おもな乗法的微積分による結果とその解釈

$$ \lim_{h\to 0}\ln\left({f(x+h)\over{f(x)}}\right)^{1\over{h}}=\lim_{h\to 0}{1\over{h}}\ln(f(x+h)-f(x))={d\over{dx}}\ln f(x) $$ ところで$$ \lim_{h\to 0}a^{1\over{h}}=\lim_{h\to 0}e^{\frac{\ln a}{h}}=e^{\lim_{h\to 0}\frac{\ln a}{h}}$$だから、 $$ \lim_{h \to 0}{ \left({f(x+h)\over{f(x)}}\right)^{1\over{h}} }=e^{\frac{d}{dx}\ln f(x)} $$

$$ (fg)^* = e^{\frac{d}{dx}\ln fg}=e^{\frac{d}{dx}\ln f+\frac{d}{dx}\ln g}=f^*g^* $$

$$ \bigg(\frac{f}{g}\bigg)^* = e^{\frac{d}{dx}\ln \frac{f}{g}}=e^{\frac{d}{dx}\ln f-\frac{d}{dx}\ln g}=\frac{f^*}{g^*} $$

$$ \sqrt[dx]{da}=e^{\frac{d}{dx}\ln a}=e^0=1 $$

$$ \sqrt[dx]{dx}=e^{\frac{d}{dx}\ln x}=e^{1\over{x}}=\sqrt[x]{e} $$

$$ \sqrt[dx]{d(ax+b)}=e^{\frac{d}{dx}\ln (ax+b)}=e^{\frac{a}{ax+b}} $$

$$ \sqrt[dx]{d\Bigl(\frac 1x\Bigr)}=e^{\frac{d}{dx}\ln \Bigl(\frac 1x\Bigr)}=e^{-\frac{1}{x}}=\frac{1}{\sqrt[x]{e}} $$

$$ \sqrt[dx]{dx^a}=e^{\frac{d}{dx}\ln (x^a)}=e^{\frac ax} $$

$$ \sqrt[dx]{da^x}=e^{\frac{d}{dx}(x\ln a)}=e^{\ln a}=a $$

$$ \sqrt[dx]{d(\sqrt[x]{a})}=e^{\frac{d}{dx}(\frac{1}{x}\ln a)}=e^{-\frac{1}{x^2}\ln a}=a^{-\frac{1}{x^2}} $$

$$ \sqrt[dx]{d(\log_a x)}=e^{\frac{d}{dx}(\ln( \ln x -\ln a))}=e^{\frac{\frac{1}{x}}{ \ln x -\ln a}}=e^{\frac{1}{x( \ln x -\ln a)}} $$

$$ \sqrt[dx]{dx^x}=e^{\frac{d}{dx}(x\ln x)}=e^{\ln x+1}=ex $$

$$ \sqrt[dx]{dx^x}=e^{\frac{d}{dx}(\ln \Gamma(x))}=e^{\frac{\Gamma(x)\psi(x)}{\Gamma(x)}}=e^{\psi(x)} $$

$$ \sqrt[dx]{d(\sin(ax))}=e^{\frac{d}{dx}(\ln (\sin(ax))}=e^{\frac{a\cos(ax)}{\sin(ax)}}=e^{a\cot(ax)} $$

$$ \int f^*(x)^{dx}=\exp\Biggl[\int\ln f^*(x)dx\Biggl]=\exp\Biggl[\int\ln e^{\frac{d}{dx}\ln f(x)}dx\Biggl]=\exp\Biggl[\int\ln e^{\frac{f'(x)}{f(x)}}dx\Biggl] $$
$$ =\exp\Biggl[\int \frac{f'(x)}{f(x)} dx\Biggl]=\exp[\ln(f(x))+\widetilde{C}]=Cf(x) $$
（また、これが成り立っていることから$\displaystyle\exp\Biggl[\int\ln f(x)dx\Biggl] $が乗法的微分の逆の操作となっていることがわかる。）

$$ \int {(f(x)^k)^{dx}}=e^{\int k\ln f(x)dx}=\bigg(e^{\int \ln f(x)dx}\bigg)^k=\bigg(\int{f(x)^{dx}}\bigg)^k $$

