IS Math-2022-02

题目来源：Problem 2 日期：2024-06-26 题目主题: Math-微积分-定积分

具体题目

Consider the following integral $I_n(\alpha )$ for $\alpha \ge 1$ and $n>0$.

$$I_n(\alpha )={\int }_{\frac{1}{n}}^n\frac{f(\alpha x)-f(x)}{x}dx$$

Assume that a real-valued function $f(x)$ is continuous and differentiable on $x\ge 0$, its derivative is continuous, and ${\lim }_{x\to \infty }f(x)=0$. Answer the following questions.

1. Define $J_n(\alpha )=\frac{d{I}_n(\alpha )}{d\alpha }$. Show that $J_n(\alpha )=\frac{1}{\alpha }\left(f(\alpha n)-f\left(\frac{\alpha }{n}\right)\right)$.

   You can use the fact that the integration and the differentiation commute in this context.

2. Define $I(\alpha )={\lim }_{n\to \infty }{I}_n(\alpha )$. Show that ${\lim }_{n\to \infty }{J}_n(\beta )$ exists for any $\beta \in [1,\alpha ]$ and it uniformly converges on $[1,\alpha ]$, and show that

$$I(\alpha )={\int }_1^{\alpha }\left(\underset{n\to \infty }{\lim }{J}_n(\beta )\right)d\beta .$$

1. Obtain $I(\alpha )$.

2. Calculate the following integral. Note that $p>q>0$.

$${\int }_0^{\infty }\frac{e^{-px}\cos (px)-e^{-qx}\cos (qx)}{x}dx$$

正确解答

1. Showing $J_n(\alpha )=\frac{1}{\alpha }\left(f(\alpha n)-f\left(\frac{\alpha }{n}\right)\right)$

To find $J_n(\alpha )$, we start by differentiating $I_n(\alpha )$ with respect to $\alpha$:

$$J_n(\alpha )=\frac{d{I}_n(\alpha )}{d\alpha }=\frac{d}{d\alpha }{\int }_{\frac{1}{n}}^n\frac{f(\alpha x)-f(x)}{x}dx.$$

Given the conditions, we can interchange the order of differentiation and integration:

$$J_n(\alpha )={\int }_{\frac{1}{n}}^n\frac{\partial }{\partial \alpha }\left(\frac{f(\alpha x)-f(x)}{x}\right)dx.$$

Since $f(x)$ is differentiable:

$$\frac{\partial }{\partial \alpha }\left(\frac{f(\alpha x)-f(x)}{x}\right)=\frac{x{f}^{\prime }(\alpha x)}{x}=f^{\prime }(\alpha x).$$

Therefore:

$$J_n(\alpha )={\int }_{\frac{1}{n}}^n{f}^{\prime }(\alpha x)dx.$$

Performing the substitution $u=\alpha x$:

$$J_n(\alpha )={\int }_{\frac{\alpha }{n}}^{\alpha n}{f}^{\prime }(u)\frac{1}{\alpha }du=\frac{1}{\alpha }{\int }_{\frac{\alpha }{n}}^{\alpha n}{f}^{\prime }(u)du.$$

By the Fundamental Theorem of Calculus:

$$\frac{1}{\alpha }{\int }_{\frac{\alpha }{n}}^{\alpha n}{f}^{\prime }(u)du=\frac{1}{\alpha }\left(f(\alpha n)-f\left(\frac{\alpha }{n}\right)\right).$$

Hence,

$$J_n(\alpha )=\frac{1}{\alpha }\left(f(\alpha n)-f\left(\frac{\alpha }{n}\right)\right).$$

2. Showing the existence of ${\lim }_{n\to \infty }{J}_n(\beta )$ and the expression for $I(\alpha )$

特殊技巧

First, we consider ${\lim }_{n\to \infty }{J}_n(\beta )$:

$$J_n(\beta )=\frac{1}{\beta }\left(f(\beta n)-f\left(\frac{\beta }{n}\right)\right).$$

Given ${\lim }_{x\to \infty }f(x)=0$, as $n\to \infty$, we have $f(\beta n)\to 0$ and $f\left(\frac{\beta }{n}\right)\to f(0)$.

