広島大学 先進理工系科学研究科 情報科学プログラム 2019年8月実施 専門科目I 問題2

Author

samparker

Description

\(D(R) = \{(x,y) \in \mathbb{R}^2 : 1 \leq x^2 + y^2 \leq (R+1)^2, 0 \leq y \leq x\}\), \(R \geq 0\) とする。

(1) 実数 \(\alpha\) に対して，\(G(R) = \iint_{D(R)} x^\alpha y \ dxdy\) を求めよ。

(2) \(\alpha = -3\) とするとき，

\[ \lim_{R \to +0} \frac{G(R) - \frac{R}{2} + \frac{R^2}{4}}{R^3} \]

を求めよ。

(3) \(\lim_{R \to +\infty} G(R)\) が有限な値に収束する実数 \(\alpha\) の範囲を定めよ。

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Let \(D(R) = \{(x,y) \in \mathbb{R}^2 : 1 \leq x^2 + y^2 \leq (R+1)^2, 0 \leq y \leq x\}\), where \(R \geq 0\).

(1) Calculate the integral \(G(R) = \iint_{D(R)} x^\alpha y \, dxdy\) for a real number \(\alpha\).

(2) Find the limit

\[ \lim_{R \to +0} \frac{G(R) - \frac{R}{2} + \frac{R^2}{4}}{R^3} \]

when \(\alpha = -3\).

(3) Determine the range of the real number \(\alpha\) on which the limit \(\lim_{R \to +\infty} G(R)\) converges.

Kai

(1)

Let

\[ x = r \cos \theta, y = r \sin \theta \]

Then the region \(D(R)\) becomes

\[ 1 \leq r \leq R+1, 0 \leq \theta \leq \frac{\pi}{4} \]

The integral becomes

\[ \begin{aligned} \int_0^{\frac{\pi}{4}} \int_{1}^{R+1} r (r \cos \theta)^{\alpha} r \sin \theta \ drd\theta &= \int_0^{\frac{\pi}{4}} \int_{1}^{R+1} r^{\alpha+2} \cos^{\alpha}\theta \sin \theta \ drd\theta \end{aligned} \]

\[ \int_{1}^{R+1} r^{\alpha +2} dr = \frac{r^{\alpha + 3}}{\alpha + 3} \bigg |_{1}^{R+1} = \frac{(R+1)^{\alpha+3} - 1}{\alpha+3} \]

\[ \int_{0}^{\frac{\pi}{4}} \cos^{\alpha}\theta \sin \theta \ d\theta = \frac{-\cos^{\alpha+1} \theta}{\alpha + 1} \bigg |_{0}^{\frac{\pi}{4}} = \frac{1 - 2^{-\frac{\alpha+1}{2}}}{\alpha + 1} \]

Hence

\[ G(R) = \frac{((R+1)^{\alpha+3}-1)(1 - 2^{-\frac{\alpha+1}{2}})}{(\alpha + 3)(\alpha + 1)} \]

(2)

When \(\alpha = -3\), we have

\[ G(R) = \int_0^{\frac{\pi}{4}} \int_{1}^{R+1} r^{-1} \cos^{-3}\theta \sin \theta \ drd\theta = \ln (R+1) \cdot C \]

where \(C = \int_0^{\frac{\pi}{4}} \cos^{-3}\theta \sin \theta \ d\theta = \frac{1}{2}\). Then

\[ \begin{aligned} \lim_{R \to +0} \frac{G(R) - \frac{R}{2} + \frac{R^2}{4}}{R^3} &=\lim_{R \to +0} \frac{\frac{1}{2}\ln (R+1) - \frac{R}{2} + \frac{R^2}{4}}{R^3} \\ &= \lim_{R \to +0} \frac{\frac{1}{2}(R - \frac{R^2}{2} + \frac{R^3}{3} + O(R^4)) - \frac{R}{2} + \frac{R^2}{4}}{R^3} \\ &= \frac{1}{6} \end{aligned} \]

(3)

To determine the range of \(\alpha\) for which \(\lim_{R \to +\infty} G(R)\) converges, observe that as \(R\) becomes large, the integral primarily depends on the behavior of \((R+1)^{\alpha + 3}\).

Since \(\lim_{R \to +\infty} (R+1)^{\alpha + 3}\) converges when \(\alpha \leq -3\), and by (2) we know that \(\lim_{R \to +\infty} G(R)\) diverges when \(\alpha = -3\). Therefore, \(\alpha < -3\).