integration – Integral from Mathematica’s documentation: $int_0^1 frac{log left(frac{1}{2} left(1+sqrt{4 x+1}right)right)}{x} , dx = frac{pi^2}{15} $


I like to puruse Mathematica’s documentation and look at the ‘Neat Examples’: this is one I managed to figure out. Apparently it’s due to Ramanujan:
$$
I=int_0^1 frac{log left(frac{1}{2} left(1+sqrt{4 x+1}right)right)}{x}
, dx = frac{pi^2}{15}
$$


Here are the steps for my solution:

  1. Make the substitution $x=y^2-y$, yielding
    $$
    I= int _{1}^{phi}frac{log(y)(2y-1)}{y(y-1)},dy,
    $$
    where $displaystyle{phi = frac{1+sqrt{5}}{2}}$ is the golden ratio.
  2. Factor out the $log(y)$ term and use partial fractions to write
    $$I = underbrace{int _{1}^{phi}frac{log(y)}{y},dy}_{I_1} + underbrace{int _{1}^{phi}frac{log(y)}{y-1},dy}_{I_2}
    $$
    $I_1$ can be evaluated using a simple substitution, yielding $displaystyle{I_1 = frac{log ^2(phi )}{2}}$.
  3. Use the Taylor series for $log(y)$ centered at $y=1$ and interchange the sum and integral to show
    $$
    I_2 = -sum_{k=1}^{infty} frac{(1-phi)^{k}}{k^2}= -sum_{k=1}^{infty} frac{(-phi^{-1})^{k}}{k^2}= – text{Li}_2(-phi^{-1})
    $$
  4. $text{Li}_2$ has the following properties:
  • $text{Li}_2(x) + text{Li}_2(-x) = frac{1}{2}text{Li}_2(x^2)$
  • $text{Li}_2(x) + text{Li}_2(1-x) = zeta(2) – log(x)log(1-x)$
  • $text{Li}_2(1-x) + text{Li}_2(1-x^{-1}) = -frac{1}{2}log^2(x)$

Put $x=phi^{-1}$ and use $phi^2=phi+1$; this gives:
$$
text{Li}_2(phi^{-1}) + text{Li}_2(-phi^{-1}) = frac{1}{2}text{Li}_2(1-phi^{-1})
$$

$$
text{Li}_2(phi^{-1}) + text{Li}_2(1-phi^{-1}) = zeta(2) -2 log^2(phi)
$$

$$
text{Li}_2(1-phi^{-1}) + text{Li}_2(-phi^{-1}) =-frac{1}{2}log^2(phi)
$$

5. Relabel for clarity. Let $A=text{Li}_2(phi^{-1})$, $B=text{Li}_2(-phi^{-1})$, $C=text{Li}_2(1-phi^{-1})$, and $L= log^2(phi)$. This gives the system
$$
begin{cases}
A+ B & = frac{1}{2}C\
A+ C&= zeta(2)- 2L\
C+B &= -frac{1}{2}L
end{cases}
$$
Solving gives $B=-I_2=displaystyle{frac{1}{2}L-frac{2}{5}Z}$, whence $displaystyle{I = frac{pi^2}{15}}.$


I’d be curious to see if there are any other methods of proof, perhaps involving simpler substitutions than the ones I used.