IS Math-2024-02

题目来源：Problem 2 日期：2024-08-13 题目主题：Math-Calculus-Integral Calculus

解题思路

这道题涉及一些高级的积分技巧和不等式证明，具体需要掌握伽马函数、积分换元以及利用不等式进行积分估计。此外，还涉及导数计算和随机变量的方差计算。

Solution

Question 1: Find the value of $f(1)$

We start by evaluating $f(1)$:

$$f(1)={\int }_0^{\infty }{t}^{1-1}\exp (-t)\mathrm{d}t={\int }_0^{\infty }\exp (-t)\mathrm{d}t.$$

This integral is well-known and is the Laplace transform of a constant function $1$. Evaluating the integral:

$${\int }_0^{\infty }\exp (-t)\mathrm{d}t={\left[-\exp (-t)\right]}_0^{\infty }=\left(0-(-1)\right)=1.$$

So, $f(1)=1$.

Question 2: Inequality proofs

Part 1: Show that ${\int }_0^{\infty }{t}^{s-1}\exp (-t)\mathrm{d}t<\frac{1}{s}$ for positive real numbers $s$

To show this inequality, we start by considering the definition of $f(s)$:

$$f(s)={\int }_0^{\infty }{t}^{s-1}\exp (-t)\mathrm{d}t.$$

The problem gives us the inequality $\exp (t)>\frac{t^n}{n!}$ for any positive real number $t$ and non-negative integer $n$. Taking the reciprocal and considering the exponential function in the integrand:

$$\exp (-t)<\frac{n!}{t^n}.$$

Substituting this into the integral, we obtain:

$$f(s)={\int }_0^{\infty }{t}^{s-1}\exp (-t)\mathrm{d}t<{\int }_0^{\infty }{t}^{s-1}\frac{n!}{t^n}\mathrm{d}t=n!{\int }_0^{\infty }{t}^{s-n-1}\mathrm{d}t.$$

The integral ${\int }_0^{\infty }{t}^{s-n-1}\mathrm{d}t$ converges when $s-n>0$. Evaluating this integral:

$${\int }_0^{\infty }{t}^{s-n-1}\mathrm{d}t=\frac{1}{s-n}.$$

Thus, the inequality becomes:

$$f(s)<\frac{n!}{s-n}.$$

Now, by setting $n=1$, we obtain:

$$f(s)<\frac{1!}{s-1}=\frac{1}{s}.$$

Therefore, we have shown that:

$${\int }_0^{\infty }{t}^{s-1}\exp (-t)\mathrm{d}t<\frac{1}{s}.$$

Part 2: Show that ${\int }_1^c{t}^{s-1}\exp (-t)\mathrm{d}t<\frac{n!}{n-s}$ for $n>s>0$ and $c>1$

We are given that $n>s>0$ and $c>1$, and we need to prove the inequality:

$${\int }_1^c{t}^{s-1}\exp (-t)\mathrm{d}t<\frac{n!}{n-s}.$$

Using the same inequality $\exp (-t)<\frac{n!}{t^n}$, we substitute it into the integral:

$${\int }_1^c{t}^{s-1}\exp (-t)\mathrm{d}t<{\int }_1^c{t}^{s-1}\frac{n!}{t^n}\mathrm{d}t=n!{\int }_1^c{t}^{s-n-1}\mathrm{d}t.$$

Next, we evaluate the integral:

$$n!{\int }_1^c{t}^{s-n-1}\mathrm{d}t=n!{\left[\frac{t^{s-n}}{s-n}\right]}_1^c=\frac{n!}{n-s}\left(1-\frac{1}{c^{n-s}}\right).$$

Since $c>1$, the term $\frac{1}{c^{n-s}}$ is less than $1$, which means:

$$1-\frac{1}{c^{n-s}}<1.$$

Thus:

$${\int }_1^c{t}^{s-1}\exp (-t)\mathrm{d}t<\frac{n!}{n-s}.$$

Question 3: Find the function $g(t,s)$

Given:

$$\frac{{\mathrm{d}}^{2}f(s)}{\mathrm{d}{s}^2}={\int }_0^{\infty }g(t,s)\exp (-t)\mathrm{d}t,$$

We first need to compute the second derivative of $f(s)$:

$$f(s)={\int }_0^{\infty }{t}^{s-1}\exp (-t)\mathrm{d}t.$$

First derivative:

$$\frac{\mathrm{d}f(s)}{\mathrm{d}s}={\int }_0^{\infty }\frac{\partial }{\partial s}\left(t^{s-1}\right)\exp (-t)\mathrm{d}t={\int }_0^{\infty }{t}^{s-1}\log (t)\exp (-t)\mathrm{d}t.$$

Second derivative:

$$\frac{{\mathrm{d}}^{2}f(s)}{\mathrm{d}{s}^2}={\int }_0^{\infty }\frac{\partial }{\partial s}\left(t^{s-1}\log (t)\right)\exp (-t)\mathrm{d}t={\int }_0^{\infty }{t}^{s-1}{\log }^2(t)\exp (-t)\mathrm{d}t.$$

Thus, $g(t,s)=t^{s-1}{\log }^2(t)$.

