廣義積分習題課————計算下列廣義積分的值

1

\[\int_0^{+\infty} \frac{dx}{(1+5x^2)\sqrt{1+x^2}} \xlongequal{x=\tan t} \int_0^{\frac{\pi}{2}} \frac{\sec t dt}{1+5\tan^2 t} \\ =\int_0^{\frac{\pi}{2}} \frac{\sec t dt}{\sec^2 t(1+4\sin^2 t)}=\int_0^\frac{\pi}{2} \frac{d\sin t}{1+4\sin^2 t}=\frac{1}{2}\arctan (2\sin t) \bigg|_0^{\frac{\pi}{2}}=\frac{\arctan 2}{2} \]

2

\[\int_1^{+\infty} \frac{\arctan x}{x^2}dx=-\frac{\arctan x}{x}\bigg|_1^{+\infty}+\int_1^{+\infty} \frac{1}{x(1+x^2)}dx \\ = \frac{\pi}{4}+\lim_{A \to +\infty} \int_1^A \left( \frac{1}{x}-\frac{x}{1+x^2} \right)dx=\frac{\pi}{4} + \lim_{A \to +\infty} \left( \ln x-\frac{1}{2}\ln(1+x^2) \right) \bigg|_1^{A} \\ =\frac{\pi}{4}+\lim_{A \to +\infty} \left(\ln A-\frac{\ln(1+A^2)}{2}+\frac{1}{2}\ln 2\right) \\ =\frac{\pi}{4}+\frac{\ln 2}{2}+\lim_{A \to +\infty} \ln \frac{A}{\sqrt{1+A^2}}=\frac{\pi}{4}+\frac{\ln2 }{2} \]

3

\[\int_0^{+\infty} \frac{xe^{-x}}{(1+e^{-x})^2}dx \xlongequal{t=e^{x}}\int_1^{+\infty} \frac{\ln t}{t(1+\frac{1}{t})^2} \frac{dt}{t}=\int_1^{+\infty} \frac{\ln t dt}{(1+t)^2} \\ =-\int_1^{+\infty} \ln t d\frac{1}{1+t}=\lim_{A \to +\infty} \left(-\frac{\ln t}{1+t}\bigg|_1^{A}+\int_1^{A}\frac{dt}{t(1+t)} \right) \\ =\lim_{A \to +\infty} \left( -\frac{\ln A}{1+A}+\ln \frac{t}{1+t} \bigg|_1^{A} \right) \\ =\lim_{A \to +\infty}\left(-\frac{\ln A}{1+A}+\ln \frac{A}{1+A}-\ln \frac{1}{2}\right)=\ln 2 \]

4

\[\int_1^{+\infty} \frac{dx}{\sqrt{e^{2x}-1}} \xlongequal{t=e^x}\int_e^{+\infty}\frac{dt}{t\sqrt{t^2-1}}=\int_e^{+\infty} \frac{d\sqrt{x^2-1}}{1+(\sqrt{x^2-1})^2} \\ =\arctan\sqrt{x^2-1}\bigg|_e^{+\infty}=\frac{\pi}{2}-\arctan \sqrt{e^2-1} \]

5

\[I=\int_a^{b}\frac{dx}{\sqrt{(x-a)(x-b)}}=\int_a^b \frac{dx}{\sqrt{\left(x-\frac{a+b}{2}\right)^2+ab-\frac{(a+b)^2}{4}}} \\ \xlongequal{u=x-\frac{a+b}{2},k=ab-\frac{(a+b)^2}{4}}\int_{\frac{a-b}{2}}^{\frac{b-a}{2}} \frac{du}{\sqrt{u^2+k}} \\ =\ln \left| u+\sqrt{u^2+k} \right|\bigg|_\frac{a-b}{2}^\frac{b-a}{2} \\ =\ln \left| \frac{\frac{b-a}{2}}{\frac{a-b}{2}} \right|=0 \]

