nice-calculus-please-prove-0-x-1-2-x-2-2x-5-dx-pi-where-is-Golden-ratio-

Question Number 124201 by mnjuly1970 last updated on 01/Dec/20

::: n i c e c a l c u l u s :::

p l e a s e p r o v e :::

Ω = \int_{0}^{\infty} \frac{x^{\frac{1}{2}}}{x^{2} + 2 x + 5} d x = \frac{π}{\sqrt{φ}}

w h e r e φ i s G o l d e n r a t i o \dots

Answered by ajfour last updated on 02/Dec/20

I = \int \frac{\sqrt{x} d x}{{(x + 1)}^{2} + {(2 φ - 1)}^{2}}

l e t \sqrt{x} = t \Rightarrow \sqrt{x} d x = 2 t^{2} d t

I = \int \frac{2 t^{2} d t}{{(t^{2} + 1)}^{2} + {(2 φ - 1)}^{2}}

l e t t = \frac{1}{z} \Rightarrow d t = - \frac{d z}{z^{2}}

I = \int \frac{2 d z}{{(1 + z^{2})}^{2} + {(2 φ - 1)}^{2} z^{4}}

= \int \frac{2 d z}{6 z^{4} + 2 z^{2} + 1}

= \frac{1}{3} \int \frac{d z}{{(z^{2} + \frac{1}{6})}^{2} + {(\frac{\sqrt{5}}{6})}^{2}}

\dots .

Commented by mindispower last updated on 02/Dec/20

s i r a j f o u r y o u r m e t h o d e

w o r c k s s i n c e 6 z^{4} + 2 z^{2} + 1 c a n b e e f a c t o r i s e d

i n p o l y n o m s o f d e g r e e 2

a n d \int \frac{d x}{x^{2} + a x + b} c a n b e e s o l v e d

Commented by mnjuly1970 last updated on 03/Dec/20

t h a n k i n g f o r y o u r e f f o r t

s i r a j f o r . .

Commented by mnjuly1970 last updated on 03/Dec/20

Answered by mindispower last updated on 02/Dec/20

Δ = \int_{- \infty}^{\infty} \frac{z^{\frac{1}{2}}}{z^{2} + 2 z + 5} d z = \int_{- \infty}^{0} \frac{z^{\frac{1}{2}}}{z^{2} + 2 z + 5} d z + Ω

= i \int_{0}^{\infty} \frac{z^{\frac{1}{2}}}{z^{2} - 2 z + 5} d z + Ω = Ω + i t, Ω, t \in R

Δ = 2 i π R e s (\frac{z^{\frac{1}{2}}}{z^{2} + 2 z + 5}, z = - 1 + 2 i)

. = 2 i π \frac{{(- 1 + 2 i)}^{\frac{1}{2}}}{2 (- 1 + 2 i) + 2} = \frac{π}{2} {(- 1 + 2 i)}^{\frac{1}{2}}

{(- 1 + 2 i)}^{\frac{1}{2}} = x + i y

\Rightarrow x y = 1, x^{2} + y^{2} = \sqrt{5}, x^{2} - y^{2} = - 1

x = {(\frac{\sqrt{5} - 1}{2})}^{\frac{1}{2}}, y = \frac{2}{\sqrt{5} - 1} = {(\frac{\sqrt{5} + 1}{2})}^{\frac{1}{2}}

x = \sqrt{φ - 1}, y = \sqrt{φ}

Δ = \frac{π}{2} \sqrt{φ - 1} + \frac{π}{2} i \sqrt{φ}

Ω = R e (Δ) = \frac{π}{2} \sqrt{φ - 1}

φ^{2} = φ + 1 \Rightarrow φ (φ - 1) = 1

\Rightarrow \sqrt{φ - 1} = \frac{1}{\sqrt{φ}}

Ω = \frac{π}{2} \sqrt{φ - 1} = \frac{π}{2 \sqrt{φ}}

Commented by mnjuly1970 last updated on 02/Dec/20

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