Формулы Фруллани

Если ${\displaystyle f(x)\in C[0,+\infty )\ }$ и ${\displaystyle \ \forall A>0\ \exists \int \limits _{A}^{\infty }{\frac {f(x)}{x}}\,dx}$, то справедлива следующая формула:

${\displaystyle \int \limits _{0}^{\infty }{\frac {f(\alpha x)-f(\beta x)}{x}}\,dx=f(0)\ln {\biggl (}{\frac {\beta }{\alpha }}{\biggr )}\ \ (\alpha >0,\beta >0)\ }$

Доказательство:

Стоит отметить, что в этом и доказательствах ниже подразумевается ${\displaystyle F'(x)={\frac {f(x)}{x}}}$, а не ${\displaystyle F'(x)=f(x)}$.

${\displaystyle \lim \limits _{A\to +0}{\Biggl (}{\int \limits _{A}^{\infty }{\frac {f(\alpha x)-f(\beta x)}{x}}\,dx}{\Biggr )}=\lim \limits _{A\to +0}{\Biggl (}{{\int \limits _{A}^{\infty }{\frac {f(\alpha x)}{x}}\,dx}-{\int \limits _{A}^{\infty }{\frac {f(\beta x)}{x}}\,dx}}{\Biggr )}=}$

${\displaystyle =\left\{\forall A>0\ \exists \int \limits _{A}^{\infty }{\frac {f(x)}{x}}\,dx\ \Rightarrow {\int \limits _{A}^{\infty }{\frac {f(x)}{x}}\,dx=F(\infty )-F(A)}\Rightarrow {\int \limits _{A}^{\infty }{\frac {f(\alpha x)}{x}}\,dx=\int \limits _{\alpha A}^{\infty }{\frac {f(x)}{x}}\,dx}=F(\infty )-F(\alpha A)\right\}}$^[1] ${\displaystyle =}$

${\displaystyle =\lim \limits _{A\to +0}{\Biggl (}F(\infty )-F(\alpha A)-F(\infty )+F(\beta A){\Biggr )}=\lim \limits _{A\to +0}{\Biggl (}F(\beta A)-F(\alpha A){\Biggr )}=\lim \limits _{A\to +0}{\Biggl (}\int \limits _{\alpha }^{\beta }{\frac {f(Ax)}{x}}\,dx{\Biggr )}=}$

${\displaystyle =\lim \limits _{A\to +0}{\Biggl (}f(A\xi )\int \limits _{\alpha }^{\beta }{\frac {1}{x}}\,dx{\Biggr )}}$^[2] ${\displaystyle =\lim \limits _{A\to +0}{\Biggl (}f(A\xi ){\biggl (}\ln(\beta )-\ln(\alpha ){\biggr )}{\Biggr )}}$^[3] ${\displaystyle =\lim \limits _{A\to +0}{\biggl (}f(A\xi ){\biggr )}\ln {\biggl (}{\frac {\beta }{\alpha }}{\biggr )}=}$

${\displaystyle =\left\{\xi \in [\alpha ,\beta ]\Rightarrow \lim \limits _{A\to +0}{A\xi }=0,f(x)\in C[0,+\infty )\Rightarrow \lim \limits _{A\to +0}{f(A\xi )=f(0)}\right\}=f(0)\ln {\biggl (}{\frac {\beta }{\alpha }}{\biggr )}.}$

Если ${\displaystyle f(x)\in C[0,+\infty )}$ и ${\displaystyle \exists \lim \limits _{x\to +\infty }f(x)<+\infty \ }$, то справедлива следующая формула:

${\displaystyle \int \limits _{0}^{\infty }{\frac {f(\alpha x)-f(\beta x)}{x}}\,dx=(f(0)-f(+\infty ))\ln {\biggl (}{\frac {\beta }{\alpha }}{\biggr )}\ \ (\alpha >0,\beta >0)}$

Доказательство:

${\displaystyle \int \limits _{0}^{\infty }{\frac {f(\alpha x)-f(\beta x)}{x}}\,dx=\lim \limits _{\epsilon \to 0,\Delta \to \infty }{\Biggl (}\int \limits _{\epsilon }^{A}{\frac {f(\alpha x)-f(\beta x)}{x}}\,dx+\int \limits _{A}^{\Delta }{\frac {f(\alpha x)-f(\beta x)}{x}}\,dx{\Biggr )}}$^[4]${\displaystyle =}$

${\displaystyle =\left\{\rho {\bigl (}\epsilon ,A{\bigr )}<\infty ,{\frac {f(x)}{x}}\in C[\epsilon ,A]\Rightarrow \int \limits _{\epsilon }^{A}{\frac {f(x)}{x}}\,dx=F(A)-F(\epsilon )\Rightarrow \int \limits _{\epsilon }^{A}{\frac {f(\alpha x)}{x}}\,dx=F(\alpha A)-F(\alpha \epsilon )\right\}}$^[1]${\displaystyle =}$

${\displaystyle =\lim \limits _{\epsilon \to 0,\Delta \to +\infty }{\Biggl (}F(\alpha A)-F(\alpha \epsilon )-F(\beta A)+F(\beta \epsilon )+\int \limits _{A}^{\Delta }{\frac {f(\alpha x)-f(\beta x)}{x}}\,dx{\Biggr )}=}$

