UA MATH523A 實(shí)分析3 積分理論例題 Fubini定理計(jì)算重積分的極限

例 求
$$\underset{k\to \infty }{\lim }{\int }_0^{\infty }{k}^{3/2}{e}^{?kx}{\int }_0^x\frac{\sin t}{t^{3/2}}dtdx$$

解
積分部分難點(diǎn)在于它積不出來，因?yàn)?span id="ze8trgl8bvbq" class="katex--inline">$\frac{\sin t}{t^{3/2}}$的積分找不到，所以我們用Taylor級(jí)數(shù)+Gamma函數(shù)的技巧來做。首先寫出$\frac{\sin t}{t^{3/2}}$的Taylor級(jí)數(shù)，
$$\frac{\sin t}{t^{3/2}}=\sum _{n=0}^{\infty }\frac{(?1{)}^n{t}^{2n?\frac{1}{2}}}{(2n+1)!}$$

因此
$$\begin{gathered} {\int }_0^x\frac{\sin t}{t^{3/2}}dt={\int }_0^x\sum _{n=0}^{\infty }\frac{(?1{)}^n{t}^{2n?\frac{1}{2}}}{(2n+1)!}dt \\ =\sum _{n=0}^{\infty }\frac{(?1{)}^n}{(2n+1)!}{\int }_0^x{t}^{2n?1}dt=\sum _{n=0}^{\infty }\frac{(?1{)}^n{x}^{2n+\frac{1}{n}}}{(2n+\frac{1}{2})(2n+1)!} \end{gathered}$$

下面我們考慮累次積分
$$\begin{gathered} {\int }_0^{\infty }{k}^{3/2}{e}^{?kx}{\int }_0^x\frac{\sin t}{t^{3/2}}dtdx \\ ={\int }_0^{\infty }{k}^{3/2}{e}^{?kx}\sum _{n=0}^{\infty }\frac{(?1{)}^n{x}^{2n+\frac{1}{2}}}{(2n+\frac{1}{2})(2n+1)!}dx \\ =\sum _{n=0}^{\infty }\frac{(?1{)}^n{k}^{3/2}}{(2n+\frac{1}{2})(2n+1)!}{\int }_0^{\infty }{e}^{?kx}{x}^{2n+\frac{1}{2}}dx \end{gathered}$$

這里的積分${\int }_0^{\infty }{e}^{?kx}{x}^{2n+\frac{1}{2}}dx$很明顯就是gamma函數(shù)的構(gòu)造，
$${\int }_0^{\infty }{e}^{?kx}{x}^{2n+\frac{1}{2}}dx=\frac{\Gamma (2n+\frac{3}{2})}{k^{2n+\frac{3}{2}}}$$

所以
$$\begin{gathered} {\int }_0^{\infty }{k}^{3/2}{e}^{?kx}{\int }_0^x\frac{\sin t}{t^{3/2}}dtdx \\ =\sum _{n=0}^{\infty }\frac{(?1{)}^n{k}^{3/2}}{(2n+\frac{1}{2})(2n+1)!}\frac{\Gamma (2n+\frac{3}{2})}{k^{2n+\frac{3}{2}}} \\ =\Gamma (\frac{1}{2})+\sum _{n=1}^{\infty }\frac{(?1{)}^n{k}^{3/2}}{(2n+\frac{1}{2})(2n+1)!}\frac{\Gamma (2n+\frac{3}{2})}{k^{2n+\frac{3}{2}}}\le \Gamma (\frac{1}{2})+O(k^{?2}) \end{gathered}$$

因此$k\to \infty$,
$${\int }_0^{\infty }{k}^{3/2}{e}^{?kx}{\int }_0^x\frac{\sin t}{t^{3/2}}dtdx\to \Gamma (1/2)=\sqrt{\pi }$$

評(píng)注
$$\begin{gathered} \sum _{n=1}^{\infty }\frac{(?1{)}^n{k}^{3/2}}{(2n+\frac{1}{2})(2n+1)!}\frac{\Gamma (2n+\frac{3}{2})}{k^{2n+\frac{3}{2}}}=\sum _{n=1}^{\infty }\frac{(?1{)}^n\Gamma (2n+\frac{1}{2})}{(2n+1)!}\frac{1}{k^{2n}} \\ \le \sum _{n=1}^{\infty }\frac{1}{k^{2n}}=\frac{1}{k^2?1}=O(k^{?2}) \end{gathered}$$

總結(jié)

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