2重積分反復積分領域、対数関数、三角関数(正弦と余弦)、対数関数

学習環境

続解析入門 (原書第2版) (S.ラング(著)、松坂和夫(翻訳)、片山孝次(翻訳)、岩波書店)の第7章(2重積分)、2(反復積分)の練習問題8の解答を求めてみる。

\begin{array}{l} \int_{1}^{2} \int_{0}^{x} \frac{1}{x + y} dy dx \\ = \int_{1}^{2} {[\log (x + y)]}_{0}^{x} dx \\ = \int_{1}^{2} (\log (2 x) - \log x) dx \\ = \int_{1}^{2} \log 2 dx \\ = {[x]}_{1}^{2} \log 2 \\ = \log 2 \end{array}

\begin{array}{l} x^{2} - y^{2} \geq 0 \\ y^{2} \leq x^{2} \\ - \sqrt{x^{2}} \leq y^{2} \leq \sqrt{x^{2}} \end{array}

求める積分。

\begin{array}{l} 2 \int_{0}^{1} \int_{0}^{\sqrt{x^{2}}} (x^{2} - y^{2}) dy dx \\ = 2 \int_{0}^{1} {[x^{2} y - \frac{1}{3} y^{3}]}_{0}^{\sqrt{x^{2}}} dx \\ = 2 \int_{0}^{1} (x^{3} - \frac{1}{3} x^{3}) dx \\ = \frac{4}{3} \int_{0}^{1} x^{3} dx \\ = \frac{1}{3} {[x^{4}]}_{0}^{1} \\ = \frac{1}{3} \end{array}

\begin{array}{l} \int_{0}^{π} \int_{0}^{1} x \sin (x y) dy dx \\ = - \int_{0}^{π} {[\cos (x y)]}_{0}^{1} dx \\ = \int_{0}^{π} (1 - \cos x) dx \\ = {[x - \sin x]}_{0}^{π} \\ = π \end{array}

\begin{array}{l} \int_{- 1}^{0} \int_{- x}^{1} (x^{2} - y^{2}) dy dx + \int_{0}^{1} \int_{x}^{1} (x^{2} - y^{2}) dy dx \\ = \int_{- 1}^{0} {[x^{2} y - \frac{1}{3} y^{3}]}_{- x}^{1} dx + \int_{0}^{1} {[x^{2} y - \frac{1}{3} y^{3}]}_{x}^{1} dx \\ = \int_{- 1}^{0} (x^{2} - \frac{1}{3} + x^{3} - \frac{1}{3} x^{3}) dx + \int_{0}^{1} (x^{2} - \frac{1}{3} - x^{3} + \frac{1}{3} x^{3}) dx \\ = \int_{- 1}^{0} (- \frac{1}{3} + x^{2} + \frac{2}{3} x^{3}) dx + \int_{0}^{1} (- \frac{1}{3} + x^{2} - \frac{2}{3} x^{3}) dx \\ = {[- \frac{1}{3} x + \frac{1}{3} x^{3} + \frac{1}{6} x^{4}]}_{- 1}^{0} + {[- \frac{1}{3} x + \frac{1}{3} x^{3} - \frac{1}{6} x^{4}]}_{0}^{1} \\ = (- \frac{1}{3} + \frac{1}{3} - \frac{1}{6}) - \frac{1}{6} \\ = - \frac{1}{3} \end{array}

\begin{array}{l} \int_{0}^{1} \int_{0}^{1} \frac{1}{x + y + 1} dy dx \\ = \int_{0}^{1} {[\log (x + y + 1)]}_{0}^{1} dx \\ = \int_{0}^{1} (\log (x + 2) - \log (x + 1)) d x \end{array}

\begin{array}{l} \int \log (x + 2) dx \\ = x \log (x + 2) - \int \frac{x}{x + 2} dx \\ = x \log (x + 2) - \int (1 - \frac{2}{x + 2}) dx \\ = x \log (x + 2) - x + 2 \log (x + 2) \\ \int \log (x + 1) dx \\ = x \log (x + 1) - x + \log (x + 1) \end{array}

\begin{array}{l} {[x \log (x + 2) - x \log (x + 1) + 2 \log (x + 2) - \log (x + 1)]}_{0}^{1} \\ = \log 3 - \log 2 + 2 \log 3 - \log 2 - 2 \log 2 \\ = 3 \log 3 - 4 \log 2 \end{array}