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

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)

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

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