\int_{0}^{1}\frac{x^a-1}{lnx}dx含参数变量积分

$${\int }_0^1\frac{x^a-1}{lnx}dx=ln(a+1)$$

求导 积分 再积分

$$\begin{gathered} F(a)={\int }_0^1\frac{x^a-1}{lnx}dx \\ \frac{dF(a)}{da}={\int }_0^1\frac{\partial \frac{x^a-1}{lnx}}{\partial a}dx \\ a{\int }_0^1\frac{x^{a-1}}{lnx}dx!!!!!!!!!!!! \\ 含参变量的积分最重要的一点就是不要把指数和对数函数的求导弄混了 \\ ={\int }_0^1{x}^{a}dx=\frac{1}{1+a}{x}^{a+1}dx{|}_0^1=\frac{1}{1+a} \end{gathered}$$
$$\begin{gathered} 则F(a)={\int }_0^1\frac{1}{1+a}da=ln(1+a)+c \\ 当a=0:F(0)=0=ln(0+1)+c,则c=0 \\ 原式=ln(a+1) \end{gathered}$$

二重积分交换积分顺序

$$\begin{gathered} ={\int }_0^1{\int }_0^a{x}^{y}dydx \\ ={\int }_0^a\frac{x^{y+1}{|}_0^1}{(y+1)}dy \\ ={\int }_0^a\frac{1}{(y+1)}dy \\ =ln(y+1){|}_0^a \\ =ln(a+1) \end{gathered}$$