${\displaystyle =\left\{\rho {\bigl (}A,\Delta {\bigr )}<\infty ,{\frac {f(x)}{x}}\in C[A,\Delta ]\Rightarrow \int \limits _{A}^{\Delta }{\frac {f(x)}{x}}\,dx=F(\Delta )-F(A)\Rightarrow \int \limits _{A}^{\Delta }{\frac {f(\alpha x)}{x}}\,dx=F(\alpha \Delta )-F(\alpha A)\right\}=}$

${\displaystyle =\lim \limits _{\epsilon \to +0,\Delta \to +\infty }{\Biggl (}F(\alpha A)-F(\alpha \epsilon )-F(\beta A)+F(\beta \epsilon )+F(\alpha \Delta )-F(\alpha A)-F(\beta \Delta )+F(\beta A){\Biggr )}=}$

${\displaystyle =\lim \limits _{\epsilon \to +0}{\biggl (}F(\beta \epsilon )-F(\alpha \epsilon ){\biggr )}-\lim \limits _{\Delta \to +\infty }{\biggl (}F(\beta \Delta )-F(\alpha \Delta ){\biggr )}=\lim \limits _{\epsilon \to +0}{\Biggl (}\int \limits _{\alpha }^{\beta }{\frac {f(\epsilon x)}{x}}\,dx{\Biggr )}-\lim \limits _{\Delta \to +\infty }{\Biggl (}\int \limits _{\alpha }^{\beta }{\frac {f(\Delta x)}{x}}\,dx{\Biggr )}=}$

${\displaystyle =\lim \limits _{\epsilon \to +0}{\Biggl (}f(\epsilon \eta )\int \limits _{\alpha }^{\beta }{\frac {1}{x}}\,dx{\Biggr )}-\lim \limits _{\Delta \to +\infty }{\Biggl (}f(\Delta \mu )\int \limits _{\alpha }^{\beta }{\frac {1}{x}}\,dx{\Biggr )}}$^[2] ${\displaystyle ={\biggl (}\lim \limits _{\epsilon \to +0}f(\epsilon \eta )-\lim \limits _{\Delta \to +\infty }f(\Delta \mu ){\biggr )}{\biggl (}\ln(\beta )-\ln(\alpha ){\biggr )}}$^[3] ${\displaystyle =}$

${\displaystyle =\left\{\eta ,\mu \in [\alpha ,\beta ]\Rightarrow \lim \limits _{\epsilon \to +0}\epsilon \eta =0,\lim \limits _{\Delta \to +\infty }\Delta \mu =+\infty ,f(x)\in C[0,+\infty ]\Rightarrow \lim \limits _{\epsilon \to +0}f(\epsilon \eta )=f(0),\lim \limits _{\Delta \to +\infty }f(\Delta \mu )=f(+\infty )\right\}=}$

${\displaystyle ={\biggl (}f(0)-f(+\infty ){\biggr )}\ln {\biggl (}{\frac {\beta }{\alpha }}{\biggr )}.}$

Если ${\displaystyle f(x)\in C(0,+\infty )\ }$ и ${\displaystyle \ \forall A>0\ \exists \int \limits _{0}^{A}{\frac {f(x)}{x}}\,dx}$ и ${\displaystyle \exists \lim \limits _{x\to +\infty }f(x)<+\infty \ }$, то справедлива следующая формула:

${\displaystyle \int \limits _{0}^{\infty }{\frac {f(\alpha x)-f(\beta x)}{x}}\,dx=f(+\infty )\ln {\biggl (}{\frac {\alpha }{\beta }}{\biggr )}\ \ (\alpha >0,\beta >0)\ }$

• ${\displaystyle \int \limits _{0}^{\infty }{\frac {{\frac {\sin(\alpha x)}{\alpha x}}-{\frac {\sin(\beta x)}{\beta x}}}{x}}\,dx\,=\,\ln \left({\frac {\beta }{\alpha }}\right)}$

• ${\displaystyle \int \limits _{0}^{\infty }\!{\frac {\sin \left(\alpha x+m\right)-\sin \left(\beta x+m\right)}{x}}{dx}=\sin(m)\cdot \ln \left({\frac {\beta }{\alpha }}\right)}$

• ${\displaystyle \int \limits _{0}^{\infty }\!{\frac {\cos \left(\alpha x+m\right)-\cos \left(\beta x+m\right)}{x}}{dx}=\cos(m)\cdot \ln \left({\frac {\beta }{\alpha }}\right)}$

• ${\displaystyle \int \limits _{0}^{\infty }{\frac {{\frac {m}{n+\alpha x}}-{\frac {m}{n+\beta x}}}{x}}\,dx\,=\,{{\frac {m}{n}}\,\ln \left({\frac {\beta }{\alpha }}\right)}}$

• ${\displaystyle \int \limits _{0}^{\infty }\!\,{\frac {{\frac {arctg\left(-\alpha \,x\right)}{\alpha \,x}}-{\frac {arctg\left(-\beta \,x\right)}{\beta \,x}}}{x}}{dx}\,={\,\ln \left({\frac {\alpha }{\beta }}\right)}}$