Closed form for $\int \limits _{1}^{\infty} \dfrac{1}{x-1} \cdot \dfrac{x^{\frac{1}{n}}-x^{\frac{-1}{n}}}{x^n} \mathrm{dx} \ \ ,\ \ n \in \mathbb{N} $ I have to find the closed form of
$\displaystyle \tag*{} \int \limits _{1}^{\infty} \dfrac{1}{x-1} \cdot \dfrac{x^{\frac{1}{n}}-x^{\frac{-1}{n}}}{x^n} \mathrm{dx} \ \ ,\ \ n \in \mathbb{N} $
I tried to express the integral in the form of $I_n (a)$ to do differentiation under integral. No matter where I place the $a$, the derivative becomes very complicated and hence difficult to solve. Please give me some hints or let me know whether there any other methods to solve. Thanks.
 A: Substituting $x=e^t$ gives the following:
$$\begin{align}I_n&=\int_1^{\infty} \frac{1}{x-1}\frac{x^{1/n}-x^{-1/n}}{x^n}~dx\\&=\int_0^{\infty} \frac{e^{t/n}-e^{-t/n}}{e^{tn}(e^t-1)}\cdot e^t~dt\\&=\int_0^{\infty} \frac{e^{-t(n-1/n)}-e^{-t(n+1/n)}}{1-e^{-t}}~dt. \end{align}$$
Now compare with integral representation for the digamma function (due to Gauss)
$$\psi(z)=\int_0^{\infty} \left(\frac{e^{-t}}{t}-\frac{e^{-zt}}{1-e^{-t}}\right)~dt$$
to obtain the same result as provided by @MariuszIwaniuk
$$I_n=\psi\left(n+\frac{1}{n}\right)-\psi\left(n-\frac{1}{n}\right).$$
Since $n$ is an integer with $n>1$ (the integral doesn't converge for $n=1$), one can simplify the above result to a form without non-elementary functions. To do so, note that by using the recurrence for the digamma function $\psi(z+1)=\psi(z)+1/z$, one obtains inductively
$$\psi\left(n+\frac{1}{n}\right)=\psi\left(\frac{1}{n}\right)+\sum_{k=0}^{n-1} \frac{1}{k+1/n},$$
$$\psi\left(n-\frac{1}{n}\right)=\psi\left(1-\frac{1}{n}\right)+n+\sum_{k=0}^{n-1} \frac{1}{k-1/n}.$$
Therefore, one has the following closed form
$$\begin{align} I_n&=\psi\left(\frac{1}{n}\right)-\psi\left(1-\frac{1}{n}\right)-n+\sum_{k=0}^{n-1} \left(\frac{1}{k+1/n}-\frac{1}{k-1/n}\right)\\&=-\pi\cot(\pi/n)-n+\sum_{k=0}^{n-1} \frac{2n}{1-k^2 n^2}\\&=\left[n+\sum_{k=1}^{n-1} \frac{2n}{1-k^2 n^2}\right]-\pi\cot(\pi/n), \end{align}$$
where we have used the reflection formula $\psi(1-z)+\psi(z)=\pi\cot(\pi z)$.
A: A nice and short method has been already provided by @projectilemotion. But what if we don't know about the integral representation of digamma function? My answer will give an alternate approach to reach the closed form.
We start with the substitution $x=1/t$. This transforms our integral to
$$I_n = \int_0^1 x^{n-1}(x^{-1/n}-x^{1/n})(1-x)^{-1}\,\mathrm dx$$
Now, we normalise the integral.
$$ I _n = \lim_{t \to0} \underbrace{\int_0^1 x^{n-1}(x^{-1/n}-x^{1/n})(1-x)^{t-1}\,\mathrm dx}_{\equiv J_n(t)}$$
It's is trivial to compute $J_n(t)$ using the beta and gamma function.
$$\begin{align}J_n(t) &= \int_0^1 x^{n-1/n-1}(1-x)^{t-1}-x^{n+1/n-1}(1-x)^{t-1}\,\mathrm dt \\ &= \mathcal B(n-1/n,t)-\mathcal B(n+1/n,t)\\ J_n(t) &= \frac{\Gamma (n-1/n)\Gamma(t)}{\Gamma(n-1/n+t)} -\dfrac{\Gamma(n+1/n)\Gamma (t)}{\Gamma(n+1/n+t)} \\ &= \frac { \Gamma(n-1/n)\Gamma(n+1/n+t)-\Gamma(n+1/n)\Gamma(n-1/n+t)}{\Gamma(n+1/n+t)\Gamma(n-1/n+t)\cdot \frac1{\Gamma(t)}} \end{align}$$
Passing the limit,
$$\begin{align}I_n &=\lim_{t\to0} \frac { \Gamma(n-1/n)\Gamma(n+1/n+t)-\Gamma(n+1/n)\Gamma(n-1/n+t)}{\Gamma(n+1/n+t)\Gamma(n-1/n+t)\cdot \frac1{\Gamma(t)}} \\ &= \frac1{\Gamma(n+1/n)\Gamma(n-1/n)} \cdot\underbrace{\lim_{t\to0} \frac{\Gamma(n-1/n)\Gamma(n+1/n+t)-\Gamma(n+1/n)\Gamma(n-1/n+t)}{\frac1{\Gamma(t)}} }_{=L}\end{align}$$
Noting that $L$ has the indeterminate form $\frac00$, we use L'Hopital's rule.
$$\begin{align}L &= \lim_{t\to0} \frac{\Gamma(n-1/n)\Gamma(n+1/n+t)\psi(n+1/n+t)-\Gamma(n+1/n)\Gamma(n-1/n+t)\psi(n-1/n+t)}{-\frac{\psi(t)}{\Gamma(t)}} \\ &= \Gamma(n-1/n)\Gamma(n+1/n)(\psi(n+1/n)-\psi(n-1/n)) \lim_{t\to0} \frac{\Gamma(t)}{-\psi(t)} \end{align}$$
The worries are finished!
As the Taylor expansion of gamma and digamma function is
$$\begin{align}\Gamma(z) &= \frac1z+o(z^0) \\ \psi(z) &= -\frac1z+o(z^0)\end{align}$$
the later limit just equals $1$. Thus
$$L = \Gamma(n-1/n)\Gamma(n+1/n)(\psi(n+1/n)-\psi(n-1/n))$$
And thus, we conclude that
$$\boxed{\boxed{ \int_1^\infty \frac{x^{1/n}-x^{-1/n}}{x^n(x-1)} \,\mathrm dx = \psi\Big(n+\frac1n\Big)-\psi\Big(n-\frac1n\Big)}}$$
