# Thread: Prove the values of the improper integrals.

1. ## Prove the values of the improper integrals.

Prove that

$\displaystyle \int_{0}^{\infty}x\frac{ ln(1+x^3)}{1+x^3}dx = ln\ 3 \ \frac{\pi}{\sqrt{3}} + \left(\frac{\pi}{3}\right)^2$

$\displaystyle \int_{0}^{\infty} \frac{ln(1+x^3)}{1+x^3} dx = ln\ 3 \ \frac{\pi}{\sqrt{3}} - \left(\frac{\pi}{3}\right)^2$

EDIT: Yep, I know the first one is a duplicate of the improper integral I posted last time, but what about the second one?

2. Originally Posted by mathwizard
Prove that

$\displaystyle \int_{0}^{\infty}x\frac{ ln(1+x^3)}{1+x^3}dx = ln\ 3 \ \frac{\pi}{\sqrt{3}} + \left(\frac{\pi}{3}\right)^2$

$\displaystyle \int_{0}^{\infty} \frac{ln(1+x^3)}{1+x^3} dx = ln\ 3 \ \frac{\pi}{\sqrt{3}} - \left(\frac{\pi}{3}\right)^2$

EDIT: Yep, I know the first one is a duplicate of the improper integral I posted last time, but what about the second one?
Let $\displaystyle J\left( \theta \right) = \int_0^\infty {\tfrac{{\ln \left( {1 + \theta ^3 \cdot x^3 } \right)}} {{1 + x^3 }}dx}$ differentiating by Leibniz rule: $\displaystyle J'\left( \theta \right) = 3 \cdot \theta ^{2} \cdot \int_0^\infty {\tfrac{{x^3 }} {{\left( {1 + \theta ^3\cdot x^3 } \right) \cdot \left( {1 + x^3 } \right)}}dx}$

Remember that: $\displaystyle \int_0^\infty {\tfrac{{dx}} {{1 + x^3 }}} = \tfrac{{2 \cdot \pi }} {{3 \cdot \sqrt 3 }}$ , calculate the derivative and integrate (considering that $\displaystyle J\left( 0 \right) = 0$) to get $\displaystyle J(1)$