广义积分习题课————判断敛散性
1
\[\lim_{x \to +\infty} \frac{\frac{x\ln x}{\sqrt{x^5+1}}}{\frac{1}{x^{\frac{5}{4}}}}=0,\int _1^{+\infty} \frac{dx}{x^{\frac{5}{4}}} 收敛 \Rightarrow 收敛 \]
2
\[\int_0^\pi \frac{dx}{\sqrt{\sin x}}=2\int_0^{\frac{\pi}{2}} \frac{dx}{\sqrt{\sin x}},\lim_{x \to 0^+} \frac{\frac{1}{\sqrt{\sin x}}}{\frac{1}{x^{\frac{1}{2}}}} =1,\int_0^{\frac{\pi}{2}} \frac{dx}{\sqrt{x}}收敛 \Rightarrow 收敛 \]
3
\[\int_0^{+\infty} \frac{\arctan x}{x^p} dx=\int_0^1 \frac{\arctan x}{x^p}dx+\int_1^{+\infty} \frac{\arctan x}{x^p}dx \\ \lim_{x \to +\infty} \frac{\frac{\arctan x}{x^p}}{\frac{1}{x^p}}=\frac{\pi}{2},\int_1^{+\infty} \frac{dx}{x^p} 收敛当且仅当p>1 \\ p>1:\lim_{x \to 0^+} \frac{\frac{\arctan x}{x^p}}{\frac{1}{x^{p-1}}}=1,\int_0^{1}\frac{dx}{x^{p-1}} 收敛当且仅当 p-1<1 \Rightarrow p<2 \\ 所以 1<p<2 \]
4
\[\int_0^{+\infty} \left[ \ln(1+\frac{1}{x})-\frac{1}{1+x} \right]dx \sim \int_0^1 \frac{\ln(1-x)+x }{(1+x)^2}dx \sim \int_0^{1} \ln xdx=(x\ln x-x)\bigg|_0^{1}=-1 收敛 \Rightarrow 收敛 \]
5
\[\int_0^{\frac{\pi}{2}} \frac{dx}{\sin^px\cos^qx} \sim \left(\int_0^{\frac{\pi}{2}} \frac{dx}{\sin^px} 和 \int_0^{\frac{\pi}{2}} \frac{dx}{\cos^qx}\right) \sim \left(\int_0^{\frac{\pi}{2}} \frac{dx}{\sin^px} 和 \int_0^{\frac{\pi}{2}} \frac{dx}{\sin^qx}\right) \\ \lim_{x \to 0^+} \frac{\frac{1}{\sin^p x}}{\frac{1}{x^p}}=1 \Rightarrow \int_0^{\frac{\pi}{2}}\frac{dx}{\sin^px}\sim \int_0^{\frac{\pi}{2}} \frac{dx}{x^p} 收敛当且仅当p<1 \Rightarrow p<1 \land q<1 \]
6
\[\int_0^{+\infty} \frac{\ln(1+x)}{x^p}dx=\int_0^{1} \frac{\ln(1+x)}{x^p}dx+\int_1^{+\infty} \frac{\ln(1+x)}{x^p}dx \\ \lim_{x \to 0^+} \frac{\frac{\ln(1+x)}{x^p}}{\frac{1}{x^{p-1}}}=1,\int_0^{1} \frac{dx}{x^{p-1}} 收敛当且仅当p-1<1 \Rightarrow p<2 \\ \int_1^{+\infty} \frac{\ln(1+x)}{x^p}dx \sim \int_1^{+\infty} \frac{\ln x}{x^p}dx \sim \int_1^{+\infty} \frac{\ln \frac{1}{x}}{x^{p}}dx\sim \int_0^1 \frac{\ln x}{x^{-p}} \frac{dx}{x^2}=\int_0^1 \frac{\ln x}{x^{2-p}}dx \\ \int_0^1 \frac{\ln x}{x^u}dx:当 u<1时,取 v\in(u,1),\lim_{x \to 0^+} \frac{x^{-u}\ln x}{x^{-v}}=0,\int_0^1 \frac{\ln x}{x^v}dx收敛;当u \ge 1时,\int_0^{e^{-1}} \frac{\ln x}{x^udx} \sim \int_0^{e^{-1}} \frac{dx}{x^u}发散 \\ 所以 2-p <1 \Rightarrow p>1 \\ 所以1<p<2 \]
7
\[\int_0^1 \frac{\ln x}{\sqrt{x}(1-x)^2}dx \sim \left( \int_0^\frac{1}{2} \frac{\ln x}{\sqrt{x}}dx 和 \int_\frac{1}{2}^1 \frac{\ln x}{(1-x)^2}dx \right) \\ \int_0^1 \frac{\ln x}{\sqrt{x}}dx \sim \int_0^1 \ln xdx 收敛 \\ \int_\frac{1}{2}^1 \frac{\ln x}{(1-x)^2}dx \sim \int_0^{\frac{1}{2}} \frac{\ln (1-x)}{x^2}dx \sim \int_0^{\frac{1}{2}} \frac{dx}{x} 发散 \\ 所以发散 \]
8
