费曼积分法

Leibniz Integral Rule

\(\newcommand{\d}{\text{ d}}\)假设二元函数\(f(x,y)\)连续，可以证明\(I(y)=\displaystyle\int_a^{b}f(x,y)\d x\)连续。连续意味着：\(\lim\limits_{y\to y_0}I(y)=I(y_0)\)，所以\(\lim\limits_{y\to y_0}I(y)=\displaystyle\int_a^b f(x,y_0)\d x=\displaystyle\int_a^b \left[\lim\limits_{y\to y_0}f(x,y)\right]\d x\)。

由此，我们可以证明\(\dfrac{\d}{\d y}I(y)=\displaystyle\int_a^b \left[\dfrac{\part}{\part y}f(x,y)\right]\d x\)：对于任意的\(\Delta y>0\)，我们有\(\dfrac{I(y+\Delta y)-I(y)}{\Delta y}=\dfrac{\displaystyle\int_a^{b}f(x,y+\Delta y)\d x-\displaystyle\int_a^{b}f(x,y)\d x}{\Delta y}\)\(=\dfrac{\displaystyle\int_a^{b}f(x,y+\Delta y)-f(x,y)\d x}{\Delta y}\)。根据一元函数的微分中值定理，存在\(0<\xi<1\)使得\(f(x,y+\Delta y)-f(x,y)=\Delta y \cdot \dfrac{\part}{\part y}f(x,y+\xi\cdot\Delta y)\)。于是\(\dfrac{I(y+\Delta y)-I(y)}{\Delta y}=\displaystyle\int_a^{b}\dfrac{\part}{\part y}f(x,y+\xi\cdot \Delta y)\d x\)。两边同时令\(\Delta y\to 0\)，得到\(\lim\limits_{\Delta y\to 0}\dfrac{I(y+\Delta y)-I(y)}{\Delta y}=\dfrac{\d}{\d y}I(y)=\lim\limits_{\Delta y\to 0}\displaystyle\int_a^{b}\dfrac{\part}{\part y}f(x,y+\xi\cdot \Delta y)\d x=\)\(\displaystyle\int_a^b \left[\dfrac{\part}{\part y}f(x,y)\right]\d x\)。

可见，对于连续函数，积分外的求导符号可以移到积分内：\(\dfrac{\d}{\d y}\displaystyle\int_a^b f(x,y)\d x=\displaystyle\int_a^b \left[\dfrac{\part}{\part y}f(x,y)\right]\d x\)。

Feynman's Trick

上面的定理可以帮助我们求解积分。

当求解定积分\(\displaystyle\int_0^1 \dfrac{x^2-1}{\ln x}\d x\)时，我们首先可以把它看作含参变量的定积分\(I(t)=\displaystyle\int_0^1 \dfrac{x^t-1}{\ln x}\d x\)，那么只需解出\(I(2)\)。两边同时对\(t\)求导，得\(I'(t)=\dfrac{\d}{\d t}\displaystyle\int_0^1 \dfrac{x^t-1}{\ln x}\d x\)，由Leibniz Integral Rule，右式等于\(\displaystyle\int_0^1 \dfrac{\part}{\part t}\dfrac{x^t-1}{\ln x}\d x=\displaystyle\int_0^1 \dfrac{x^{t}\ln x}{\ln x}\d x=\displaystyle\int_0^1 x^{t}\d x=\dfrac{1}{t+1}\)。所以\(I(t)=\displaystyle\int \dfrac{1}{t+1}\d t=\ln(t+1)+C\)。当\(t=0\)时，积分恒为0，因此\(C=0\)。代入\(t=2\)，得到原积分的值为\(\ln 3\)。