Question 4: Find the value of $D$

We need to find the value of the expression

$$D={\int }_0^{\infty }(\log t{)}^2\exp (-t)\mathrm{d}t-{\left({\int }_0^{\infty }(\log t)\exp (-t)\mathrm{d}t\right)}^2.$$

This problem requires us to calculate two integrals: one involving $(\log t{)}^2$ and another involving $\log t$. We are also given the hint that the following relation holds:

$$\frac{{\mathrm{d}}^2\log f(s)}{\mathrm{d}{s}^2}{|}_{s=1}=\frac{{\pi }^2}{6}.$$

Step 1: Expressing the Second-Order Derivative of $f(s)$

From Question 3, we know that:

$$\frac{{\mathrm{d}}^{2}f(s)}{\mathrm{d}{s}^2}={\int }_0^{\infty }{t}^{s-1}{\log }^2(t)\exp (-t)\mathrm{d}t.$$

Setting $s=1$:

$$\frac{{\mathrm{d}}^{2}f(1)}{\mathrm{d}{s}^2}={\int }_0^{\infty }{t}^{1-1}{\log }^2(t)\exp (-t)\mathrm{d}t={\int }_0^{\infty }{\log }^2(t)\exp (-t)\mathrm{d}t.$$

This integral represents the first term in $D$, which is:

$${\int }_0^{\infty }(\log t{)}^2\exp (-t)\mathrm{d}t.$$

Thus, we have:

$${\int }_0^{\infty }(\log t{)}^2\exp (-t)\mathrm{d}t=\frac{{\mathrm{d}}^{2}f(1)}{\mathrm{d}{s}^2}.$$

Step 2: Calculating the First Integral

The value of the second derivative of the logarithm of $f(s)$ at $s=1$ is given as:

$$\frac{{\mathrm{d}}^2\log f(s)}{\mathrm{d}{s}^2}{|}_{s=1}=\frac{{\pi }^2}{6}.$$

We know that:

$$\frac{{\mathrm{d}}^2\log f(s)}{\mathrm{d}{s}^2}=\frac{f^{\prime \prime }(s)f(s)-{\left(f^{\prime }(s)\right)}^2}{{\left(f(s)\right)}^2}.$$

At $s=1$, $f(1)=1$, $f^{\prime }(1)$ is the first moment (which is ${\int }_0^{\infty }\log t\exp (-t)\mathrm{d}t$), and $f^{\prime \prime }(1)$ is the second moment (which is ${\int }_0^{\infty }{\log }^{2}t\exp (-t)\mathrm{d}t$).

We can express:

$$\frac{{\mathrm{d}}^2\log f(1)}{\mathrm{d}{s}^2}=f^{\prime \prime }(1)-{\left(f^{\prime }(1)\right)}^2.$$

Given:

$$\frac{{\mathrm{d}}^2\log f(s)}{\mathrm{d}{s}^2}{|}_{s=1}=\frac{{\pi }^2}{6},$$

Thus:

$$D=\frac{{\pi }^2}{6}.$$

Question 5: Find the value of $S$

We are asked to find the value of $S$, defined as:

$$S={\int }_0^{\infty }(\log r{)}^{2}p(r)\mathrm{d}r-{\left({\int }_0^{\infty }(\log r)p(r)\mathrm{d}r\right)}^2,$$

where the function $p(r)$ is given by:

$$p(r)=\frac{r}{\alpha }\exp \left(-\frac{r^2}{2\alpha }\right).$$

Step 1: Identify the form of $p(r)$

The function $p(r)$ is a probability density function corresponding to a Rayleigh distribution, with the parameter $\alpha$. The Rayleigh distribution has the form:

$$p(r)=\frac{r}{\alpha }\exp \left(-\frac{r^2}{2\alpha }\right),$$

which is commonly used to describe the distribution of the magnitude of a two-dimensional vector with independent and identically distributed normal components.

Step 2: Calculation of the first moment $\mathbb{E}[\log r]$

We need to compute the expected value of $\log r$ under this distribution, given by:

$$\mathbb{E}[\log r]={\int }_0^{\infty }(\log r)p(r)\mathrm{d}r={\int }_0^{\infty }\log r\cdot \frac{r}{\alpha }\exp \left(-\frac{r^2}{2\alpha }\right)\mathrm{d}r.$$

We perform a substitution to simplify this integral:

Let $u=\frac{r^2}{2\alpha }$, hence $\mathrm{d}u=\frac{r\mathrm{d}r}{\alpha }$.

