5

How do prove this triple integral? $$\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\frac{1}{2-zx^{2}-zy^{2}}dxdydz=\ln(2^{G})$$ where G is Catalan's constant.

As my try I only reach to this hard single integral: $$\int _0^1\frac{\operatorname{Li}_2\left(\frac{1+x^2}{2}\right)}{1+x^2}dx=G\ln \left(2\right)$$

Quanto
  • 97,352
asmatqatea
  • 142
  • 2
    Welcome to [math.se] SE. Take a [tour]. You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an [edit]): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance. – Another User Nov 25 '22 at 19:34
  • 3
    @Invisible Changing the order of integral does not make any sense. because all 3 integral have same orders. – asmatqatea Nov 25 '22 at 19:41
  • 1
    It does make sense as the integrand is not symmetric. It may be fruitful to first integrate wrt $z$ – FShrike Nov 25 '22 at 19:53

1 Answers1

4

Integrate by parts

$$I=\int _0^1\frac{\operatorname{Li}_2\left(\frac{1+x^2}{2}\right)}{1+x^2}dx = \frac{\pi^3}{24}+\int_0^1\frac{2x\tan^{-1}x\ln\frac{1-x^2}2}{1+x^2}dx$$ Then, utilize $\int_0^1 \frac{x \tan^{-1} x \ln \left( 1-x^2\right)}{1+x^2}dx= -\frac{\pi^3}{48}-\frac{\pi}{8}\ln^2 2 +G\ln 2$ and $\int_0^1 \frac{x\tan^{-1}x}{1+x^2}dx = \frac12G-\frac\pi2\ln2$ to obtain $I= G\ln2$.

Quanto
  • 97,352