UA MATH523A 实分析3 积分理论例题 Fubini定理计算重积分的极限
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
lim?k→∞∫0∞k3/2e?kx∫0xsin?tt3/2dtdx\lim_{k \to \infty}\int_0^{\infty} k^{3/2}e^{-kx}\int_0^x\frac{\sin t}{t^{3/2}}dtdxk→∞lim?∫0∞?k3/2e?kx∫0x?t3/2sint?dtdx
解
積分部分難點(diǎn)在于它積不出來，因?yàn)?span id="ozvdkddzhkzd" class="katex--inline">sin?tt3/2\frac{\sin t}{t^{3/2}}t3/2sint?的積分找不到，所以我們用Taylor級數(shù)+Gamma函數(shù)的技巧來做。首先寫出sin?tt3/2\frac{\sin t}{t^{3/2}}t3/2sint?的Taylor級數(shù)，
sin?tt3/2=∑n=0∞(?1)nt2n?12(2n+1)!\frac{\sin t}{t^{3/2}}=\sum_{n=0}^{\infty}\frac{(-1)^nt^{2n-\frac{1}{2}}}{(2n+1)!}t3/2sint?=n=0∑∞?(2n+1)!(?1)nt2n?21??
因此
∫0xsin?tt3/2dt=∫0x∑n=0∞(?1)nt2n?12(2n+1)!dt=∑n=0∞(?1)n(2n+1)!∫0xt2n?1dt=∑n=0∞(?1)nx2n+1n(2n+12)(2n+1)!\int_0^x\frac{\sin t}{t^{3/2}}dt = \int_0^x\sum_{n=0}^{\infty}\frac{(-1)^nt^{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)^nx^{2n+\frac{1}{n}}}{(2n+\frac{1}{2})(2n+1)!}∫0x?t3/2sint?dt=∫0x?n=0∑∞?(2n+1)!(?1)nt2n?21??dt=n=0∑∞?(2n+1)!(?1)n?∫0x?t2n?1dt=n=0∑∞?(2n+21?)(2n+1)!(?1)nx2n+n1??
下面我們考慮累次積分
∫0∞k3/2e?kx∫0xsin?tt3/2dtdx=∫0∞k3/2e?kx∑n=0∞(?1)nx2n+12(2n+12)(2n+1)!dx=∑n=0∞(?1)nk3/2(2n+12)(2n+1)!∫0∞e?kxx2n+12dx\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)^nx^{2n+\frac{1}{2}}}{(2n+\frac{1}{2})(2n+1)!}dx \\ =\sum_{n=0}^{\infty} \frac{(-1)^nk^{3/2}}{(2n+\frac{1}{2})(2n+1)!}\int_0^{\infty} e^{-kx}x^{2n+\frac{1}{2}}dx∫0∞?k3/2e?kx∫0x?t3/2sint?dtdx=∫0∞?k3/2e?kxn=0∑∞?(2n+21?)(2n+1)!(?1)nx2n+21??dx=n=0∑∞?(2n+21?)(2n+1)!(?1)nk3/2?∫0∞?e?kxx2n+21?dx
這里的積分∫0∞e?kxx2n+12dx\int_0^{\infty} e^{-kx}x^{2n+\frac{1}{2}}dx∫0∞?e?kxx2n+21?dx很明顯就是gamma函數(shù)的構(gòu)造，
∫0∞e?kxx2n+12dx=Γ(2n+32)k2n+32\int_0^{\infty} e^{-kx}x^{2n+\frac{1}{2}}dx=\frac{\Gamma(2n+\frac{3}{2})}{k^{2n+\frac{3}{2}}}∫0∞?e?kxx2n+21?dx=k2n+23?Γ(2n+23?)?
所以
∫0∞k3/2e?kx∫0xsin?tt3/2dtdx=∑n=0∞(?1)nk3/2(2n+12)(2n+1)!Γ(2n+32)k2n+32=Γ(12)+∑n=1∞(?1)nk3/2(2n+12)(2n+1)!Γ(2n+32)k2n+32≤Γ(12)+O(k?2)\int_0^{\infty} k^{3/2}e^{-kx}\int_0^x\frac{\sin t}{t^{3/2}}dtdx \\ =\sum_{n=0}^{\infty} \frac{(-1)^nk^{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)^nk^{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})∫0∞?k3/2e?kx∫0x?t3/2sint?dtdx=n=0∑∞?(2n+21?)(2n+1)!(?1)nk3/2?k2n+23?Γ(2n+23?)?=Γ(21?)+n=1∑∞?(2n+21?)(2n+1)!(?1)nk3/2?k2n+23?Γ(2n+23?)?≤Γ(21?)+O(k?2)
因此k→∞k \to \inftyk→∞,
∫0∞k3/2e?kx∫0xsin?tt3/2dtdx→Γ(1/2)=π\(zhòng)int_0^{\infty} k^{3/2}e^{-kx}\int_0^x\frac{\sin t}{t^{3/2}}dtdx \to \Gamma(1/2)=\sqrt{\pi}∫0∞?k3/2e?kx∫0x?t3/2sint?dtdx→Γ(1/2)=π?
評注
∑n=1∞(?1)nk3/2(2n+12)(2n+1)!Γ(2n+32)k2n+32=∑n=1∞(?1)nΓ(2n+12)(2n+1)!1k2n≤∑n=1∞1k2n=1k2?1=O(k?2)\sum_{n=1}^{\infty} \frac{(-1)^nk^{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})n=1∑∞?(2n+21?)(2n+1)!(?1)nk3/2?k2n+23?Γ(2n+23?)?=n=1∑∞?(2n+1)!(?1)nΓ(2n+21?)?k2n1?≤n=1∑∞?k2n1?=k2?11?=O(k?2)
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