作业 29 广义积分

标签：

1. (1)
\[
    \mbox{原式}= \lim_{b\to +\infty} \int_1^b \frac{e^2}{e^{2x}+e^2}dx= \frac1e \lim_{b\ to+\infty} \arctan(e^x/e) \bigg|_{1}^b=\frac1e (\lim_{b\to +\infty}\arctan \frac{e^b}{e} -\frac{\pi}{4})=\frac{\pi}{4 e}.
\]



(2)
\[
\begin{aligned}
     \mbox{原式}= \lim_{b\to +\infty} \frac{\arctan x}{x^2} dx =-\lim_{b\to +\infty} \frac{\arctan x}{x}\bigg|_{1}^{b} +\lim_{b\to +\infty} \int_1^b \frac{1}{x(1+x^2)} dx
\\=\frac{\pi}{4} +\lim_{b\to+\infty}(\ln|x|-\frac12 \ln|1+x^2|)\bigg|_{1}^b
= \frac \pi 4 +\frac12 \ln 2 +\lim_{b\to +\infty} (\ln b-\frac12 \ln(1+b^2)) = \frac \pi 4 + \frac12 \ln 2.
\end{aligned}
\]


(3)
\[
 \mbox{原式}=\lim_{b\to +\infty}\int_0^b e^{-ax}\sin x dx
\]
由分部积分容易得到
\[
\int_0^b e^{-ax}\sin x dx = \frac{1-e^{-ab} \cos b-ae^{-ab}\sin b}{1+a^2},
\]
因此
\[
 \mbox{原式}=\lim_{b\to +\infty}\int_0^b e^{-ax}\sin x dx=\lim_{b\to +\infty} \frac{1-e^{-ab} \cos b-ae^{-ab}\sin b}{1+a^2}
=\frac{1}{1+a^2}.
\]


(4)
\[
   \mbox{原式}=\lim_{\varepsilon \to 0^+}  \int_0^{1-\varepsilon} \frac{x}{\sqrt{1-x^2}} dx
= \lim_{\varepsilon \to 0^+} (-\sqrt{1-x^2})\bigg|_0^{1-\varepsilon} =1.
\]



(5)
\[
     \mbox{原式}= \lim_{\varepsilon \to 0^+} \int_{1+\varepsilon}^2 \frac{x}{\sqrt{x-1}} dx
=  \lim_{\varepsilon \to 0^+}  \int_{1+\varepsilon}^2 (\sqrt{x-1}+\frac{1}{\sqrt{x-1}}) dx=
 \lim_{\varepsilon \to 0^+}  (\frac23 (x-1)^{3/2}+2\sqrt{x-1} )\bigg|_{1+\varepsilon}^2 =\frac83.
\]


(6)
\[
   \mbox{原式}=  \int_{-\frac \pi 4}^{\frac \pi 2 } \sec^2 x dx + \int^{\frac {3\pi} 4}_{\frac \pi 2 } \sec^2 x dx =
\lim_{\varepsilon_1 \to 0^+} \int_{-\frac \pi 4}^{\frac \pi 2-\varepsilon_1} \sec^2 x dx +\lim_{\varepsilon_2 \to 0^+} \int^{\frac {3\pi} 4}_{\frac \pi 2+\varepsilon_2} \sec^2 x dx
\]
其中
\[
  \lim_{\varepsilon_1 \to 0^+} \int_{-\frac \pi 4}^{\frac \pi 2-\varepsilon_1} \sec^2 x dx =\lim_{\varepsilon_1 \to 0^+} \tan x\bigg|_{-\frac \pi 4}^{\frac \pi 2 -\varepsilon_1}
=  +\infty,
\]
因此不收敛，同理可证 $\lim\limits_{\varepsilon_2 \to 0^+} \int^{\frac {3\pi} 4}_{\frac \pi 2+\varepsilon_2} \sec^2 x dx$ 不收敛, 所以原积分不收敛。




(7)
\[
   \mbox{原式}= \lim_{\varepsilon \to 0^+} \int_1^{e-\varepsilon} \frac{dx}{x\sqrt{1-(\ln x)^2 }}=  \lim_{\varepsilon \to 0^+} \arcsin(\ln x) \bigg|_1^{e-\varepsilon}
= \frac \pi 2.
\]

(8)
\[
    \mbox{原式}= \int_{\frac12}^{1} \frac{dx}{\sqrt{x-x^2}}+ \int_{1}^{\frac 32} \frac{dx}{\sqrt{x^2-x}}
= \lim_{\varepsilon_1 \to 0^+}\int_{\frac12}^{1-\varepsilon_1} \frac{dx}{\sqrt{x-x^2}}+ \lim_{\varepsilon_2 \to 0^+}\int_{1+\varepsilon_2}^{\frac 32} \frac{dx}{\sqrt{x^2-x}}
\]
其中
\[
  \lim_{\varepsilon_1 \to 0^+}\int_{\frac12}^{1-\varepsilon_1} \frac{dx}{\sqrt{x-x^2}}= \lim_{\varepsilon_1 \to 0^+}\arcsin(2x-1)\bigg|_{\frac12}^{1-\varepsilon_1}=\frac \pi 2.
\]
而
\[
  \lim_{\varepsilon_2 \to 0^+}\int_{1+\varepsilon_2}^{\frac 32} \frac{dx}{\sqrt{x^2-x}}= \lim_{\varepsilon_2 \to 0^+} \ln( x-\frac12 +\sqrt{x^2-x} )\bigg|_{1+\varepsilon_2}^{\frac32}=
   \ln(2+\sqrt 3)
\]
因此
\[
  \mbox{原式}= \frac{ \pi }{2} +\ln(2+\sqrt 3).
\]

作业 29 广义积分

标签：

原文地址：http://www.cnblogs.com/mmmmmm6m/p/4199327.html