Question Number 212461 by MrGaster last updated on 14/Oct/24
$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{prove}: \\ $$$$\:\:\int_{\mathrm{0}} ^{+\infty} \frac{{dx}}{\:\sqrt{{x}^{\mathrm{4}} +\mathrm{25}{x}^{\mathrm{2}} +\mathrm{160}}}=\int_{\mathrm{0}} ^{+\infty} \frac{{dx}}{\:\sqrt{{x}^{\mathrm{4}} −\mathrm{95}{x}^{\mathrm{2}} +\mathrm{2560}}} \\ $$
Commented by Ghisom last updated on 19/Oct/24
![do you know the solution? I tried this but... ∫_0 ^∞ (dx/( (√(x^4 +ax^2 +b))))= [t=2arctan (x/( (b)^(1/4) )) → dx=((x^2 +(√b))/(2(b)^(1/4) ))dt] =(√2)∫_0 ^(2π) (dt/( (√((6(√b)+a)+(2(√b)−a)cos 2t))))= =2(√2)∫_0 ^π (dt/( (√((6(√b)+a)+(2(√b)−a)cos 2t))))= [let c=((6(√b)+a)/(2(√b)−a))] =((√(c+1))/( (b)^(1/4) ))∫_0 ^π (dt/( (√(c+cos 2t))))=(1/( (b)^(1/4) ))∫_0 ^π (dt/( (√(1−((2sin^2 t)/(c+1))))))= =(1/( (b)^(1/4) ))[F (t∣(2/(c+1)))]_0 ^π = =(1/( (b)^(1/4) ))[F (t∣(1/2)−(a/(4(√b))))]_0 ^π = =(1/( (b)^(1/4) ))F (π∣(1/2)−(a/(4(√b)))) now the question is (1/(2((10))^(1/4) ))F (π∣((16−5(√(10)))/(32))) =^(???) (1/(4((10))^(1/4) ))F (π∣((64+19(√(10)))/(128))) ⇔ 2F (π∣((16−5(√(10)))/(32))) =^(???) F (π∣((64+19(√(10)))/(128)))](https://www.tinkutara.com/question/Q212565.png)
$$\mathrm{do}\:\mathrm{you}\:\mathrm{know}\:\mathrm{the}\:\mathrm{solution}? \\ $$$$\mathrm{I}\:\mathrm{tried}\:\mathrm{this}\:\mathrm{but}… \\ $$$$ \\ $$$$\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{{dx}}{\:\sqrt{{x}^{\mathrm{4}} +{ax}^{\mathrm{2}} +{b}}}= \\ $$$$\:\:\:\:\:\left[{t}=\mathrm{2arctan}\:\frac{{x}}{\:\sqrt[{\mathrm{4}}]{{b}}}\:\rightarrow\:{dx}=\frac{{x}^{\mathrm{2}} +\sqrt{{b}}}{\mathrm{2}\sqrt[{\mathrm{4}}]{{b}}}{dt}\right] \\ $$$$=\sqrt{\mathrm{2}}\underset{\mathrm{0}} {\overset{\mathrm{2}\pi} {\int}}\frac{{dt}}{\:\sqrt{\left(\mathrm{6}\sqrt{{b}}+{a}\right)+\left(\mathrm{2}\sqrt{{b}}−{a}\right)\mathrm{cos}\:\mathrm{2}{t}}}= \\ $$$$=\mathrm{2}\sqrt{\mathrm{2}}\underset{\mathrm{0}} {\overset{\pi} {\int}}\frac{{dt}}{\:\sqrt{\left(\mathrm{6}\sqrt{{b}}+{a}\right)+\left(\mathrm{2}\sqrt{{b}}−{a}\right)\mathrm{cos}\:\mathrm{2}{t}}}= \\ $$$$\:\:\:\:\:\left[\mathrm{let}\:{c}=\frac{\mathrm{6}\sqrt{{b}}+{a}}{\mathrm{2}\sqrt{{b}}−{a}}\right] \\ $$$$=\frac{\sqrt{{c}+\mathrm{1}}}{\:\sqrt[{\mathrm{4}}]{{b}}}\underset{\mathrm{0}} {\overset{\pi} {\int}}\frac{{dt}}{\:\sqrt{{c}+\mathrm{cos}\:\mathrm{2}{t}}}=\frac{\mathrm{1}}{\:\sqrt[{\mathrm{4}}]{{b}}}\underset{\mathrm{0}} {\overset{\pi} {\int}}\frac{{dt}}{\:\sqrt{\mathrm{1}−\frac{\mathrm{2sin}^{\mathrm{2}} \:{t}}{{c}+\mathrm{1}}}}= \\ $$$$=\frac{\mathrm{1}}{\:\sqrt[{\mathrm{4}}]{{b}}}\left[\mathrm{F}\:\left({t}\mid\frac{\mathrm{2}}{{c}+\mathrm{1}}\right)\right]_{\mathrm{0}} ^{\pi} = \\ $$$$=\frac{\mathrm{1}}{\:\sqrt[{\mathrm{4}}]{{b}}}\left[\mathrm{F}\:\left({t}\mid\frac{\mathrm{1}}{\mathrm{2}}−\frac{{a}}{\mathrm{4}\sqrt{{b}}}\right)\right]_{\mathrm{0}} ^{\pi} = \\ $$$$=\frac{\mathrm{1}}{\:\sqrt[{\mathrm{4}}]{{b}}}\mathrm{F}\:\left(\pi\mid\frac{\mathrm{1}}{\mathrm{2}}−\frac{{a}}{\mathrm{4}\sqrt{{b}}}\right) \\ $$$$ \\ $$$$\mathrm{now}\:\mathrm{the}\:\mathrm{question}\:\mathrm{is} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}\sqrt[{\mathrm{4}}]{\mathrm{10}}}\mathrm{F}\:\left(\pi\mid\frac{\mathrm{16}−\mathrm{5}\sqrt{\mathrm{10}}}{\mathrm{32}}\right)\:\:\overset{???} {=}\:\frac{\mathrm{1}}{\mathrm{4}\sqrt[{\mathrm{4}}]{\mathrm{10}}}\mathrm{F}\:\left(\pi\mid\frac{\mathrm{64}+\mathrm{19}\sqrt{\mathrm{10}}}{\mathrm{128}}\right) \\ $$$$\Leftrightarrow \\ $$$$\mathrm{2F}\:\left(\pi\mid\frac{\mathrm{16}−\mathrm{5}\sqrt{\mathrm{10}}}{\mathrm{32}}\right)\:\:\overset{???} {=}\:\mathrm{F}\:\left(\pi\mid\frac{\mathrm{64}+\mathrm{19}\sqrt{\mathrm{10}}}{\mathrm{128}}\right) \\ $$