6

\[\int_0^{+\infty} \frac{dx}{1+x^3} =\int_0^{+\infty} \frac{dx}{(x+1)(x^2-x+1)} \\ =\frac{1}{3}\int_0^{+\infty}\left( \frac{1}{x+1}+\frac{-x+2}{x^2-x+1}\right)dx \\ =\frac{1}{3}\lim_{A \to +\infty}\left( \ln(A+1)+\int_0^{A} \frac{-x+2}{(x-\frac{1}{2})^2+\frac{3}{4}}dx \right) \\ =\frac{1}{3}\lim_{A \to +\infty}\left( \ln(A+1)+\frac{4}{3}\int_0^{A} \frac{ -\frac{3}{8}d \left(\frac{x-\frac{1}{2}}{\frac{\sqrt{3}}{2}} \right)^2 +\frac{3\sqrt{3}}{4} d \frac{x-\frac{1}{2}}{\frac{\sqrt{3}}{2}} }{\left( \frac{x-\frac{1}{2}}{\frac{\sqrt{3}}{2}} \right)^2+1} \right) \\ =\frac{1}{3}\lim_{A \to +\infty} \left( \ln(A+1) -\frac{1}{2} \left(\ln \left( \left( \frac{2x-1}{\sqrt{3}} \right)^2+1\right) \bigg|_0^A\right) +\sqrt{3}\left(\arctan \frac{2x-1}{\sqrt{3}} \bigg|_0^A \right) \right) \\ =\frac{1}{3}\left( \ln(A+1) -\ln \left(\sqrt{\frac{(2A-1)^2}{3}+1} \cdot \frac{\sqrt{3}}{2}\right) +\sqrt{3}\left(\frac{\pi}{2}+\frac{\pi}{6}\right) \right) \\ =\frac{2\sqrt{3}\pi}{9} \]

7

\[I=\int_0^{+\infty} \frac{dx}{(1+x^2)(1+x^a)}=\int_{+\infty}^0 \frac{d\frac{1}{x}}{(1+\frac{1}{x^2})(1+\frac{1}{x^a})}=\int_0^{+\infty} \frac{x^adx}{(1+x^2)(1+x^a)} \\ 2I=\int_0^{+\infty} \frac{1+x^a}{(1+x^2)(1+x^a)}dx=\int_0^{+\infty} \frac{dx}{1+x^2}=\arctan x\bigg|_0^{+\infty}=\frac{\pi}{2} \\ I=\frac{\pi}{4} \]\[或者考慮三角換元\\ I=\int_0^{+\infty}\frac{dx}{(1+x^2)(1+x^a)} \xlongequal{x=\tan t}\int_0^{\frac{\pi}{2}} \frac{\sec^2 tdt}{\sec^2 t(1+\tan^a t)}=\int_0^{\frac{\pi}{2}}\frac{dt}{1+\tan^at} \\ =\int_{0}^{\frac{\pi}{2}}\frac{d(\frac{\pi}{2}-t)}{1+\cot^at}=\int_{0}^{\frac{\pi}{2}}\frac{\tan^a tdt}{1+\tan^at} \\ 2I=\int_0^{\frac{\pi}{2}}\frac{1+\tan^at}{1+\tan^a t}dt=\frac{\pi}{2} \Rightarrow I=\frac{\pi}{4} \]

8

\[\int_0^{+\infty} \frac{1+x^2}{1+x^4}dx=\int_0^{+\infty} \frac{d(x-\frac{1}{x})}{(x-\frac{1}{x})^2+2} \\ \xlongequal{u=x-\frac{1}{x}}\int_{-\infty}^{+\infty} \frac{du}{u^2+2}=\frac{1}{\sqrt{2}}\arctan \frac{u}{\sqrt{2}} \bigg|_{-\infty}^{+\infty}=\frac{\pi}{\sqrt{2}} \]

9

\[\int_0^{+\infty} \frac{\sin x\cos x}{x}dx=\frac{1}{2}\int_0^{+\infty} \frac{\sin 2x}{2x}d(2x)=\frac{\pi}{4} \\ \int_0^{+\infty} \frac{\sin^2 x}{x^2}dx=-\int_0^{+\infty}\sin^2 xd\frac{1}{x}=-\frac{\sin^2 x}{x}\bigg|_0^{+\infty}+\int_0^{+\infty}\frac{2\sin x\cos x}{x}dx=0+2\cdot \frac{\pi}{4}=\frac{\pi}{2} \]