0$
11) $\int_0^{+\infty}\frac{\ln x}{\sqrt{x}(x+1)^2}dx$, 12) $v.p.\int_0^{+\infty}\frac{\ln^2 x}{\sqrt{x}(x-2)}dx$

назад