広島大学 先進理工系科学研究科 情報科学プログラム 2017年8月実施 専門科目I 問題2

标签：

• Hiroshima-University
• Local-Extrema
• Improper-Integral
• Gamma-Function
• Limit

Author

samparker

Description

(1) $\alpha > 0$ とする時、関数 $f(x) = x^{\alpha} e^{-x}$ の $[0, \infty)$ における最大値を求めよ。

(2) 以下の広義積分が $x > 0$ において収束することを示せ。

$$\Gamma(x) = \int_0^{\infty} t^{x-1} e^{-t} dt$$

(3) 極限 $\lim_{x \to +0} x \Gamma(x)$ を求めよ。

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(1) For $\alpha > 0$, find the maximum of the function $f(x) = x^{\alpha} e^{-x}$ on $[0, \infty)$.

(2) Show the following improper integral converges on $x > 0$:

$$\Gamma(x) = \int_0^{\infty} t^{x-1} e^{-t} dt$$

(3) Find the limit:

$$\lim_{x \to +0} x \Gamma(x)$$

Kai

(1)

$$\begin{aligned} &f'(x) = 0 \\ &\Rightarrow \alpha x^{\alpha - 1} e^{-x} - x^{\alpha} e^{-x} = 0 \\ &\Rightarrow e^{-x} (\alpha x^{\alpha - 1} - x^{\alpha}) = 0 \\ &\Rightarrow x = \alpha \end{aligned}$$

Note that $f'(x) > 0$ when $x < \alpha$ and $f'(x) < 0$ when $x > \alpha$. Therefore, the maximum of $f(x)$ is $f(\alpha) = \alpha^{\alpha} e^{-\alpha}$.

(2)

Let’s divide the integral in a sum of two terms,

$$\Gamma(x) = \int_0^{1} t^{x-1} e^{-t} dt + \int_1^{\infty} t^{x-1} e^{-t} dt$$

For the first term, since the function $e^{-t}$ is decreasing, it's maximum on the interval $[0, 1]$ is attained at $t = 0$, hence

$$\int_0^{1} t^{x-1} e^{-t} dt < \int_0^1 t^{x-1} dt.$$

But for $x > 0$, this last integral converges to $1/x$.

For the second term, since the exponential grows faster than any polynomial, for every $x$ we can take $N \in \mathbb{N}$ so big that $t \geq N \Rightarrow e^{t/2} > t^{x-1}$ so

$$\begin{aligned} \int_1^{\infty} t^{x-1} e^{-t} dt &= \int_1^{N} t^{x-1} e^{-t} dt + \int_N^{\infty} t^{x-1} e^{-t} dt \\ &< \int_1^{N} t^{x-1} e^{-t} dt + \int_N^{\infty} e^{t/2} e^{-t} dt \\ &= \int_1^{N} t^{x-1} e^{-t} dt + \int_N^{\infty} e^{-t/2} dt \\ &< \infty \end{aligned}$$

which completes the proof.

(3)

$$\begin{aligned} \Gamma(x+1) &= \int_0^{\infty} t^{x} e^{-t} dt \\ &= \left[- e^{-t} t^x \right]_0^{\infty} + x \int_{0}^{\infty} t^{x-1} e^{-t} dt \\ &= x \int_{0}^{\infty} t^{x-1} e^{-t} dt \\ &= x \Gamma(x) \end{aligned}$$

$$\lim_{x \to +0} x \Gamma(x) = \lim_{x \to +0} \Gamma(x+1) = \Gamma(1) = 1$$