MHB Five, Phi and Pi in one integral = −5ϕπ

  • Thread starter Thread starter Tony1
  • Start date Start date
  • Tags Tags
    Integral Phi Pi
AI Thread Summary
The integral $$\int_{0}^{\pi/2} \frac{\ln(\sin^2 x)}{\sin(2x)} \cdot \sqrt[5]{\tan(x)} \, dx$$ is proposed to equal $$-5\phi\pi$$, where $\phi$ is the golden ratio. Some participants question the validity of this result, suggesting an alternative outcome of $$-\frac{5\pi}{\phi}$$. A key point of discussion revolves around the relationship between sine values and the golden ratio, specifically noting that $$\sin(\pi/10) = \frac{1}{2\phi}$$. The conversation emphasizes the need for clarity on the integral's evaluation and the potential error in the proposed result.
Tony1
Messages
9
Reaction score
0
$$\int_{0}^{\pi\over 2}{\ln(\sin^2 x)\over \sin(2x)}\cdot \sqrt[5]{\tan(x)}\mathrm dx=-5\phi \pi$$

$\phi$ is the golden ratio

Make $u=\tan x$

$${1\over 2}\int_{0}^{\infty}u^{-{4\over 5}}\ln\left({1+u^2\over u^2}\right)\mathrm du$$

hmmm, too complicate to continue.
Any help, please. Thank you!
 
Last edited:
Mathematics news on Phys.org
Is there an error in the result? I'm getting

$$I = \int_{0}^{\pi\over 2}{\ln(\sin^2 x)\over \sin(2x)}\cdot \sqrt[5]{\tan(x)}\mathrm dx = -\frac{5 \pi}{\phi}$$.
 
Theia said:
Is there an error in the result? I'm getting

$$I = \int_{0}^{\pi\over 2}{\ln(\sin^2 x)\over \sin(2x)}\cdot \sqrt[5]{\tan(x)}\mathrm dx = -\frac{5 \pi}{\phi}$$.

Taking your substitution $$u = \tan x$$, one obtains

$$I = -\frac{1}{2}\int _0^{\infty}u^{-4/5}\ln \left( \frac{1+u^2}{u^2} \right) \mathrm d u$$.

Now taking $$\frac{1+u^2}{u^2} = q$$, the integral becomes

$$I = -\frac{1}{4}\int _1 ^{\infty} \ln q \ (q - 1)^{-11/10} \mathrm dq.$$

Let's then integrate by parts ($$ \mathrm d f = \ln q$$) and one obtains

$$I = -\frac{5}{2}\int _1 ^{\infty} q^{-1} \ (q - 1)^{-1/10} \mathrm dq.$$

One more substitution: $$1 - \dfrac{1}{q} = p$$ and one obtains

$$I = -\frac{5}{2}\int _0 ^1 p^{-1/10} \ (1 - p)^{-9/10} \mathrm dp.$$

By definition of the Beta function one reads this as

$$I =-\frac{5}{2} \beta \left( \frac{9}{10}, \frac{1}{10} \right) = -\frac{5}{2} \frac{\pi}{\sin \tfrac{\pi}{10}} = -\frac{5\pi}{\phi}$$.
 
Last edited:
Theia said:
Taking your substitution $$u = \tan x$$, one obtains

$$I = -\frac{1}{2}\int _0^{\infty}u^{-4/5}\ln \left( \frac{1+u^2}{u^2} \right) \mathrm d u$$.

Now taking $$\frac{1+u^2}{u^2} = q$$, the integral becomes

$$I = -\frac{1}{4}\int _1 ^{\infty} \ln q \ (q - 1)^{-11/10} \mathrm dq.$$

Let's then integrate by parts ($$ \mathrm d f = \ln q$$) and one obtains

$$I = -\frac{5}{2}\int _1 ^{\infty} q^{-1} \ (q - 1)^{-1/10} \mathrm dq.$$

One more substitution: $$1 - \dfrac{1}{q} = p$$ and one obtains

$$I = -\frac{5}{2}\int _0 ^1 p^{-1/10} \ (1 - p)^{-9/10} \mathrm dp.$$

By definition of the Beta function one reads this as

$$I =-\frac{5}{2} \beta \left( \frac{9}{10}, \frac{1}{10} \right) = -\frac{5}{2} \frac{\pi}{\sin \tfrac{\pi}{10}} = -\frac{5\pi}{\phi}$$.

Note $$\sin(\pi/10)={1\over 2\phi}$$

$$-{5\over 2}\cdot {\pi\over \sin(\pi/10)}=-5\pi\phi$$
 
Suppose ,instead of the usual x,y coordinate system with an I basis vector along the x -axis and a corresponding j basis vector along the y-axis we instead have a different pair of basis vectors ,call them e and f along their respective axes. I have seen that this is an important subject in maths My question is what physical applications does such a model apply to? I am asking here because I have devoted quite a lot of time in the past to understanding convectors and the dual...
Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Back
Top