nice-calculus-prove-that-0-1-x-1-ln-x-2-dx-5-ln-m-n-

Question Number 122882 by mnjuly1970 last updated on 20/Nov/20

\dots n i c e c a l c u l u s \dots

p r o v e t h a t :::

Ω = \int_{0}^{1} {(\frac{x^{φ} - 1}{l n (x)})}^{2} d x = \sqrt{5} l n (φ)

. m . n .

Answered by TANMAY PANACEA last updated on 20/Nov/20

f (φ) = \int_{0}^{1} \frac{x^{φ} - 1}{l n x} d x

\frac{d f}{d φ} = \int_{0}^{1} \frac{x^{φ} l n x}{l n x} d x =∣ \frac{x^{φ + 1}}{φ + 1} ∣_{0}^{1} = \frac{1}{φ + 1}

d f = \frac{d φ}{φ + 1}

f = l n (φ + 1) + C

w h e n φ = 0 f (φ) = 0 C = 0

f (φ) = l n (φ + 1)

w a i t \dots

Commented by mnjuly1970 last updated on 21/Nov/20

t h a n k y o u m r t a n m a y f o r y o u r

e f f o r t

w i t h y o u r p e r m i s s i o n

i p r e s e n t t h e s o l u t i o n .

n o t e : l n (N) = \int_{0}^{1} \frac{x^{N - 1} - 1}{l n (x)} d x

f (a) = \int_{0}^{1} \frac{{(x^{a} - 1)}^{2}}{{l n}^{2} (x)} d x

f^{'} (a) = \int_{0}^{1} \frac{\partial}{\partial a} [\frac{{(x^{a} - 1)}^{2}}{{l n}^{2} (x)}]

= \int_{0}^{1} \frac{2 (x^{a} - 1) x^{a} l n (x)}{{l n}^{2} (x)} d x = 2 \int_{0}^{1} \frac{x^{2 a} - x^{a}}{l n (x)} d x

= 2 \int_{0}^{1} \frac{x^{2 a} - 1}{l n (x)} + 2 \int_{0}^{1} \frac{1 - x^{a}}{l n (x)} d x

= 2 l n (2 a + 1) - 2 l n (a + 1)

f (a) = 2 \int^{} l n (2 a + 1) d a - 2 \int l n (a + 1) d a + C

= (2 a + 1) l n (2 a + 1) - (2 a + 1)

- 2 (a + 1) l n (a + 1) + 2 (a + 1) + C

f (0) = 1 + C \Rightarrow C = - 1

f (a) = (2 a + 1) l n (2 a + 1) - 2 (a + 1) l n (a + 1) + 1 - 1

Ω = f (φ) = (2 φ + 1) l n (2 φ + 1) - 2 (φ + 1) l n (φ + 1)

φ : i s g o l d e n r a t i o

a n d w e k n o w : φ^{2} = φ + 1

= (φ^{2} + φ) l n (φ^{2} + φ) - 2 φ^{2} l n (φ^{2})

= φ^{3} l n (φ^{3}) - 4 φ^{2} l n (φ)

= (3 φ^{3} - 4 φ^{2}) l n (φ) = (3 φ^{2} + 3 φ - 4 φ^{2}) l n (φ)

= (- φ^{2} + 3 φ) l n (φ) = (- φ - 1 + 3 φ) l n (φ)

= (2 φ - 1) l n (φ) = (2 (\frac{1 + \sqrt{5}}{2}) - 1) l n (φ)

= \sqrt{5} l n (φ) ✓ ✓ ✓