一般地
看到 f′(ξ)+f(ξ)g(ξ)f'(\xi) + f(\xi)g(\xi)f′(ξ)+f(ξ)g(ξ)，应该想到 f(x)e∫g(x)dxf(x)e^{\int g(x) \mathrm{d}x}f(x)e∫g(x)dx，因为
[f(x)e∫g(x)dx]′=[f′(x)+f(x)g(x)‾]e∫g(x)dx\left[f(x)e^{\int g(x) \mathrm{d}x}\right]'=[\underline{f'(x)+f(x)g(x)}]e^{\int g(x) \mathrm{d}x}[f(x)e∫g(x)dx]′=[f′(x)+f(x)g(x)]e∫g(x)dx
具体地
看到 mf(ξ)+nf′(ξ)mf(\xi)+n f'(\xi)mf(ξ)+nf′(ξ)，应该想到 f(x)emnxf(x)e^{ {m\over n}x}f(x)enmx，因为
[f(x)emnx]′=[f′(x)+mnf(x)]emnx=1n[mf(x)+nf′(x)‾]emnx\left[ f(x)e^{ {m \over n}x} \right]'=\left[ f'(x)+{m \over n}f(x) \right]e^{ {m \over n}x} = {1 \over n} [\underline{m f(x) + n f'(x)}]e^{ {m \over n}x }[f(x)enmx]′=[f′(x)+nmf(x)]enmx=n1[mf(x)+nf′(x)]enmx
看到 mf(ξ)+nf′(ξ)mf(\xi)+n f'(\xi)mf(ξ)+nf′(ξ)，应该想到 xmfn(x)x^{m} f^{n} (x)xmfn(x)，因为
[xmfn(x)]′=[mf(x)+nxf′(x)‾]xm−1fn−1(x)[x^{m}f^{n}(x)]'=[\underline{mf(x)+nxf'(x)}]x^{m-1}f^{n-1} (x)[xmfn(x)]′=[mf(x)+nxf′(x)]xm−1fn−1(x)
看到 mf(ξ)−nξf′(ξ)mf(\xi)-n\xi f'(\xi)mf(ξ)−nξf′(ξ)，应该想到 fn(x)xm\displaystyle {f^{n}(x)\over x^{m} }xmfn(x) 或 xmfn(x)\displaystyle {x^{m} \over f^{n} (x)}fn(x)xm，因为
[fn(x)xm]′=fn−1(x)xm+1[nxf′(x)−mf(x)‾]\left[ {f^{n}(x) \over x^{m} } \right]' = {f^{n-1}(x) \over x^{m+1}} [\underline{nxf'(x)-mf(x)}][xmfn(x)]′=xm+1fn−1(x)[nxf′(x)−mf(x)]
[xmfn(x)]′=xm−1fn+1(x)[mf(x)−nxf′(x)‾]\left[{ x^{m} \over f^{n} (x) }\right]' = {x ^ {m-1} \over f^{n+1} (x)}[\underline{mf(x)-nxf'(x)}][fn(x)xm]′=fn+1(x)xm−1[mf(x)−nxf′(x)]
看到 nf′(ξ)f(1−ξ)−mf(ξ)f′(1−ξ)nf'(\xi)f(1-\xi)-mf(\xi)f'(1-\xi)nf′(ξ)f(1−ξ)−mf(ξ)f′(1−ξ)，应该想到 fn(x)fm(1−x)f^{n}(x)f^{m}(1-x)fn(x)fm(1−x)，因为
[nf′(ξ)f(1−ξ)−mf(ξ)f′(1−ξ)]′=[nf′(x)(1−x)−mf(x)f′(1−x)‾]fn−1(x)fm−1(1−x)\left [nf'(\xi)f(1-\xi)-mf(\xi)f'(1-\xi) \right]' = [\underline{ nf'(x)(1-x) - mf(x)f'(1-x) }]f^{n-1}(x)f^{m-1}(1-x)[nf′(ξ)f(1−ξ)−mf(ξ)f′(1−ξ)]′=[nf′(x)(1−x)−mf(x)f′(1−x)]fn−1(x)fm−1(1−x)
看到 mf′(x)g(x)+nf(x)g′(x)\displaystyle{ m f ^{\prime} \left( x \right) g \left( x \right) + n f \left( x \right) g ^{\prime} \left( x \right) }mf′(x)g(x)+nf(x)g′(x)，应该想到 fm(x)gn(x)f^{m}(x)g^{n}(x)fm(x)gn(x)，因为
