prove-0-dx-x-4-25x-2-160-0-dx-x-4-95x-2-2560-

Question Number 212461 by MrGaster last updated on 14/Oct/24

prove :

\int_{0}^{+ \infty} \frac{d x}{\sqrt{x^{4} + 25 x^{2} + 160}} = \int_{0}^{+ \infty} \frac{d x}{\sqrt{x^{4} - 95 x^{2} + 2560}}

Commented by Ghisom last updated on 19/Oct/24

do you know the solution ?

I tried this but \dots

\underset{0}{\int^{\infty}} \frac{d x}{\sqrt{x^{4} + {a x}^{2} + b}} =

[t = 2 \arctan \frac{x}{\sqrt[4]{b}} \to d x = \frac{x^{2} + \sqrt{b}}{2 \sqrt[4]{b}} d t]

= \sqrt{2} \underset{0}{\int^{2 π}} \frac{d t}{\sqrt{(6 \sqrt{b} + a) + (2 \sqrt{b} - a) \cos 2 t}} =

= 2 \sqrt{2} \underset{0}{\int^{π}} \frac{d t}{\sqrt{(6 \sqrt{b} + a) + (2 \sqrt{b} - a) \cos 2 t}} =

[let c = \frac{6 \sqrt{b} + a}{2 \sqrt{b} - a}]

= \frac{\sqrt{c + 1}}{\sqrt[4]{b}} \underset{0}{\int^{π}} \frac{d t}{\sqrt{c + \cos 2 t}} = \frac{1}{\sqrt[4]{b}} \underset{0}{\int^{π}} \frac{d t}{\sqrt{1 - \frac{{2 \sin}^{2} t}{c + 1}}} =

= \frac{1}{\sqrt[4]{b}} {[F (t ∣ \frac{2}{c + 1})]}_{0}^{π} =

= \frac{1}{\sqrt[4]{b}} {[F (t ∣ \frac{1}{2} - \frac{a}{4 \sqrt{b}})]}_{0}^{π} =

= \frac{1}{\sqrt[4]{b}} F (π ∣ \frac{1}{2} - \frac{a}{4 \sqrt{b}})

now the question is

\frac{1}{2 \sqrt[4]{10}} F (π ∣ \frac{16 - 5 \sqrt{10}}{32}) \overset{? ? ?}{=} \frac{1}{4 \sqrt[4]{10}} F (π ∣ \frac{64 + 19 \sqrt{10}}{128})

\Leftrightarrow

2 F (π ∣ \frac{16 - 5 \sqrt{10}}{32}) \overset{? ? ?}{=} F (π ∣ \frac{64 + 19 \sqrt{10}}{128})

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