Therefore:

$$\underset{n\to \infty }{\lim }{J}_n(\beta )=\frac{1}{\beta }\left(0-f(0)\right)=-\frac{f(0)}{\beta }.$$

This limit exists and is uniform on $[1,\alpha ]$ since it is independent of $n$.

Next, we evaluate $I(\alpha )$:

$$I(\alpha )=\underset{n\to \infty }{\lim }{I}_n(\alpha ).$$

Apparently $I_n(1)=0$, and since we have the definition $J_n(\alpha )=\frac{d{I}_n(\alpha )}{d\alpha }$:

$$I(\alpha )=\underset{n\to \infty }{\lim }{I}_n(\alpha )=\underset{n\to \infty }{\lim }(I_n(\alpha )-I_n(1))=\underset{n\to \infty }{\lim }{\int }_1^{\alpha }{J}_n(\beta )d\beta .$$

Since ${\lim }_{n\to \infty }{J}_n(\beta )=-\frac{f(0)}{\beta }$, using the fact that integration and limits commute:

$$I(\alpha )={\int }_1^{\alpha }\underset{n\to \infty }{\lim }{J}_n(\beta )d\beta .$$

3. Obtaining $I(\alpha )$

The integral of $\frac{1}{\beta }$ is $\ln (\beta )$:

$$I(\alpha )={\int }_1^{\alpha }\underset{n\to \infty }{\lim }{J}_n(\beta )d\beta ={\int }_1^{\alpha }\left(-\frac{f(0)}{\beta }\right)d\beta =-f(0){\left[\ln (\beta )\right]}_1^{\alpha }=-f(0)(\ln (\alpha )-\ln (1)).$$

Since $\ln (1)=0$, we have:

$$I(\alpha )=-f(0)\ln (\alpha ).$$

4. Calculating the integral ${\int }_0^{\infty }\frac{e^{-px}\cos (px)-e^{-qx}\cos (qx)}{x}dx$

Consider the integral:

$$I={\int }_0^{\infty }\frac{e^{-px}\cos (px)-e^{-qx}\cos (qx)}{x}dx$$

We can use the properties of $I(\alpha )$ derived earlier to evaluate this integral.

Define $F(x)=e^{-x}\cos (x)$. Using the result of $I(\alpha )$ with $f(x)=F(x)$:

$$I_n(\alpha )={\int }_{\frac{1}{n}}^n\frac{e^{-\alpha x}\cos (\alpha x)-e^{-x}\cos (x)}{x}dx$$

As we previously derived:

$$I(\alpha )=-\ln (\alpha )$$

where $f(0)=1$ since $e^0\cos (0)=1$.

We want to calculate:

$${\int }_0^{\infty }\frac{e^{-px}\cos (px)-e^{-qx}\cos (qx)}{x}dx$$

Using the result from $I(\alpha )$:

Set $\alpha =\frac{p}{q}$:

$$I\left(\frac{p}{q}\right)=-\ln \left(\frac{p}{q}\right)$$

Thus, we have:

$${\int }_0^{\infty }\frac{e^{-px}\cos (px)-e^{-qx}\cos (qx)}{x}dx=-\ln \left(\frac{p}{q}\right)$$

Therefore, the final result is:

$$\boxed{{\int }_0^{\infty }\frac{e^{-px}\cos (px)-e^{-qx}\cos (qx)}{x}dx=-\ln \left(\frac{p}{q}\right)}$$

知识点

定积分反常积分极限函数序列

解题技巧和信息

当处理涉及无穷区间上的积分时，可以考虑使用换元法和积分与极限的交换法则。对于指数函数和三角函数的积分，熟悉它们在复数域上的性质可以帮助简化计算。

重点词汇

integral 积分

limit 极限

change of variables 换元法

uniform convergence 一致收敛