$$ \int {(c^{f(x)})^{dx}}=e^{\int f(x)\ln c~dx}=(e^{\ln c})^{\int f(x)dx}= c^{\int f(x)dx} $$

$$\int a^{dx}=\exp\Biggl[\int\ln a ~dx\Biggl]=e^{x\ln a+\widetilde{C} }=Ca^x$$

$$\int x^{dx}=\exp\Biggl[\int\ln x ~dx\Biggl]=e^{x\ln x-x+\widetilde{C} }=C\frac{x^x}{e^x}$$

$$\int (ax+b)^{dx}=\exp\Biggl[\int\ln (ax+b) ~dx\Biggl]=e^{\bigl(\frac{b}{a}+x\bigr)\ln (ax+b)+\widetilde{C} }=C\frac{(b+a x)^{\frac{b}{a}+x}}{e^x}$$

$$\int \Bigl(\frac 1x\Bigr)^{dx}=\exp\Biggl[\int\ln \Bigl(\frac 1x\Bigr) ~dx\Biggl]=\exp\Biggl[\int-\ln x ~dx\Biggl]=e^{x-x\ln x+\widetilde{C} }=C\frac{e^x}{x^x}$$

$$\int (x^a)^{dx}=\exp\Biggl[\int \ln x^a ~dx\Biggl]=\exp\Biggl[\int a\ln x ~dx\Biggl]=e^{a(x\ln x-x)+\widetilde{C} }=C\frac{x^{ax}}{e^{ax}}$$

$$\int (a^x)^{dx}=\exp\Biggl[\int \ln a^x ~dx\Biggl]=\exp\Biggl[\int x\ln a ~dx\Biggl]=e^{\frac{x^2}{2}\ln a+\widetilde{C} }=Ca^{\frac{x^2}{2}}$$

$$\int (\sqrt[x]{a})^{dx}=\exp\Biggl[\int \ln a^{\frac{1}{x}} ~dx\Biggl]=\exp\Biggl[\int \frac{1}{x}\ln a ~dx\Biggl]=e^{(\ln a\ln|x|)+\widetilde{C} }=Ca^{\ln x}$$

$$\int (\log_a x)^{dx}=\exp\Biggl[\int \ln \log_a x ~dx\Biggl]$$
いま、$$\int \ln \log_a x ~dx=\int \ln \ln x -\ln \ln a ~dx=x\ln \ln x-\int x \cdot \frac{1}{x \ln x}dx-\int\ln \ln a~dx $$
$$=x\ln \ln x-\int \frac{1}{\ln x}dx-\int\ln \ln a~dx=x( \ln\ln x -\ln\ln a )-\operatorname{li}(x)+\widetilde{C}$$
だから、
$$\int (\log_a x)^{dx}=\exp\Biggl[\int \ln \log_a x ~dx\Biggl]=e^{x( \ln\ln x -\ln\ln a )-\operatorname{li}(x)+\widetilde{C}}=C\frac{(\log_a x)^x}{e^{\operatorname{li}(x)}}$$

$$\int (\sqrt[x]{a})^{dx}=\exp\Biggl[\int \ln x^x ~dx\Biggl]=\exp\Biggl[\int x\ln x ~dx\Biggl]=\exp\Biggl[\frac{1}{2}x^2\ln x-\frac{1}{2}\int x ~dx\Biggl]=e^{\frac{1}{4}x^2(2\ln x-1)+\widetilde{C}}=Ce^{\frac{1}{4}x^2(2\ln x-1)}$$

$$\int (\Gamma(x))^{dx}=\exp\Biggl[\int \ln \Gamma(x) ~dx\Biggl]$$
いま、
$$\int\int\psi(u)~dudt=\psi^{(-2)}(x)$$によって
$$\int x\psi(x)~dx= x\ln\Gamma(x)-\int\int\psi(u)~dudt=x\ln\Gamma(x)-\psi^{(-2)}(x)$$
によって
$$\int \ln \Gamma(x) ~dx=x\ln\Gamma(x)-\int x\psi(x)~dx=\psi^{(-2)}(x)+\widetilde{C} $$
によって
$$\int (\Gamma(x))^{dx}=e^{\psi^{(-2)}(x)+\widetilde{C}}=Ce^{\psi^{(-2)}(x)} $$