Assignment 29 Generalized integrals

1. (1)
\[
\mbox{original}= \lim_{b\to +\infty} \int_1^b \frac{e^2}{e^{2x}+e^2}dx= \frac1e \lim_{b\ to+\infty} \arctan (e^x/e) \bigg|_{1} ^B=\FRAC1E (\lim_{b\to +\infty}\arctan \frac{e^b}{e}-\frac{\pi}{4}) =\frac{\pi}{4 e}.
\]



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


(3)
\[
\mbox{Original}=\lim_{b\to +\infty}\int_0^b e^{-ax}\sin x DX
\]
Easy to get by partial integration
\[
\int_0^b e^{-ax}\sin x dx = \frac{1-e^{-ab} \cos b-ae^{-ab}\sin b}{1+a^2},
\]
So
\[
\mbox{Original}=\lim_{b\to +\infty}\int_0^b e^{-ax}\sin x dx=\lim_{b\to +\infty} \frac{1-e^{-ab} \cos b-ae^{-ab}\sin b}{1+a^2 }
=\FRAC{1}{1+A^2}.
\]


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



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


(6)
\[
\mbox{original}= \int_{-\frac \pi 4}^{\frac \pi 2} \sec^2 x dx + \int^{\frac {3\pi} 4}_{\frac \pi 2} \sec^2 x dx =
\lim_{\varepsilon_1 \to 0^+} \int_{-\frac \pi 4}^{\frac \pi 2-\varepsilon_1} \sec^2 x dx +\lim_{\varepsilon_2 \to 0^+} \in T^{\frac {3\pi} 4}_{\frac \pi 2+\varepsilon_2} \sec^2 x dx
\]
which
\[
\lim_{\varepsilon_1 \to 0^+} \int_{-\frac \pi 4}^{\frac \pi 2-\varepsilon_1} \sec^2 x dx =\lim_{\varepsilon_1 \to 0^+} \ta n X\bigg|_{-\frac \pi 4}^{\frac \pi 2-\varepsilon_1}
= +\infty,
\]
So do not converge, the same can be $\lim\limits_{\varepsilon_2 \to 0^+} \int^{\frac {3\pi} 4}_{\frac \pi 2+\varepsilon_2} \sec^2 x dx$ does not converge, so the original integral is not Convergence.




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

(8)
\[
\mbox{original}= \int_{\frac12}^{1} \frac{dx}{\sqrt{x-x^2}}+ \int_{1}^{\frac} \frac{dx}{\sqrt{x^2-x}}
= \lim_{\varepsilon_1 \to 0^+}\int_{\frac12}^{1-\varepsilon_1} \frac{dx}{\sqrt{x-x^2}}+ \lim_{\varepsilon_2 \to 0^+}\ Int_{1+\varepsilon_2}^{\frac} \frac{dx}{\sqrt{x^2-x}}
\]
which
\[
\lim_{\varepsilon_1 \to 0^+}\int_{\frac12}^{1-\varepsilon_1} \frac{dx}{\sqrt{x-x^2}}= \lim_{\varepsilon_1 \to 0^+}\ Arcsin (2x-1) \bigg|_{\frac12}^{1-\varepsilon_1}=\frac \pi 2.
\]
and
\[
\lim_{\varepsilon_2 \to 0^+}\int_{1+\varepsilon_2}^{\frac) \frac{dx}{\sqrt{x^2-x}}= \lim_{\varepsilon_2 \to 0^+} \ ln (X-\FRAC12 +\sqrt{x^2-x}) \bigg|_{1+\varepsilon_2}^{\frac32}=
\ln (2+\sqrt 3)
\]
So
\[
\mbox{original}= \frac{\pi}{2} +\ln (2+\sqrt 3).
\]

Assignment 29 Generalized integrals