integral from 0 to 1 of 5(e^{1/x})/(x^3)

解

\int_{0}^{1} 5 \frac{e ^{\frac{1}{x}}}{x ^{3}} d x

は発散する

解答ステップ

\int_{0}^{1} 5 \cdot \frac{e ^{\frac{1}{x}}}{x ^{3}} d x

定数を除く:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 5 \cdot \int_{0}^{1} \frac{e ^{\frac{1}{x}}}{x ^{3}} d x

部分積分を適用する

= 5 [\frac{e ^{\frac{1}{x}}}{x ^{2}} - \int - \frac{e ^{\frac{1}{x}} ( 3 x + 1 )}{x ^{4}} d x]_{0}^{1}

\int - \frac{e ^{\frac{1}{x}} ( 3 x + 1 )}{x ^{4}} d x = \frac{3 e ^{\frac{1}{x}}}{x} - e^{\frac{1}{x}} + \frac{e ^{\frac{1}{x}}}{x ^{2}} - \frac{2 e ^{\frac{1}{x}}}{x}

= 5 [\frac{e ^{\frac{1}{x}}}{x ^{2}} - (\frac{3 e ^{\frac{1}{x}}}{x} - e^{\frac{1}{x}} + \frac{e ^{\frac{1}{x}}}{x ^{2}} - \frac{2 e ^{\frac{1}{x}}}{x})]_{0}^{1}

簡素化

5 [\frac{e ^{\frac{1}{x}}}{x ^{2}} - (\frac{3 e ^{\frac{1}{x}}}{x} - e^{\frac{1}{x}} + \frac{e ^{\frac{1}{x}}}{x ^{2}} - \frac{2 e ^{\frac{1}{x}}}{x})]_{0}^{1} : 5 [\frac{- e ^{\frac{1}{x}} + e ^{\frac{1}{x}} x}{x}]_{0}^{1}

= 5 [\frac{- e ^{\frac{1}{x}} + e ^{\frac{1}{x}} x}{x}]_{0}^{1}

限度を計算する:

は発散する

= は発散する