[fm(x)gn(x)]′=[mf′(x)g(x)+nf(x)g′(x)‾]fm−1(x)gn−1(x)\left[f^{m}(x)g^{n}(x)\right]' = [\underline{mf'(x)g(x)+nf(x)g'(x)}]f^{m-1}(x)g^{n-1}(x)[fm(x)gn(x)]′=[mf′(x)g(x)+nf(x)g′(x)]fm−1(x)gn−1(x)
看到 mf′(x)g(x)−nf(x)g′(x)\displaystyle{ m f ^{\prime} \left( x \right) g \left( x \right) - n f \left( x \right) g ^{\prime} \left( x \right) }mf′(x)g(x)−nf(x)g′(x)，应该想到 fm(x)gn(x)\displaystyle {f^{m}(x) \over g^{n}(x)}gn(x)fm(x)，因为
[fm(x)gn(x)]′=fm−1(x)gn+1(x)[mf′(x)g(x)−nf(x)g′(x)‾]\left[ {f^{m}(x) \over g^{n}(x)} \right]' = {f^{m-1}(x) \over g^{n+1}(x)} [\underline{ mf'(x)g(x)-nf(x)g'(x) }][gn(x)fm(x)]′=gn+1(x)fm−1(x)[mf′(x)g(x)−nf(x)g′(x)]
看到 f(ξ)g′′(ξ)−g(ξ)f′′(ξ)f(\xi)g''(\xi)-g(\xi)f''(\xi)f(ξ)g′′(ξ)−g(ξ)f′′(ξ)，应该想到 f′(x)g(x)−f(x)g′(x)\displaystyle{ f ^{\prime} \left( x \right) g \left( x \right) - f \left( x \right) g ^{\prime} \left( x \right) }f′(x)g(x)−f(x)g′(x)，因为
[f′(x)g(x)−f(x)g′(x)]′=f(x)g′′(x)−g(x)f′′(x)‾[f'(x)g(x)-f(x)g'(x)]' = \underline{ f(x)g''(x)-g(x)f''(x) }[f′(x)g(x)−f(x)g′(x)]′=f(x)g′′(x)−g(x)f′′(x)
伽马函数
Γ(r)=∫0+∞xr−1e−xdx,r>0\displaystyle \Gamma{\left( r\right)}=\int_{0}^{+\infty} x^{r- 1}\text{e}^{- x}\text{d} x, r> 0Γ(r)=∫0+∞xr−1e−xdx,r>0
特别的
Γ(12)=π\displaystyle \Gamma{\left(\frac{1}{2}\right)}=\sqrt{\pi}Γ(21)=π
递归性质有
Γ(x+1)=xΓ(x)\displaystyle \Gamma{\left( x+ 1\right)}= x\Gamma{\left( x\right)}Γ(x+1)=xΓ(x)
它是阶乘的一个延拓
Γ(n+1)=n!\displaystyle \Gamma{\left( n+ 1\right)}={n!}Γ(n+1)=n!
如果令 x=t\displaystyle \sqrt{x}= tx=t，则
Γ(r)=∫0+∞t2r−2e−t2d(t2)=2∫0+∞t2r−1e−t2dt\begin{aligned}\displaystyle \Gamma{\left( r\right)}&=\int_{0}^{+\infty} t^{2 r- 2}\text{e}^{- t^{2} }\text{d}{\left( t^{2}\right)} \\ \displaystyle &= 2\int_{0}^{+\infty} t^{2 r- 1}\text{e}^{- t^{2} }\text{d} t\end{aligned}Γ(r)=∫0+∞t2r−2e−t2d(t2)=2∫0+∞t2r−1e−t2dt
特别的
Γ(1)=2∫0+∞te−t2dt=1Γ(2)=2∫0+∞t3e−t2dt=1\begin{aligned}\displaystyle \Gamma{\left( 1\right)}&= 2\int_{0}^{+\infty} t\text{e}^{- t^{2} }\text{d} t= 1 \\ \displaystyle \Gamma{\left( 2\right)}&= 2\int_{0}^{+\infty} t^{3}\text{e}^{- t^{2} }\text{d} t= 1\end{aligned}Γ(1)Γ(2)=2∫0+∞te−t2dt=1=2∫0+∞t3e−t2dt=1