An integral with fractional part function

I'll take the general case, the integral is : $I_n= \sum_{k=1}^{\infty}\int_k^{k+1} \frac{1}{x(x-1)} \sqrt[n]{\frac{x-k}{1+k-x}} \ \mathrm{d}x \overset{t=x-k}{=} \sum_{k=1}^{\infty}\int_0^1 \sqrt[n] {\frac{t}{1-t}} \left(\frac{1}{t+k-1} - \frac{1}{t+k} \right) \ \mathrm{d}t$ Now, we interchange the sum integral (this is allowed by MCT since all summands are positive and the limit is integrable) to get : $I_n= \int_0^1 t^{\frac{1}{n} - 1 } (1-t)^{\frac{-1}{n}} \ \mathrm{d}t= \text{B}\left(\frac{1}{n}, 1- \frac{1}{n} \right)$ Where $\text{B}$ is the Beta function , now use the Beta-Gamma relation and the reflection formula to get : $I_n = \Gamma\left(\frac{1}{n} \right) \Gamma\left(1-\frac{1}{n} \right) = \boxed{\pi \csc \left(\frac{\pi}{n} \right)}$ In this case, $n=3$ gives us $\frac{2\pi}{\sqrt{3}}$ , then the answer is $\boxed{7}$ .

Remark : There is more elegant methods to prove the reflection formula using real and/or complex analysis, try to prove it alone.