Question Number 212626 by Ghisom last updated on 19/Oct/24
$$\mathrm{let}\:{f}\left({x}\right)=\frac{\mathrm{1}}{\:\sqrt{\left({x}−{a}\right)\left({x}−{b}\right)\left({x}−{c}\right)}} \\ $$$$\mathrm{let}\:{a},\:{b},\:{c}\:\in\mathbb{R}\:\wedge{a}<{b}<{c} \\ $$$$\Rightarrow\:{D}\left({f}\left({x}\right)\right)=\left({a},\:{b}\right)\cup\left({c},\:\infty\right) \\ $$$$\mathrm{prove}\:\underset{{a}} {\overset{{b}} {\int}}{f}\left({x}\right){dx}=\underset{{c}} {\overset{\infty} {\int}}{f}\left({x}\right){dx} \\ $$
Answered by MrGaster last updated on 02/Nov/24
$$\int_{{a}} ^{{b}} {f}\left({x}\right){dx}=\int_{{a}} ^{{b}} \frac{\mathrm{1}}{\:\sqrt{\left({x}−{a}\right)\left({x}−{b}\right)\left({x}−{c}\right)}}{dx} \\ $$$$\int_{{c}} ^{\infty} {f}\left({x}\right){dx}=\int_{{c}} ^{\infty} \frac{\mathrm{1}}{\:\sqrt{\left({x}−{a}\right)\left({x}−{b}\right)\left({x}−{c}\right)}}{dxx}= \\ $$$${a}+\left({b}−{a}\right){tdx}=\left({b}−{a}\right){dt} \\ $$$$\int_{{a}} ^{{b}} \frac{\mathrm{1}}{\:\sqrt{\left({x}−{a}\left({x}−{b}\left({x}−{c}\right)\right.\right.}}{dx}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\left({b}−{a}\right)}{\:\sqrt{\left({b}−{a}\right)\left({b}−{a}\right){t}\left({b}−{a}\right)\left(\mathrm{1}−{t}\right)\left({b}−{a}\right)\left(\mathrm{1}−\frac{{a}+\left({b}−{a}\right){t}−{c}}{{b}−{a}}\right)}}{dt} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\:\sqrt{{t}\left(\mathrm{1}−{t}\right)\left(\mathrm{1}−\frac{{a}−{c}}{{b}−{a}}−{t}\right)}}{dtx}={c}+\left({x}−{c}\right){t}'{dx}=\left({x}−{c}\right){dt}' \\ $$$$\int_{{c}} ^{\infty} \frac{\mathrm{1}}{\:\sqrt{\left({x}−{a}\right)\left({x}−{b}\right)\left({x}−{c}\right)}}{dx}=\int_{\mathrm{0}} ^{\infty} \frac{\left({x}−{c}\right)}{\:\sqrt{\left({x}−{c}\right)\left({x}−{c}\right){t}'\left({x}−{c}\right)\left(\mathrm{1}−{t}'\right)\left({x}−{c}\right)}}{dt}' \\ $$$$\: \\ $$$$=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\:\sqrt{{t}'\left(\mathrm{1}−{t}'\right)\left(\mathrm{1}+\frac{{x}−{c}}{{c}−{a}}−{t}'\right)}}{dt}' \\ $$$${t}'=\mathrm{1}−{tdt}'=−{dt} \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\:\sqrt{{t}'\left(\mathrm{1}−{t}'\right)\left(\mathrm{1}+\frac{{x}−{c}}{{c}−{a}}−{t}'\right)}}{dt}'=\int_{\mathrm{0}} ^{−\infty} \frac{−\mathrm{1}}{\:\sqrt{\left(\mathrm{1}−{t}\right)\left({t}\right)\left(\mathrm{1}+\frac{{x}−{c}}{{c}−{a}}−\mathrm{1}+{t}\right)}}{dt} \\ $$$$=\int_{−\infty} ^{\mathrm{1}} \frac{\mathrm{1}}{\:\sqrt{\left(\mathrm{1}−{t}\right)\left({t}\right)\left(\mathrm{1}+\frac{{x}−{c}}{{c}−{a}}−\mathrm{1}+{t}\right)}}{dt}=\int_{−\infty} ^{\mathrm{1}} \frac{\mathrm{1}}{\:\sqrt{\left(\mathrm{1}−{t}\right)\left({t}\right)\left({t}+\frac{{x}−{c}}{{c}−{a}}\right)}}{dt}=\int_{−\infty} ^{\mathrm{1}} \frac{\mathrm{1}}{\:\sqrt{\left(\mathrm{1}−{t}\right)\left({t}\right)\left({t}+\frac{{b}−{c}}{{b}−{a}}\right)}}{dt}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\:\sqrt{{t}\left(\mathrm{1}−{t}\right)\left(\mathrm{1}−\frac{{a}−{c}}{{b}−{a}}\right)}}{dt} \\ $$$$\int_{{a}} ^{{b}} {f}\left({x}\right){dx}=\int_{{c}} ^{\infty} {f}\left({x}\right){dx} \\ $$$$\: \\ $$
Commented by Ghisom last updated on 02/Nov/24
$$\mathrm{you}\:\mathrm{write} \\ $$$${a}+\left({b}−{a}\right){tdx}=\left({b}−{a}\right){dt} \\ $$$$\mathrm{but}\:\mathrm{this}\:\mathrm{makes}\:\mathrm{no}\:\mathrm{sense}.\:\mathrm{please}\:\mathrm{correct} \\ $$$$\mathrm{and}\:\mathrm{clarify}\:\mathrm{your}\:\mathrm{answer}. \\ $$