integration - How to evaluate this improper integral $int_{0}^{infty}frac{1-x}{1-x^{n}},mathrm dx$?

$\def\d{\mathrm{d}}$How to evaluate this improper integral?
$$\int_{0}^{\infty}\frac{1-x}{1-x^{n}}\,\d x.$$

What I tried is a substitution i.e $x^{n}=t$, but then things got complicated, and I'm stuck.

Answer

Here are two methods,using real analysis

Method : 1 : using some special functions :
$$I=\displaystyle \int_{0}^{\infty}\dfrac{1-x}{1-x^{n}}\,dx=\underbrace{\displaystyle \int_{0}^{1}\dfrac{1-x}{1-x^{n}}\,dx}_{I_{1}(n)}+\underbrace{\displaystyle \int_{1}^{\infty}\dfrac{1-x}{1-x^{n}}\,dx}_{I_{2}(n)} $$
$$I_{1}(n)=\displaystyle \int_{0}^{1}\dfrac{1-x}{1-x^{n}}\,dx$$
Now lets make a substitution , $x^{n}=t \implies dx=\dfrac{1}{n}t^{\frac{1}{n}-1}dt$ , so $I_{1}(n)$ becomes
$$I_{1}(n)=\dfrac{1}{n}\displaystyle \int_{0}^{1}\dfrac{1-t^{\frac{1}{n}}}{1-t}.t^{\frac{1}{n}-1}\,dt= \dfrac{1}{n}\left[\displaystyle
\int_{0}^{1}\dfrac{t^{\frac{1}{n}-1}}{1-t}\,dt-\displaystyle \int_{0}^{1}\dfrac{{t^{\frac{2}{n}-1}}}{1-t}\,dt \right]$$
$$I_{1}(n)= \dfrac{1}{n}\displaystyle \lim_{m \rightarrow 0}\left[ \beta\left(m,\dfrac{1}{n}\right)-\beta \left(m,\dfrac{2}
{n}\right)\right] $$
$$\large \boxed{I_{1}(n)=\dfrac{1}{n} \left[\psi\left(\dfrac{2}{n}\right)-\psi \left(\dfrac{1}{n}\right)\right]}$$

now lets take $I_{2}(n)$
$$I_{2}(n)=\displaystyle \int_{1}^{\infty}\dfrac{1-x}{1-x^{n}}\,dx$$
Now for this one substitute $x=\dfrac{1}{t} \implies dx=-\dfrac{1}{t^{2}} dt$
$$I_{2}(n)=\displaystyle \int_{0}^{1}\dfrac{1-t}{1-t^{n}}t^{n-3}\,dt$$
Now make another substitution $t^{n}=u \implies dt=\dfrac{1}{n}u^{\frac{1}{n}-1}du$ , so the integral becomes
$$I_{2}(n)=\dfrac{1}{n}\displaystyle \int_{0}^{1}\dfrac{1-u^{\frac{1}{n}}}{1-u}\left(u^{\frac{1}{n}-1}\right)\left(u^{1-\frac{3}{n}}\right)\,du$$
$$I_{2}(n)= \dfrac{1}{n}\left[\displaystyle \int_{0}^{1}\dfrac{u^{-\frac{2}{n}}}{1-u}\,du-\displaystyle \int_{0}^{1}\dfrac{{u^{-\frac{1}{n}}}}{1-u}\,du \right]$$
$$I_{2}(n)= \dfrac{1}{n}\displaystyle \lim_{m \rightarrow 0}\left[ \beta\left(m,1-\dfrac{2}{n}\right)-\beta \left(m,1-\dfrac{1}{n}\right)\right]$$
$$\large \boxed{I_{2}(n)= \dfrac{1}{n}\left[ \psi\left(1-\dfrac{1}{n}\right)-\psi \left(1-\dfrac{2}{n}\right)\right]}$$
Now we just have to add $I_{1}(n)$ and $I_{2}(n)$

