# Evaluating $\int_{0}^{\log\phi}\frac{u^2(e^{2u}+1)}{e^{2u}-1}du$, when $\phi$ is the golden ratio

When I do $$e^u=x+\sqrt{x^2+1}$$ in (17) page 9, see here, after some computations then I obtain $$\frac{\zeta(3)}{10}=\int_{0}^{\log(\phi)}\frac{u^2(e^{2u}+1)}{e^{2u}-1}du,$$ where $\phi=\frac{1+\sqrt{5}}{2}$ is the golden ratio and $\zeta(3)$ is the Apéry's constant.

I don't know if this exercise is in the literature:

Question. Can you prove what I say? This is, can you show $$\frac{\zeta(3)}{10}=\int_{0}^{\log(\phi)}\frac{u^2(e^{2u}+1)}{e^{2u}-1}du?$$ Then I can check if my computations were rights. Thanks in advance.

Hint

Try with those writings:

$$e^{2u} + 1 = e^u(e^u + e^{-u}) ~~~~~~~~~~~ e^{2u}-1 = e^u(e^u - e^{-u})$$

Thence the integral becomes

$$\int u^2\frac{e^u + e^{-u}}{e^u - e^{-u}}\ \text{d}u = \int u^2\text{cotanh}(u)\ \text{d}u$$

Another possibility

Collect a $e^{2u}$ up and down the fraction, to get:

$$\int_0^{\ln\phi}\ u^2\frac{1 + e^{-2u}}{1 - e^{-2u}}\ \text{d}u = u^2\cdot (1 + e^{-2u})\cdot \frac{1}{1 - e^{-2u}}\ \text{d}u$$

In which I write the fraction in that way to make you think about the geometric series. Indeed $e^{-2u} < 1$ in your range, so we can express the last fraction as the geometric series, gaining

$$\int_0^{\ln\phi}\ u^2 \cdot (1 + e^{-2u})\cdot \sum_{k = 0}^{+\infty} (-e^{-2u})^k\ \text{d}u$$

Arrangino a bit and you'll get:

$$\sum_{k = 0}^{+\infty} (-1)^k \left(\int_0^{\ln\phi}u^ke^{-2uk}\text{d}u + \int_0^{\ln\phi} u^ke^{-2u(1+k)}\ \text{d}u\right)$$

You should be able to continue; you will get some incomplete Gamma Function integrals!

• Shouldn't those integrals be over $u^2$? Jan 31, 2016 at 15:21