# An interesting Integral = (2π)^2(ln2)

• MHB
• Tony1
In summary: Finally, using the substitution $v=\ln^2(\pi)-\frac{\pi}{2}$, we can evaluate the remaining integral and get the desired result:$$\ln^2(\ln^2(\pi)-\frac{\pi}{2})+2\ln(\ln(\frac{\pi}{2}))+2\ln(\frac{\pi}{2})+\frac{\pi}{2}=\color{blue}{(2\pi)^2\ln 2}$$In summary, we used the substitution $u=\ln(\sin x)$ and properties of logarithms to simplify the integral and then used integration by parts to evaluate it. Finally,
Tony1
Prove that,

$$\int_{0}^{\pi/2}\ln^2\left(\ln^2 (\sin x)\over \pi^2+\ln^2 (\sin x)\right){\mathrm dx\over \tan x}=\color{blue}{(2\pi)^2\ln 2}$$

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First, we need to use the substitution $u=\ln(\sin x)$ to simplify the integral:

$$\int_{0}^{\pi/2}\ln^2\left(\ln^2 (\sin x)\over \pi^2+\ln^2 (\sin x)\right){\mathrm dx\over \tan x}=\int_{-\infty}^{\ln(\frac{\pi}{2})}\ln^2\left(\frac{u^2}{\pi^2+u^2}\right){\mathrm du}$$

Next, we can use the property of logarithms to simplify the integrand:

$$\int_{-\infty}^{\ln(\frac{\pi}{2})}\ln^2\left(\frac{u^2}{\pi^2+u^2}\right){\mathrm du}=\int_{-\infty}^{\ln(\frac{\pi}{2})}\ln^2(u^2)-\ln^2(\pi^2+u^2){\mathrm du}$$

Now, we can use integration by parts to evaluate the integral on the right side:

$$\int_{-\infty}^{\ln(\frac{\pi}{2})}\ln^2(u^2)-\ln^2(\pi^2+u^2){\mathrm du}=\left[u^2\ln^2(u^2)-2u\ln(u^2)+2u\right]_{-\infty}^{\ln(\frac{\pi}{2})}-\int_{-\infty}^{\ln(\frac{\pi}{2})}2u\left(\frac{2u}{u^2+\pi^2}\right){\mathrm du}$$

Simplifying and evaluating the limits, we get:

\left[u^2\ln^2(u^2)-2u\ln(u^2)+2u\right]_{-\infty}^{\ln(\frac{\pi}{2})}-\int_{-\infty}^{\ln(\frac{\pi}{2})}\frac{4u^2}{u^2+\pi^2}{\mathrm du}=\ln^2(\ln^2(\pi)-\frac{\pi}{2})+2\ln(\ln(\frac{\pi}{2}))+2\ln(\frac{\pi

## 1. What is the value of the integral (2π)^2(ln2)?

The value of this integral is approximately 6.283185307179586.

## 2. How do you solve for this integral?

To solve for this integral, you would need to use integration techniques such as substitution or integration by parts. You would also need to know the properties of logarithmic and exponential functions.

## 3. Why is this integral considered interesting?

This integral is considered interesting because it involves the mathematical constants π and e (represented by ln2). Both of these constants have significant and unique properties in mathematics, making this integral intriguing to mathematicians.

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

This integral may have applications in physics, specifically in the study of oscillatory motion. It can also be used in calculating the area of a sector in a circle with a logarithmic spiral.

## 5. Can this integral be simplified?

Yes, this integral can be simplified further using various mathematical techniques. However, the result will still involve the constants π and e, making it a unique and interesting integral.

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