Question Number 212777 by Ghisom last updated on 23/Oct/24
![question 212462 prove that ∫_0 ^∞ (dx/( (√(x^4 +25x^2 +160))))=∫_0 ^∞ (dx/( (√(x^4 −95x^2 +2560)))) my attempt (but is it a proof?): I_b ^a :=∫_0 ^∞ (dx/( (√(x^4 +ax^2 +b))))=(2/( π(b)^(1/4) agm (1, ((√(a+2(√b)))/(2(b)^(1/4) ))))) ∫_0 ^∞ (dx/( (√(x^4 +ax^2 +b))))= [t=2arctan (x/( (b)^(1/4) ))] =(1/( (b)^(1/4) ))∫_0 ^(π/2) (dt/( (√(1−((1/2)−(a/(4(√b))))sin^2 t))))= [K (k^2 ):=∫_0 ^(π/2) (dt/( (√(1−k^2 sin^2 t))))] =(1/( (b)^(1/4) ))K ((1/2)−(a/(4(√b)))) K (k^2 ) =(2/(πagm (1, (√(1−k^2 ))))) (1/( (b)^(1/4) ))K ((1/2)−(a/(4(√b)))) =(2/( π(b)^(1/4) agm (1, ((√(a+2(√b)))/(2(b)^(1/4) ))))) u_0 =p∧v_0 =q u_(n+1) =((u_n +v_n )/2)∧v_(n+1) =(√(u_n v_n )) agm (p, q) =lim_(n→∞) u_n =lim_(n→∞) v_n I_(160) ^(25) =(1/( π((10))^(1/4) agm (1, ((√(25+8(√(10))))/(4((10))^(1/4) ))))) I_(2560) ^(−95) =(1/(2π((10))^(1/4) agm (1, ((√(−95+32(√(10))))/(8((10))^(1/4) ))))) I_(160) ^(25) =I_(2560) ^(−95) ⇔ agm (1, ((√(25+8(√(10))))/(4((10))^(1/4) )))=2agm (1, ((√(−95+32(√(10))))/(8((10))^(1/4) ))) which is true](https://www.tinkutara.com/question/Q212777.png)
$$\mathrm{question}\:\mathrm{212462} \\ $$$$\mathrm{prove}\:\mathrm{that} \\ $$$$\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{{dx}}{\:\sqrt{{x}^{\mathrm{4}} +\mathrm{25}{x}^{\mathrm{2}} +\mathrm{160}}}=\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{{dx}}{\:\sqrt{{x}^{\mathrm{4}} −\mathrm{95}{x}^{\mathrm{2}} +\mathrm{2560}}} \\ $$$$ \\ $$$$\mathrm{my}\:\mathrm{attempt}\:\left(\mathrm{but}\:\mathrm{is}\:\mathrm{it}\:\mathrm{a}\:\mathrm{proof}?\right): \\ $$$$ \\ $$$${I}_{{b}} ^{{a}} :=\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{{dx}}{\:\sqrt{{x}^{\mathrm{4}} +{ax}^{\mathrm{2}} +{b}}}=\frac{\mathrm{2}}{\:\pi\sqrt[{\mathrm{4}}]{{b}}\:\mathrm{agm}\:\left(\mathrm{1},\:\frac{\sqrt{{a}+\mathrm{2}\sqrt{{b}}}}{\mathrm{2}\sqrt[{\mathrm{4}}]{{b}}}\right)} \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{{dx}}{\:\sqrt{{x}^{\mathrm{4}} +{ax}^{\mathrm{2}} +{b}}}= \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left[{t}=\mathrm{2arctan}\:\frac{{x}}{\:\sqrt[{\mathrm{4}}]{{b}}}\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\:\sqrt[{\mathrm{4}}]{{b}}}\underset{\mathrm{0}} {\overset{\pi/\mathrm{2}} {\int}}\frac{{dt}}{\:\sqrt{\mathrm{1}−\left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{{a}}{\mathrm{4}\sqrt{{b}}}\right)\mathrm{sin}^{\mathrm{2}} \:{t}}}= \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left[{K}\:\left({k}^{\mathrm{2}} \right):=\underset{\mathrm{0}} {\overset{\pi/\mathrm{2}} {\int}}\frac{{dt}}{\:\sqrt{\mathrm{1}−{k}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \:{t}}}\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\:\sqrt[{\mathrm{4}}]{{b}}}{K}\:\left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{{a}}{\mathrm{4}\sqrt{{b}}}\right) \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:{K}\:\left({k}^{\mathrm{2}} \right)\:=\frac{\mathrm{2}}{\pi\mathrm{agm}\:\left(\mathrm{1},\:\sqrt{\mathrm{1}−{k}^{\mathrm{2}} }\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\:\sqrt[{\mathrm{4}}]{{b}}}{K}\:\left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{{a}}{\mathrm{4}\sqrt{{b}}}\right)\:=\frac{\mathrm{2}}{\:\pi\sqrt[{\mathrm{4}}]{{b}}\:\mathrm{agm}\:\left(\mathrm{1},\:\frac{\sqrt{{a}+\mathrm{2}\sqrt{{b}}}}{\mathrm{2}\sqrt[{\mathrm{4}}]{{b}}}\right)} \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:{u}_{\mathrm{0}} ={p}\wedge{v}_{\mathrm{0}} ={q} \\ $$$$\:\:\:\:\:\:\:\:\:\:{u}_{{n}+\mathrm{1}} =\frac{{u}_{{n}} +{v}_{{n}} }{\mathrm{2}}\wedge{v}_{{n}+\mathrm{1}} =\sqrt{{u}_{{n}} {v}_{{n}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{agm}\:\left({p},\:{q}\right)\:=\underset{{n}\rightarrow\infty} {\mathrm{lim}}{u}_{{n}} =\underset{{n}\rightarrow\infty} {\mathrm{lim}}{v}_{{n}} \\ $$$$ \\ $$$${I}_{\mathrm{160}} ^{\mathrm{25}} =\frac{\mathrm{1}}{\:\pi\sqrt[{\mathrm{4}}]{\mathrm{10}}\:\mathrm{agm}\:\left(\mathrm{1},\:\frac{\sqrt{\mathrm{25}+\mathrm{8}\sqrt{\mathrm{10}}}}{\mathrm{4}\sqrt[{\mathrm{4}}]{\mathrm{10}}}\right)} \\ $$$${I}_{\mathrm{2560}} ^{−\mathrm{95}} =\frac{\mathrm{1}}{\mathrm{2}\pi\sqrt[{\mathrm{4}}]{\mathrm{10}}\:\mathrm{agm}\:\left(\mathrm{1},\:\frac{\sqrt{−\mathrm{95}+\mathrm{32}\sqrt{\mathrm{10}}}}{\mathrm{8}\sqrt[{\mathrm{4}}]{\mathrm{10}}}\right)} \\ $$$$ \\ $$$$ \\ $$$${I}_{\mathrm{160}} ^{\mathrm{25}} ={I}_{\mathrm{2560}} ^{−\mathrm{95}} \\ $$$$\Leftrightarrow \\ $$$$\mathrm{agm}\:\left(\mathrm{1},\:\frac{\sqrt{\mathrm{25}+\mathrm{8}\sqrt{\mathrm{10}}}}{\mathrm{4}\sqrt[{\mathrm{4}}]{\mathrm{10}}}\right)=\mathrm{2agm}\:\left(\mathrm{1},\:\frac{\sqrt{−\mathrm{95}+\mathrm{32}\sqrt{\mathrm{10}}}}{\mathrm{8}\sqrt[{\mathrm{4}}]{\mathrm{10}}}\right) \\ $$$$\mathrm{which}\:\mathrm{is}\:\mathrm{true} \\ $$