\quad

\quad

求反常积分

∫0∞1(1+x2)(1+xα)dx\int_0^\infty\frac{1}{(1+x^2)(1+x^\alpha)}\text{d}x.

解 \quad 令x=1tx=\frac{1}{t}，则

∫0+∞1(1+x2)(1+xα)dx=∫0+∞tα(1+t2)(1+tα)dt=∫0+∞xα(1+x2)(1+xα)dx=12∫0+∞1+xα(1+x2)(1+xα)dx=12∫0+∞11+x2dx=π4.\begin{aligned} \int_0^{+\infty}\frac{1}{(1+x^2)(1+x^\alpha)}\text{d}x &= \int_0^{+\infty}\frac{t^\alpha}{(1+t^2)(1+t^\alpha)}\text{d}t \\ &= \int_{0}^{+\infty}\frac{x^\alpha}{(1+x^2)(1+x^\alpha)}\text{d}x\\ &= \frac{1}{2}\int_0^{+\infty}\frac{1+x^\alpha}{(1+x^2)(1+x^\alpha)}\text{d}x\\ &= \frac{1}{2}\int_0^{+\infty}\frac{1}{1+x^2}\text{d}x=\frac{\pi}{4}. \end{aligned}

求级数的敛散性

∑n=1∞sin⁡n2+1πn\sum_{n=1}^{\infty}\frac{\sin\sqrt{n^2+1}\pi}{\sqrt{n}}.

解

an=sin⁡[nπ+(n2+1−n)π]n=(−1)nsin⁡πn2+1+nn\begin{aligned} a_n &= \frac{\sin [n\pi+(\sqrt{n^2+1}-n)\pi]}{\sqrt{n}}\\ &=(-1)^n\frac{\sin\frac{\pi}{\sqrt{n^2+1}+n}}{\sqrt{n}} \end{aligned}

根据莱布尼茨判别法，原级数收敛。

当nn趋于无穷时，

∣an∣=sin⁡πn2+1+nn∼πn2+1+nn∼π21n32\begin{aligned} |a_n| &= \frac{\sin\frac{\pi}{\sqrt{n^2+1}+n}}{\sqrt{n}}\sim\frac{\frac{\pi}{\sqrt{n^2+1}+n}}{\sqrt{n}}\sim\frac{\pi}{2}\frac{1}{n^\frac{3}{2}} \end{aligned}

故原级数绝对收敛。

求级数∑n=1∞x2n+3n(n+1)\sum_{n=1}^{\infty}\frac{x^{2n+3}}{n(n+1)}在−1<x<1-1<x<1内的和函数

∑n=1∞x2n+3n(n+1)=∑n=1∞x2n+3n−∑n=1∞x2n+3(n+1)=x3∑n=1∞x2nn−x∑n=1∞x2n+2n+1\begin{aligned} \sum_{n=1}^{\infty}\frac{x^{2n+3}}{n(n+1)} &= \sum_{n=1}^{\infty}\frac{x^{2n+3}}{n} - \sum_{n=1}^{\infty}\frac{x^{2n+3}}{(n+1)}\\ &=x^3\sum_{n=1}^{\infty}\frac{x^{2n}}{n} - x\sum_{n=1}^{\infty}\frac{x^{2n+2}}{n+1} \end{aligned}

由ln⁡(1−x)=−∑n=1∞xnn\ln(1-x)=-\sum_{n=1}^\infty\frac{x^n}{n}得

上式=−x3ln⁡(1−x2)−x(∑n=0∞(x2)n+1n+1−x2)=−x3ln⁡(1−x2)−x(∑n=1∞x2nn−x2)=−x3ln⁡(1−x2)−x(−ln⁡(1−x2)−x2)=(x−x3)ln⁡(1−x2)+x3\begin{aligned} \text{上式} &= -x^3\ln(1-x^2) - x(\sum_{n=0}^{\infty}\frac{(x^{2})^{n+1}}{n+1}-x^2)\\ &= -x^3\ln(1-x^2) - x(\sum_{n=1}^{\infty}\frac{x^{2n}}{n}-x^2)\\ &=-x^3\ln(1-x^2)-x(-\ln(1-x^2)-x^2)\\ &=(x-x^3)\ln(1-x^2)+x^3 \end{aligned}

求级数∑n=0∞n+1n!xn\sum_{n=0}^\infty\frac{n+1}{n!}x^n的和函数

∑n=0∞n+1n!xn=x∑n=1∞xn−1(n−1)!+∑n=0∞xnn!=xex+ex\begin{aligned} \sum_{n=0}^\infty\frac{n+1}{n!}x^n&= x\sum_{n=1}^\infty\frac{x^{n-1}}{(n-1)!}+\sum_{n=0}^\infty\frac{x^n}{n!}\\ &=xe^x+e^x \end{aligned}