東京大学 情報理工学研究科 2023年8月実施 数学 第2問

标签：

• Tokyo-University
• Gamma-Function-Log-Derivatives-and-Log-Moments
• Log-Moments-of-Rayleigh-Distribution

Author

zephyr

Description

Consider a function $f(s)$ defined by the following integral for positive real numbers $s$.

$$f(s) = \int_0^\infty t^{s-1} \exp(-t) \,\mathrm{d}t.$$

Answer the following questions. You may answer without showing that the above integral converges.

(1) Find the value of $f(1)$.

(2) The inequality $\exp(t) > \frac{t^n}{n!}$ holds for any positive real number $t$ and non-negative integer $n$.

• (a) For positive real numbers $s$, show the following inequality.

$$\int_0^1 t^{s-1} \exp(-t) \,\mathrm{d}t < \frac{1}{s}.$$

• (b) When $n > s > 0$, show that the following inequality holds for any real number $c$ that satisfies $c > 1$.

$$\int_1^c t^{s-1} \exp(-t) \,\mathrm{d}t < \frac{n!}{n-s}.$$

(3) When the second-order derivative of $f(s)$ is expressed as

$$\frac{\mathrm{d}^2 f(s)}{\mathrm{d}s^2} = \int_0^\infty g(t, s) \exp(-t) \,\mathrm{d}t,$$

find a function $g(t, s)$. You may answer without showing that the order of differentiation and integration can be exchanged.

(4) Find the value of $D$ defined as

$$D = \int_0^\infty (\log t)^2 \exp(-t) \,\mathrm{d}t - \left(\int_0^\infty (\log t) \exp(-t) \,\mathrm{d}t \right)^2.$$

Here, you may use the fact that the following relation holds.

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

(5) Define a function $p(r)$ for positive real numbers $r$ and $\alpha$ as

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

Find the value of $S$ defined as

$$S = \int_0^\infty (\log r)^2 p(r) \,\mathrm{d}r - \left(\int_0^\infty (\log r) p(r) \,\mathrm{d}r\right)^2.$$

题目描述

对正实数 $s$，定义

$$f(s)=\int_0^\infty t^{s-1}e^{-t}\,\mathrm dt.$$

无需证明积分收敛。回答下列问题。

（1）求 $f(1)$。

（2）已知对 $t>0$ 和非负整数 $n$， $e^t>t^n/n!$。

• （a）对 $s>0$，证明
  $$\int_0^1t^{s-1}e^{-t}\,\mathrm dt<\frac1s.$$
• （b）当 $n>s>0$ 时，证明对任意 $c>1$，
  $$\int_1^ct^{s-1}e^{-t}\,\mathrm dt <\frac{n!}{n-s}.$$

（3）若

$$f''(s)=\int_0^\infty g(t,s)e^{-t}\,\mathrm dt,$$

求 $g(t,s)$；无需证明可交换微分和积分。

（4）求

$$D=\int_0^\infty(\log t)^2e^{-t}\,\mathrm dt -\left(\int_0^\infty(\log t)e^{-t}\,\mathrm dt\right)^2.$$

可以使用

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

（5）对 $r,\alpha>0$ 定义

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

求

$$S=\int_0^\infty(\log r)^2p(r)\,\mathrm dr -\left(\int_0^\infty(\log r)p(r)\,\mathrm dr\right)^2.$$

考点

• Gamma 函数：识别积分定义并通过指数函数下界控制端点积分。
• 参数积分的对数矩：对 $s$ 求导，使 $\log t$ 与 $(\log t)^2$ 出现在被积函数中。
• 对数 Gamma 的二阶导数：用 $(\log f)''=f''/f-(f'/f)^2$ 计算指数分布下 $\log T$ 的方差。
• Rayleigh 分布的对数矩：以 $t=r^2/(2\alpha)$ 变换为指数分布，并分析缩放常数对方差的影响。

Kai

(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, $f(1) = 1$.

(2)

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:

$$\int_1^c t^{s-1} \exp(-t) \,\mathrm{d}t < \frac{n!}{n-s}.$$

(3)

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

(4)

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:

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

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:

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

We know that:

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

At $s = 1$, $f(1) = 1$, $f'(1)$ is the first moment (which is $\int_0^\infty \log t \exp(-t) \,\mathrm{d}t$), and $f''(1)$ is the second moment (which is $\int_0^\infty \log^2 t \exp(-t) \,\mathrm{d}t$).

We can express:

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

Given:

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

Thus:

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

(5)

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 $S$.

Knowledge

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