高専の数学解説

第5章 5.39

(1)

In=∫0∞xne−x2 dxI_n = \displaystyle\int_0^\infty x^n e^{-x^2}\,dx ただし I0=∫0∞e−x2 dx=π2I_0 = \displaystyle\int_0^\infty e^{-x^2}\,dx = \dfrac{\sqrt{\pi}}{2} In=n−12In−2I_n = \dfrac{n-1}{2}I_{n-2} I0=π2I_0 = \dfrac{\sqrt{\pi}}{2} I1=12I_1 = \dfrac{1}{2}

I2=2−12I0=π4I_2 = \frac{2-1}{2}I_0 = \frac{\sqrt{\pi}}{4}

I3=3−12I1=22⋅12I_3 = \frac{3-1}{2}I_1 = \frac{2}{2}\cdot\frac{1}{2}

I4=4−12I2=32⋅12⋅π2I_4 = \frac{4-1}{2}I_2 = \frac{3}{2}\cdot\frac{1}{2}\cdot\frac{\sqrt{\pi}}{2}

I5=5−12I3=42⋅22⋅12I_5 = \frac{5-1}{2}I_3 = \frac{4}{2}\cdot\frac{2}{2}\cdot\frac{1}{2}

よって n が偶数の時 In=n−12⋅n−32⋅⋯⋅32⋅12⋅π2I_n = \frac{n-1}{2}\cdot\frac{n-3}{2}\cdot\cdots\cdot\frac{3}{2}\cdot\frac{1}{2}\cdot\frac{\sqrt{\pi}}{2}

n が奇数の時 In=n−12⋅n−32⋅⋯⋅42⋅22⋅12I_n = \frac{n-1}{2}\cdot\frac{n-3}{2}\cdot\cdots\cdot\frac{4}{2}\cdot\frac{2}{2}\cdot\frac{1}{2}

(2)

In=∫01(log⁡x)n dxI_n = \displaystyle\int_0^1 (\log x)^n\,dx In=−nIn−1I_n = -nI_{n-1} I0=1I_0 = 1

I1=−1I_1 = -1 I0=−1I_0 = -1

I2=−2⋅I1=−2⋅(−1)⋅I0=−2⋅(−1)⋅1=2I_2 = -2\cdot I_1 = -2\cdot(-1)\cdot I_0 = -2\cdot(-1)\cdot 1 = 2

I3=−3⋅I2=−6I_3 = -3\cdot I_2 = -6

よって

In=−n⋅{−(n−1)}⋅{−(n−2)}⋅⋯⋅(−3)⋅(−2)⋅(−1)⋅1=(−1)n⋅n!I_n = -n\cdot\{-(n-1)\}\cdot\{-(n-2)\}\cdot\cdots\cdot(-3)\cdot(-2)\cdot(-1)\cdot 1 = (-1)^n\cdot n!

解説: r31bn1z