Euler积分

含参积分

Γ(s)=∫0∞xs−1e−xdx,s>0\Gamma(s) = \int_{0}^\infty x^{s-1}e^{-x}\mathrm dx,\quad s > 0

B(p,q)=∫01xp−1(1−x)q−1dx,p>0,q>0\Beta(p,q) = \int_{0}^1 x^{p-1}(1-x)^{q-1}\mathrm d x,\quad p > 0,q > 0

在应用中经常出现, 它们统称为Euler积分. 其中前者被称为Gamma函数, 后者称为 Beta函数.

Gamma函数

  • Γ(s)\Gamma(s)的定义域为s>0s>0

Gamma函数Γ(s)\Gamma(s)可以写成如下两个积分的和:

Γ(s)=∫01xs−1e−xdx+∫1∞xs−1e−xdx=Φ(s)+Ψ(s)\Gamma(s) = \int_0^1 x^{s-1}e^{-x}\mathrm dx + \int_1^\infty x^{s-1}e^{-x}\mathrm dx = \Phi(s) + \Psi(s) \notag

其中Φ(s)\Phi(s)当s≥1s\geq1时是正常积分, 当0<s<10<s<1时是收敛的无界反常积分(瑕积分), 其收敛性可以用柯西判别法推得; Ψ(s)\Psi(s)当s>0s>0时是收敛的无穷限反常积分(无穷积分), 收敛性用柯西判别法推得. 所以含参积分Γ(s)\Gamma(s)在s>0s>0时是收敛的, 即Gamma函数的定义域为s>0s>0.

  • Γ(s)\Gamma(s)在定义域s>0s>0上连续可导

先给出结果, 证明待定(需要说明一致收敛性)

Γ′(s)=∂∂s∫0∞xs−1e−xdx=一致收敛∫0∞∂∂s(xs−1e−x)dx=∫0∞xs−1e−xln⁡xdx\begin{split} \Gamma'(s) &= \frac{\partial}{\partial s} \int_{0}^\infty x^{s-1}e^{-x} \mathrm dx \\ \overset{一致收敛}{=}& \int_{0}^\infty \frac{\partial}{\partial s}\left( x^{s-1}e^{-x} \right) \mathrm dx = \int_{0}^\infty x^{s-1}e^{-x}\ln{x}\mathrm dx \end{split}

Note: 需要证明 ∫0∞xs−1e−xln⁡xdx\int_{0}^\infty x^{s-1}e^{-x}\ln{x}\mathrm dx 的在 s>0s>0 上的一致收敛性, 而 Γ(s)\Gamma(s) 本身的收敛性已经在定义域中讨论了.

任意阶导数

Γ(n)(s)=∫0∞xs−1e−x(ln⁡x)ndx,s>0\Gamma^{(n)}(s) = \int_{0}^\infty x^{s-1}e^{-x}(\ln{x})^n\mathrm dx, \quad s>0

  • 递推公式Γ(s+1)=sΓ(s)\Gamma{(s+1)} = s\Gamma{(s)}

对下述积分应用分部积分法, 有

∫0Axse−xdx=−xse−x∣0A+s∫0Axs−1e−xdx=−Ase−A+s∫0Axs−1e−xdx\int_{0}^A x^{s}e^{-x} \mathrm dx = -x^s e^{-x}\bigg\vert_0^A + s\int_{0}^A x^{s-1}e^{-x} \mathrm dx = -A^se^{-A} + s\int_0^A x^{s-1}e^{-x}\mathrm dx \notag

令A→+∞A\rightarrow +\infty就得到Gamma函数的递推公式:

Γ(s+1)=sΓ(s)\Gamma(s + 1) = s\Gamma(s)

Note: lim⁡A→+∞−Ase−A=L′Hospital0\underset{A\rightarrow +\infty}{\lim} -A^se^{-A} \overset{L'Hospital}{=} 0

  • Γ(1)\Gamma(1)和Γ(12)\Gamma(\frac{1}{2})

Γ(1)=∫0∞e−xdx=−e−x∣0∞=1\Gamma(1) = \int_{0}^\infty e^{-x}\mathrm dx = -e^{-x}\bigg|_0^\infty = 1

Γ(12)=∫0∞x−12e−xdx=令x=t22∫0∞e−t2dt=∫−∞∞e−t2dt=π\Gamma(\frac{1}{2}) = \int_{0}^\infty x^{-\frac{1}{2}}e^{-x}\mathrm dx \overset{令x=t^2}{=} 2\int_{0}^{\infty}e^{-t^2}\mathrm dt = \int_{-\infty}^{\infty}e^{-t^2}\mathrm dt = \sqrt{\pi}

Note: 令 I=∫−∞∞e−x2dxI = \int_{-\infty}^{\infty}e^{-x^2}\mathrm dx , 有

I2=∫−∞∞e−x2dx⋅∫−∞∞e−y2dx=∫−∞∞∫−∞∞e−(x2+y2)dxdy=极坐标变换∫02πdθ∫0∞e−r2rdr=πI^2 = \int_{-\infty}^{\infty}e^{-x^2}\mathrm dx \cdot \int_{-\infty}^{\infty}e^{-y^2}\mathrm dx = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-(x^2+y^2)}\mathrm dx \mathrm dy \overset{极坐标变换}{=} \int_0^{2\pi}\mathrm d\theta\int_0^\infty e^{-r^2}r\mathrm dr = \pi \notag

  • Γ(n)\Gamma(n)和Γ(n+12),n=1,2,⋯\Gamma(n+\frac{1}{2}),\quad n=1,2,\cdots

特别的, 当 ss 取 nn 和 n+12n+\frac{1}{2} 时有

Γ(n)=(n−1)×⋯×2×1×Γ(1)=(n−1)!\Gamma(n) = (n-1)\times\cdots\times2\times1\times\Gamma(1) = (n-1)!

Γ(n+12)=(n−12)×(n−32)×⋯×12×Γ(12)=(2n−1)!!2nπ\Gamma{(n+\frac{1}{2})} = (n-\frac{1}{2})\times(n-\frac{3}{2})\times\cdots\times\frac{1}{2}\times\Gamma(\frac{1}{2}) = \frac{(2n-1)!!}{2^n}\sqrt{\pi}

Note: (2n−1)!!=(2n−1)×(2n−3)×⋯×3×1(2n-1)!! = (2n-1)\times(2n-3)\times\cdots\times3\times1.