Background
When I met the integral
$$\int_0^1 \frac{\left(x^\phi-1\right)^2}{\ln ^2 x} d x\\$$
where $\phi$ is the golden ratio: $\phi^2= \phi+1, $
I was surprised by its simple and decent value though it is hard to tackle. I had tried some methods such as substitutions, integration by parts etc. and failed.
Then I tried Feynman’s trick by introducing the integral parametrized by $t$
$$I(t)=
\int_0^1 \frac{\left(x^t-1\right)^2}{\ln ^2 x} d x\\
$$
As usual differentiating $I(t) $ w.r.t. $t$ once and twice yields
$$
I^{\prime}(t)=\int_0^1 \frac{2\left(x^t-1\right) x^t}{\ln x}dx
$$and
$$
\begin{aligned}
I^{\prime \prime}(t) & =\int_0^1 \left(4x^{2 t}-2x^t\right) d x \\
& =\frac{4}{2 t+1}-\frac{2}{t+1}
\end{aligned}
$$
Noticing that $I(0)=I^{\prime}(0)=0$, we can easily integrating back to $I(t)$ in two steps.
$$
I^{\prime}(t)-I^{\prime}(0)=\int_0^t I^{\prime \prime}(u) d u=\int_0^t\left(\frac{4}{2 u+1}-\frac{2}{u+1}\right) du
$$
$$
I^{\prime}(t)=2\ln (2 t+1)-2 \ln (t+1)
$$
Similarly,
$$
\begin{aligned}
I(t)-I(0) & =\int_0^t I^{\prime}(u) d u =\int_0^t[2\ln (2 u+1)-2 \ln (u+1)] d u
\end{aligned}
$$
Using the result $\int \ln x d x=x \ln x-x+C$, we have
$$
\boxed{\int_0^1 \frac{\left(x^t-1\right)^2}{\ln ^2 x} d x =(2 t+1) \ln (2 t+1)-2(t+1) \ln (t+1)}
$$
Using $\phi^2= \phi+1 $ gives
$$I=I(\phi)= (2 \phi+1) \ln \left(\phi^3\right)-2(\phi+1) \ln \left(\phi^2\right)= (2 \phi-1) \ln \phi =(2 \phi-1) \ln \phi =\boxed{\sqrt 5 \ln \phi }$$
My Questions:
- Is there any method other than Feynman’s trick?
- Can we go further with the powers higher than 2?
Best Answer
$x=e^{-u}$ is easier for guessing. Consider $$I=\int_0^{\infty}\left(\frac{1-e^{-u\phi}}{u}\right)^2e^{-u}du=\int_0^{\infty}\left(\int_0^{\phi}\int_0^{\phi}e^{-ut_1-ut_2}dt_1dt_2\right)du=\int_0^{\phi}\int_0^{\phi}\frac{dt_1dt_2}{1+t_1+t_2}=\int_0^{\phi}[\log(1+t_1+\phi)-\log(1+t_1)]dt_1=(2\phi+1)\log (2\phi+1)-2(\phi+1)\log (\phi+1)$$ and we conclude as in the previous solution.
Of course the power $2$ can be replaced by $n$: all what we need are the primitives of $x^k\log x$ for $k=0,1,n-2.$