$$\begin{equation} I(n)=\dfrac{1}{n}\left[ -\psi\left(\dfrac{1}{n}\right)+\psi\left(1-\dfrac{1}{n}\right)+\psi\left(\dfrac{2}{n}\right)-
\psi \left(1-\dfrac{2}{n}\right)\right]\\= \dfrac{\pi}{n}\left[ \cot \left(\dfrac{\pi}{n}\right)-\cot \left(\dfrac{2\pi}{n}\right)\right]\\=
\dfrac{\pi}{n}\csc\left(\dfrac{2 \pi}{n}\right)\end{equation}$$
$$\Large \displaystyle\ \bbox[10pt, border:2pt solid #06f]{I(n)=\dfrac{\pi}{n}\csc\left(\dfrac{2 \pi}{n}\right)} \tag*{}$$

hete $\psi\Rightarrow $Digamma function (https://en.wikipedia.org/wiki/Digamma_function) and $\beta\Rightarrow $Beta function

(https://en.wikipedia.org/wiki/Beta_function#Derivatives).
and the reflection formula i used to simplify is known as Euler’s reflection formula which is …
$$\boxed{\psi(1-z)-\psi(z)=\pi \cot(\pi z)} $$

Method : 2 : Mellin Transform(http://mathworld.wolfram.com/MellinTransform.html) :
$$I(n)=\displaystyle \int_{0}^{\infty}\dfrac{1-x}{1-x^{n}}\,dx$$
so lets use the same substitution again ..
$x^{n}=t \implies dx=\dfrac{1}{n}t^{\frac{1}{n}-1}dt $ so the integral becomes
$$I(n)=\dfrac{1}{n} \displaystyle \int_{0}^{\infty}\dfrac{1-t^{\frac{1}{n}}}{1-t}t^{\frac{1}{n}-1}\,dt$$
$$I(n)=\dfrac{1}{n}\left[\underbrace{\displaystyle \int_{0}^{\infty}\dfrac{t^{\frac{1}{n}-1}}{1-t}\,dt}_{F_{1}(t)}- \underbrace
{\displaystyle \int_{0}^{\infty}\dfrac{t^{\frac{2}{n}-1}}{1-t}\,dt}_{F_{2}(t)}\right]$$
So again we have two separate integrals , now using melling transform we can evaluate these two very easily , so first lets define the

standard Mellin transform of a function let's say $f(t)$ , it is given by ..
$$\mathcal{M}[f(t)]=F(s)=\displaystyle \int_{0}^{\infty}f(t)t^{s-1}dt$$
So, now if we compare $F_{1}(t)$ with the above integral then we can see that ..
$s_{1}=\dfrac{1}{n}$ and $f_{1}(t)=\dfrac{1}{1-t}$
Now Mellin transform of $f_{1}(t)$ is well known, it is ...
$$\mathcal{M}[f_{1}(t)]\bigg{|}_{s_{1}=1/n}=\pi \cot(\pi s)\bigg{|}_{s=s_{1}}=\pi \cot\left(\dfrac{\pi}{n}\right),0and for $f_{2}(t)$ at $s_{2}=\dfrac{2}{n}$ it will be
$$\mathcal{M}[f_{2}(t)]\bigg{|}_{s_{2}=2/n}=\pi \cot(\pi s)\bigg{|}_{s=s_{2}}=\pi \cot\left(\dfrac{2\pi}{n}\right),0$$I(n)=\dfrac{1}{n} \left[\mathcal{M}[f_{1}(t)]\bigg{|}_{s_{1}=2/n}- \mathcal{M}[f_{2}(t)]\bigg{|}_{s_{2}=2/n} \right]$$
$$I(n)=\dfrac{\pi}{n} \left[\cot\left(\dfrac{\pi}{n}\right)-\cot\left(\dfrac{2\pi}{n}\right)\right]$$

So, again we arrive at the same result i.e
$$\Large \displaystyle\ \bbox[10pt, border:2pt solid #06f]{I(n)=\dfrac{\pi}{n}\csc\left(\dfrac{2 \pi}{n}\right)} \tag*{}$$