The integral becomes:

$$\mathbb{E}[\log r]={\int }_0^{\infty }\log \left(\sqrt{2\alpha u}\right)\exp (-u)\mathrm{d}u.$$

This simplifies to:

$$\mathbb{E}[\log r]=\frac{1}{2}\log (2\alpha ){\int }_0^{\infty }\exp (-u)\mathrm{d}u+\frac{1}{2}{\int }_0^{\infty }\log u\exp (-u)\mathrm{d}u.$$

The first integral evaluates to 1 because it is the integral of the exponential distribution. The second integral is a well-known result:

$${\int }_0^{\infty }\log u\exp (-u)\mathrm{d}u=-\gamma ,$$

where $\gamma$ is the Euler-Mascheroni constant. Thus,

$$\mathbb{E}[\log r]=\frac{1}{2}\log (2\alpha )-\frac{\gamma }{2}.$$

Step 3: Calculation of the second moment $\mathbb{E}[(\log r{)}^2]$

Next, we need to compute the second moment:

$$\mathbb{E}[(\log r{)}^2]={\int }_0^{\infty }(\log r{)}^{2}p(r)\mathrm{d}r={\int }_0^{\infty }(\log r{)}^2\frac{r}{\alpha }\exp \left(-\frac{r^2}{2\alpha }\right)\mathrm{d}r.$$

Using the same substitution $u=\frac{r^2}{2\alpha }$:

$$\mathbb{E}[(\log r{)}^2]={\int }_0^{\infty }{\left[\log \left(\sqrt{2\alpha u}\right)\right]}^2\exp (-u)\mathrm{d}u.$$

This expands to:

$$\mathbb{E}[(\log r{)}^2]=\frac{1}{4}{\left[\log (2\alpha )\right]}^2+\frac{1}{2}\log (2\alpha ){\int }_0^{\infty }\log u\exp (-u)\mathrm{d}u+\frac{1}{4}{\int }_0^{\infty }(\log u{)}^2\exp (-u)\mathrm{d}u.$$

Using the known results:

$${\int }_0^{\infty }\log u\exp (-u)\mathrm{d}u=-\gamma ,$$

and

$${\int }_0^{\infty }(\log u{)}^2\exp (-u)\mathrm{d}u={\gamma }^2+\frac{{\pi }^2}{6},$$

we have:

$$\mathbb{E}[(\log r{)}^2]=\frac{1}{4}{\left[\log (2\alpha )\right]}^2-\frac{\gamma }{2}\log (2\alpha )+\frac{1}{4}\left({\gamma }^2+\frac{{\pi }^2}{6}\right).$$

Step 4: Calculate $S$

Finally, $S$ is the variance, which is given by:

$$S=\mathbb{E}[(\log r{)}^2]-{\left(\mathbb{E}[\log r]\right)}^2.$$

Substitute the values:

$$\mathbb{E}[\log r]=\frac{1}{2}\log (2\alpha )-\frac{\gamma }{2},$$

so:

$${\left(\mathbb{E}[\log r]\right)}^2=\frac{1}{4}{\left[\log (2\alpha )\right]}^2-\gamma \log (2\alpha )+\frac{{\gamma }^2}{4}.$$

Subtracting:

$$S=\frac{1}{4}{\left[\log (2\alpha )\right]}^2-\gamma \log (2\alpha )+\frac{1}{4}\left({\gamma }^2+\frac{{\pi }^2}{6}\right)-\left(\frac{1}{4}{\left[\log (2\alpha )\right]}^2-\gamma \log (2\alpha )+\frac{{\gamma }^2}{4}\right).$$

Simplifying:

$$S=\frac{{\pi }^2}{24}.$$

This is the final value of $S$.

知识点

Gamma函数不定积分定积分方差

解题技巧和信息

1. Gamma Function: Recognize that $f(s)$ represents the Gamma function $\Gamma (s)$.
2. Inequality Manipulation: Use known inequalities such as $\exp (t)>\frac{t^n}{n!}$ to estimate integrals.
3. Variance Calculation: The variance of logarithms of exponential and Rayleigh distributed variables often results in expressions involving $\frac{{\pi }^2}{6}$.

重点词汇

• Gamma function 伽马函数
• Inequality 不等式
• Variance 方差
• Logarithm 对数
• Rayleigh distribution 瑞利分布
• Logarithm 对数
• Euler-Mascheroni constant 欧拉-马歇罗尼常数
• Variance 方差