Tokyo-University
東京大学 情報理工学研究科 2023年8月実施 数学 第2問
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
\[
S = \int_0^\infty (\log r)^2 p(r) \,\mathrm{d}r - \left(\int_0^\infty (\log r) p(r) \,\mathrm{d}r\right)^2.
\]
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:
\[
p(r) = \frac{r}{\alpha} \exp\left(-\frac{r^2}{2\alpha}\right).
\]
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
Gamma Function : Recognize that \(f(s)\) represents the Gamma function \(\Gamma(s)\) .Inequality Manipulation : Use known inequalities such as \(\exp(t) > \frac{t^n}{n!}\) to estimate integrals.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