integral from 1 to x of log_{1/2}(t)

解答

\int_{1}^{x} lo g_{\frac{1}{2}} (t) d t

- \frac{- x + x ln ( x ) + 1}{ln ( 2 )}

求解步骤

\int_{1}^{x} lo g_{\frac{1}{2}} (t) d t

使用对数计算法则:

lo g_{\frac{1}{a}} (x) = - lo g_{a} (x)

= \int_{1}^{x} - lo g_{2} (t) d t

提出常数:

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

= - \int_{1}^{x} lo g_{2} (t) d t

使用分布积分法

= - [t lo g_{2} (t) - \int \frac{1}{ln ( 2 )} d t]_{1}^{x}

\int \frac{1}{ln ( 2 )} d t = \frac{1}{ln ( 2 )} t

= - [t lo g_{2} (t) - \frac{1}{ln ( 2 )} t]_{1}^{x}

化简

t lo g_{2} (t) - \frac{1}{ln ( 2 )} t : \frac{- t + t ln ( t )}{ln ( 2 )}

= - [\frac{- t + t ln ( t )}{ln ( 2 )}]_{1}^{x}

带入上下限计算后得:

\frac{- x + x ln ( x )}{ln ( 2 )} + \frac{1}{ln ( 2 )}

= - (\frac{- x + x ln ( x )}{ln ( 2 )} + \frac{1}{ln ( 2 )})

化简

= - \frac{- x + x ln ( x ) + 1}{ln ( 2 )}

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