乗法的不定和分

• ${\displaystyle \prod _{x}f(x)g(x)=\prod _{x}f(x)\prod _{x}g(x),}$
• ${\displaystyle \prod _{x}f(x)^{a}=\left(\prod _{x}f(x)\right)^{a},}$
• ${\displaystyle \prod _{x}a^{f(x)}=a^{\sum _{x}f(x)}.}$

周期函数 f(x) の周期 T に対して

${\displaystyle \prod _{x}f(Tx)=Cf(Tx)^{x-1}}$

が成り立つ。

基本性質から明らかに、不定和分 ∑_x の言葉を用いて

${\displaystyle \prod _{x}f(x)=\exp \left(\sum _{x}\ln f(x)\right)}$

と書くことができる。ここに exp は自然指数函数、ln は自然対数である。

• ${\displaystyle \prod _{x}a=Ca^{x},}$
• ${\displaystyle \prod _{x}x=C\Gamma (x),}$
• ${\displaystyle \prod _{x}ax=Ca^{x}\Gamma (x),}$
• ${\displaystyle \prod _{x}{\frac {x+1}{x}}=Cx,}$
• ${\displaystyle \prod _{x}{\frac {x+a}{x}}={\frac {C\Gamma (x+a)}{\Gamma (x)}},}$
• ${\displaystyle \prod _{x}x+a=C\Gamma (x+a),}$
• ${\displaystyle \prod _{x}ax+b=Ca^{x}\Gamma \left(x+{\frac {b}{a}}\right),}$
• ${\displaystyle \prod _{x}x^{2}+1=C\Gamma (x-i)\Gamma (x+i),}$
• ${\displaystyle \prod _{x}ax^{2}+bx=Ca^{x}\Gamma (x)\Gamma \left(x+{\frac {b}{a}}\right),}$
• ${\displaystyle \prod _{x}x+{\frac {1}{x}}={\frac {C\Gamma (x-i)\Gamma (x+i)}{\Gamma (x)}},}$
• ${\displaystyle \prod _{x}x^{a}=C\Gamma (x)^{a},}$
• ${\displaystyle \prod _{x}a^{x}=Ca^{{\frac {x}{2}}(x-1)},}$
• ${\displaystyle \prod _{x}a^{\frac {1}{x}}=Ca^{\frac {\Gamma '(x)}{\Gamma (x)}},}$
• ${\displaystyle \prod _{x}x^{x}=Ce^{\zeta ^{\prime }(-1,x)-\zeta ^{\prime }(-1)}=Ce^{\psi ^{(-2)}(x)+{\frac {x^{2}-x}{2}}-{\frac {x}{2}}\ln(2\pi )}=C\operatorname {\mathit {K}} (x).}$

  (ただし、K(x) はK函数である)

• ${\displaystyle \prod _{x}\Gamma (x)={\frac {C\Gamma (x)^{x-1}}{\operatorname {\mathit {K}} (x)}}=C\Gamma (x)^{x-1}e^{{\frac {x}{2}}\ln(2\pi )-{\frac {x^{2}-x}{2}}-\psi ^{(-2)}(x)}=C\operatorname {\mathit {G}} (x),}$

  (ただし、G(x) はバーンズのG函数である)

• ${\displaystyle \prod _{x}\operatorname {sexp} _{a}(x)={\frac {C(\operatorname {sexp} _{a}(x))'}{\operatorname {sexp} _{a}(x)(\ln a)^{x}}}.}$

  (ただし、sexp は超指数函数である)

• ${\displaystyle \prod _{x}\csc x\sin(x+1)=C\sin x,}$
• ${\displaystyle \prod _{x}\sec x\cos(x+1)=C\cos x,}$
• ${\displaystyle \prod _{x}\cot x\tan(x+1)=C\tan x,}$
• ${\displaystyle \prod _{x}\tan x\cot(x+1)=C\cot x.}$