Question Number 219957 by hardmath last updated on 04/May/25
$$\mathrm{if}\:\:\:\mathrm{f}:\left(\mathrm{0},\infty\right)\rightarrow\left(\mathrm{0},\infty\right) \\ $$$$\:\:\:\:\:\:\mathrm{f}\:\:\mathrm{continuous} \\ $$$$\mathrm{and}\:\:\:\mathrm{f}\left(\frac{\mathrm{1}}{\mathrm{x}}\right)\:+\:\mathrm{f}\left(\frac{\mathrm{1}}{\mathrm{y}}\right)\:=\:\mathrm{2f}\left(\frac{\mathrm{1}}{\mathrm{x}+\mathrm{y}}\right) \\ $$$$\forall\:\mathrm{x},\mathrm{y}\:>\:\mathrm{0}\:\:\:\mathrm{then}\:\:\:\forall\:\mathrm{a},\mathrm{b}\:>\:\mathrm{0}: \\ $$$$\int_{\boldsymbol{\mathrm{a}}} ^{\:\boldsymbol{\mathrm{b}}} \int_{\boldsymbol{\mathrm{a}}} ^{\:\boldsymbol{\mathrm{b}}} \int_{\boldsymbol{\mathrm{a}}} ^{\:\boldsymbol{\mathrm{b}}} \:\mathrm{f}\left(\frac{\mathrm{1}}{\mathrm{x}+\mathrm{y}+\mathrm{z}}\right)\mathrm{dxdydz}\:=\:\left(\mathrm{b}−\mathrm{a}\right)^{\mathrm{2}} \int_{\boldsymbol{\mathrm{a}}} ^{\:\boldsymbol{\mathrm{b}}} \mathrm{f}\left(\frac{\mathrm{1}}{\mathrm{x}}\right)\mathrm{dx} \\ $$
Answered by MrGaster last updated on 04/May/25
Answered by MrGaster last updated on 04/May/25
![(2):∫_a ^( b) ∫_a ^( b) ∫_a ^( b) f((1/(x+y+z)))dxdydz=(1/2)∫_a ^( b) ∫_a ^( b) ∫_a ^( b) [f((1/(x+y)))+f((1/z))]dxxydz =(1/2)[(b−a)∫_a ^b ∫_a ^b f((1/(x+y)))dydx+(b−a)^2 ∫_a ^b f((1/z))dz] ∫_a ^( b) ∫_a ^( b) f((1/(x+y)))dydx=(b−a)∫_a ^b f((1/x))dx ⇒Original formula=(1/2)[(b−a)^2 ∫_(a ) ^b ((1/x))dx+(b−a)^2 ∫_a ^b f((1/x))dx] =(b−a)^2 ∫_a ^b f((1/x))dx](https://www.tinkutara.com/question/Q219961.png)
$$\left(\mathrm{2}\right):\int_{\boldsymbol{\mathrm{a}}} ^{\:\boldsymbol{\mathrm{b}}} \int_{\boldsymbol{\mathrm{a}}} ^{\:\boldsymbol{\mathrm{b}}} \int_{\boldsymbol{\mathrm{a}}} ^{\:\boldsymbol{\mathrm{b}}} \:\mathrm{f}\left(\frac{\mathrm{1}}{\mathrm{x}+\mathrm{y}+\mathrm{z}}\right)\mathrm{dxdydz}=\frac{\mathrm{1}}{\mathrm{2}}\int_{\boldsymbol{\mathrm{a}}} ^{\:\boldsymbol{\mathrm{b}}} \int_{\boldsymbol{\mathrm{a}}} ^{\:\boldsymbol{\mathrm{b}}} \int_{\boldsymbol{\mathrm{a}}} ^{\:\boldsymbol{\mathrm{b}}} \left[{f}\left(\frac{\mathrm{1}}{{x}+{y}}\right)+{f}\left(\frac{\mathrm{1}}{{z}}\right)\right]{dxxydz} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left[\left({b}−{a}\right)\int_{{a}} ^{{b}} \int_{{a}} ^{{b}} {f}\left(\frac{\mathrm{1}}{{x}+{y}}\right){dydx}+\left({b}−{a}\right)^{\mathrm{2}} \int_{{a}} ^{{b}} {f}\left(\frac{\mathrm{1}}{{z}}\right){dz}\right] \\ $$$$\int_{\boldsymbol{\mathrm{a}}} ^{\:\boldsymbol{\mathrm{b}}} \int_{\boldsymbol{\mathrm{a}}} ^{\:\boldsymbol{\mathrm{b}}} {f}\left(\frac{\mathrm{1}}{{x}+{y}}\right){dydx}=\left({b}−{a}\right)\int_{{a}} ^{{b}} {f}\left(\frac{\mathrm{1}}{{x}}\right){dx} \\ $$$$\Rightarrow\mathrm{Original}\:\mathrm{formula}=\frac{\mathrm{1}}{\mathrm{2}}\left[\left({b}−{a}\right)^{\mathrm{2}} \int_{{a}\:} ^{{b}} \left(\frac{\mathrm{1}}{{x}}\right){dx}+\left({b}−{a}\right)^{\mathrm{2}} \int_{{a}} ^{{b}} {f}\left(\frac{\mathrm{1}}{{x}}\right){dx}\right] \\ $$$$=\left({b}−{a}\right)^{\mathrm{2}} \int_{{a}} ^{{b}} {f}\left(\frac{\mathrm{1}}{{x}}\right){dx} \\ $$
Commented by hardmath last updated on 04/May/25
$$ \\ $$Thank you, my dear, valuable magical mathematician, it's nice to see your perfect solutions.