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

• MHB
• Tony1
In summary, the conversation discusses an integral, with the result being -5πφ. The golden ratio, φ, is mentioned and an error is questioned in the result. It is then clarified that the result is correct and derived from the value of sin(π/10) being 1/2φ. The summary also includes the equation used to calculate the result.
Tony1
$$\int_{0}^{\pi\over 2}{\ln(\sin^2 x)\over \sin(2x)}\cdot \sqrt[5]{\tan(x)}\mathrm dx=-5\phi \pi$$

$\phi$ 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!

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Is there an error in the result? I'm getting

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

Theia said:
Is there an error in the result? I'm getting

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

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

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

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

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

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

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

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

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

By definition of the Beta function one reads this as

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

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Theia said:
Taking your substitution $$\displaystyle u = \tan x$$, one obtains

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

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

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

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

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

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

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

By definition of the Beta function one reads this as

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

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

$$-{5\over 2}\cdot {\pi\over \sin(\pi/10)}=-5\pi\phi$$

## 1. What is the significance of "Five, Phi and Pi" in the integral?

The numbers five (5), phi (ϕ), and pi (π) are all important mathematical constants that have been studied and used by scientists and mathematicians for centuries. The combination of these three constants in one integral likely has a deeper mathematical meaning that would require further investigation to fully understand.

## 2. How do you solve an integral with multiple constants?

Solving an integral with multiple constants involves using various techniques such as substitution, integration by parts, and partial fractions. It is important to carefully analyze the given integral and choose the appropriate method to solve it.

## 3. Is there a real-world application for this integral?

While it may not have a direct real-world application, the use of multiple mathematical constants in one integral could potentially have applications in fields such as physics, engineering, and finance where complex mathematical models are used to solve real-world problems.

## 4. Can this integral be simplified further?

Without knowing the specific integral in question, it is difficult to determine if it can be simplified further. However, in general, it is always worth exploring different methods and techniques to see if an integral can be simplified or evaluated in a more efficient way.

## 5. How can we apply this integral to other areas of mathematics?

The combination of multiple mathematical constants in one integral may have implications in other areas of mathematics such as number theory, geometry, and calculus. Exploring the connections between different mathematical concepts can lead to new discoveries and advancements in the field of mathematics.

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