\[\int_0^{+\infty} \frac{\sin x^2}{x^p}dx \sim \int_0^{+\infty} \frac{\sin x}{x^{\frac{p}{2}}} \frac{dx}{\sqrt{x}} \sim \int_0^{+\infty} \frac{\sin x}{x^{\frac{p-1}{2}}}dx \xlongequal{q=\frac{p-1}{2}}\int_0^{+\infty} \frac{\sin x}{x^q}dx \\ \xlongequal{感觉这里拆开不太对} \int_0^{1} \frac{\sin x}{x^q}dx+\int_1^{+\infty} \frac{\sin x}{x^q}dx \\ \sim \left( \int_0^{1}\frac{dx}{x^{q-1}} 和\int_1^{+\infty} \frac{\sin x}{x^q}dx \right) \\ \Rightarrow 收敛当且仅当(q-1<1 \land q>0) \Rightarrow 收敛当且仅当1<p<5 \\ \Rightarrow 前半部分绝对收敛，后半部分当q>1时绝对收敛 \\ \Rightarrow 3<p<5时绝对收敛,1<p \le 3时条件收敛 \]
9
\[\int_0^{+\infty} \frac{x^\alpha}{1+x^{\beta}} dx \\ 若\alpha=0,则发散 \\ 若\alpha>0,则\sim \left(\int_0^{1}x^\alpha dx 和 \int_1^{+\infty}\frac{dx}{x^{\beta-\alpha}+x^{-\alpha}} \right) \sim \left( \int_1^{+\infty} \frac{dx}{x^{\beta-\alpha}} \right) \sim \beta-\alpha>1 \\ \Rightarrow 0<\alpha<\beta-1,\beta>1时收敛 \\ 若 \alpha<0,则-1<\alpha<0且\max(-\alpha,\beta-\alpha)>1 \Rightarrow -1<\alpha<\min(0,\beta-1),\beta > 0 时收敛 \]
10
\[\int_0^{+\infty} \frac{\sin^2 x}{x^p}dx=\int_0^{1}\frac{\sin^2 x}{x^p}dx+\int_1^{+\infty} \frac{\sin^2x}{x^p}dx \\ \int_0^{1}\frac{\sin^2 x}{x^p}dx \sim \int_0^{1} \frac{dx}{ x^{p-2}} \Rightarrow p<3 \\ \int_1^{+\infty} \frac{\sin^2 x}{x^p}dx \sim \int_1^{+\infty} \frac{1-\cos (2x)}{x^p}dx \xlongequal{感觉这里不太对}\int_1^{+\infty} \frac{dx}{x^p}-\int_1^{+\infty} \frac{\cos (2x)}{x^p}dx \Rightarrow p>1 \\ 所以 1<p<3 \]
11
\[\int_1^{+\infty}x\cos x^3dx \sim \int_1^{+\infty} \sqrt[3]{x}\cos x d \sqrt[3]{x} \sim \int_1^{+\infty}\frac{\cos x}{x^{\frac{1}{3}}}dx 条件收敛(0<\frac{1}{3}<1) \]
12
\[\int_0^{+\infty}\frac{x^p\arctan x}{2+x^q}dx=\int_0^{+\infty}\frac{\arctan x}{2x^{-p}+x^{q-p}}dx \\ \int_0^{+\infty} \frac{\arctan x}{x^p}dx在1<p<2时收敛 \\ q>0:\int_1^{+\infty} \frac{x^p\arctan x}{2+x^q}dx \sim\int_{1}^{+\infty} \frac{\arctan x}{x^{q-p}}dx \sim \int_1^{+\infty} \frac{dx}{x^{q-p}} 当q-p>1时收敛 \\ q \le 0:\int_1^{+\infty} \frac{x^p\arctan x}{2+x^q}dx \sim \int_1^{+\infty} \frac{\arctan x}{x^{-p}}dx 当p<-1时收敛 \\ q \ge 0:\int_0^{1} \frac{x^p\arctan x}{2+x^q}dx \sim \int_0^1\frac{dx}{x^{-p-1}}当p<-2时收敛 \\ q < 0:\int_0^{1} \frac{x^p\arctan x}{2+x^q}dx \sim \int_0^1\frac{dx}{x^{q-p-1}} 当q-p>2时收敛 \\ 综上,当q\ge 0 \land p<-2时，或当q<0,p<q-2时收敛 \]
13
\[\int_0^{\frac{\pi}{2}} \left(\ln \sin x+\frac{1}{\ln \sin x} \right)dx =\int_0^{\frac{\pi}{2}} \left(\frac{\left(\ln \sin x\right)^2+1}{\ln \sin x} \right)dx \\ \sim \left( \int_{0}^{\frac{1}{2}}\ln \sin x dx 和 \int_\frac{1}{2}^{\frac{\pi}{2}}\frac{dx}{\ln \sin x} \right) \\ \int_0^\frac{1}{2} \ln \sin xdx \sim \int_0^{\frac{1}{2}}\ln xdx 收敛 \\ \int_\frac{\pi}{4}^\frac{\pi}{2} \frac{dx}{\ln \sin x}\sim\int_0^{\frac{\pi}{4}} \frac{dx}{\ln \cos x}\\ \lim_{x \to 0^+} \frac{x}{\ln \cos x}=\lim_{x \to 0^+}-\frac{1}{\frac{\sin x}{\cos x}}=-\infty,\int_0^{\frac{\pi}{4}} \frac{dx}{x} 发散,所以\int_0^{\frac{\pi}{4}}\frac{dx}{\ln \cos x} 发散 \\ 所以发散 \]