おもな乗法的微積分による結果とその解釈

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$$\int a^{dx}=\exp [\int \ln adx]=e^{x\ln a+\tilde{C}}=C{a}^x$$

$$\int x^{dx}=\exp [\int \ln xdx]=e^{x\ln x-x+\tilde{C}}=C\frac{x^x}{e^x}$$

$$\int (ax+b{)}^{dx}=\exp [\int \ln (ax+b)dx]=e^{(\frac{b}{a}+x)\ln (ax+b)+\tilde{C}}=C\frac{(b+ax{)}^{\frac{b}{a}+x}}{e^x}$$

$$\int (\frac{1}{x}{)}^{dx}=\exp [\int \ln (\frac{1}{x})dx]=\exp [\int -\ln xdx]=e^{x-x\ln x+\tilde{C}}=C\frac{e^x}{x^x}$$

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

$$\int (a^x{)}^{dx}=\exp [\int \ln a^{x}dx]=\exp [\int x\ln adx]=e^{\frac{x^2}{2}\ln a+\tilde{C}}=C{a}^{\frac{x^2}{2}}$$

$$\int (\sqrt[x]{a}{)}^{dx}=\exp [\int \ln a^{\frac{1}{x}}dx]=\exp [\int \frac{1}{x}\ln adx]=e^{(\ln a\ln |x|)+\tilde{C}}=C{a}^{\ln x}$$

$$\int ({\log }_{a}x{)}^{dx}=\exp [\int \ln {\log }_{a}xdx]$$
いま、$$\int \ln {\log }_{a}xdx=\int \ln \ln x-\ln \ln adx=x\ln \ln x-\int x\cdot \frac{1}{x\ln x}dx-\int \ln \ln adx$$
$$=x\ln \ln x-\int \frac{1}{\ln x}dx-\int \ln \ln adx=x(\ln \ln x-\ln \ln a)-\mathrm{li}(x)+\tilde{C}$$
だから、
$$\int ({\log }_{a}x{)}^{dx}=\exp [\int \ln {\log }_{a}xdx]=e^{x(\ln \ln x-\ln \ln a)-\mathrm{li}(x)+\tilde{C}}=C\frac{({\log }_{a}x{)}^x}{e^{\mathrm{li}(x)}}$$

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

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