$\def\d{\mathrm{d}}$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.

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    $\begingroup$ Do you know complex analysis at all? This seem like it'd be much easier with those methods. $\endgroup$ – Mark Apr 24 '17 at 17:14
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    $\begingroup$ yeah complex analysis will be easier , unless you want to involve some beta,digamma functions $\endgroup$ – Siddhartha Apr 24 '17 at 17:15
  • $\begingroup$ Roots of unity are spread evenly over the unit circle and we can pair them complex conjugates with de Moivre to quadratic factors and do partial fraction decomposition so it does not have to get so very dirty using real methods. $\endgroup$ – mathreadler Apr 24 '17 at 17:26
  • $\begingroup$ thanks i know a little bit of complex analysis , residue theorem @Mark $\endgroup$ – user440024 Apr 24 '17 at 17:30
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    $\begingroup$ Related: a solution using complex analysis. $\endgroup$ – Frenzy Li Apr 25 '17 at 5:57

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 ... $$\mathcal{M}[f_{1}(t)]\bigg{|}_{s_{1}=1/n}=\pi \cot(\pi s)\bigg{|}_{s=s_{1}}=\pi \cot\left(\dfrac{\pi}{n}\right),0<Re(s)<1 $$ and for $f_{2}(t)$ at $s_{2}=\dfrac{2}{n}$ it will be $$\mathcal{M}[f_{2}(t)]\bigg{|}_{s_{2}=2/n}=\pi \cot(\pi s)\bigg{|}_{s=s_{2}}=\pi \cot\left(\dfrac{2\pi}{n}\right),0<Re(s)<1$$ $$I(n)=\dfrac{1}{n} \left[\mathcal{M}[f_{1}(t)]\bigg{|}_{s_{1}=2/n}- \mathcal{M}[f_{2}(t)]\bigg{|}_{s_{2}=2/n} \right]$$ $$I(n)=\dfrac{\pi}{n} \left[\cot\left(\dfrac{\pi}{n}\right)-\cot\left(\dfrac{2\pi}{n}\right)\right]$$ So, again we arrive at the same result i.e $$\Large \displaystyle\ \bbox[10pt, border:2pt solid #06f]{I(n)=\dfrac{\pi}{n}\csc\left(\dfrac{2 \pi}{n}\right)} \tag*{}$$

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    $\begingroup$ Using the fact that $~H_n~=~\displaystyle\int_0^1\frac{1-x^n}{1-x}~dx,~$ one can then write $~\displaystyle\int_0^1\frac{x^a-x^b}{1-x}~dx~=$ $$=~\int_0^1\frac{(1-x^b)-(1-x^a)}{1-x}~dx ~=~ \int_0^1\frac{1-x^b}{1-x}~dx - \int_0^1\frac{1-x^a}{1-x}~dx ~=~ H_b-H_a,$$ where the latter connections to the digamma function are well-known. $\endgroup$ – Lucian Sep 1 '17 at 16:45

I am going to assume that $n\in\{3,4,5,6,\ldots\}$. By splitting the integration range as $(0,1)\cup(1,+\infty)$ and applying the substitution $z\mapsto\frac{1}{z}$ on the second interval, we get that

$$ I_n=\int_{0}^{+\infty}\frac{1-z}{1-z^n}\,dz = \int_{0}^{1}\frac{z^{n-3}-z^{n-2}+1-z}{1-z^n}\,dz =\int_{0}^{1}\frac{(1-z)(z^3+z^n)}{1-z^n}\,dz$$ and by performing the substitution $z=u^{1/n}$ it follows that

$$ I_n = \frac{1}{n}\int_{0}^{1}\frac{(1-u^{1/n})(u^{3/n}+u)u^{1/n-1}}{1-u}\,du $$ where the perturbated integral $$ I_n^\varepsilon = \frac{1}{n}\int_{0}^{1}\frac{(1-u^{1/n})(u^{3/n}+u)u^{1/n-1}}{(1-u)^{1-\varepsilon}}\,du $$ can be computed in terms of the $\Gamma$ function due to the integral representation for the Beta function, for any $\varepsilon>0$. By considering the limit as $\varepsilon\to 0^+$ we get an expression involving different values of the $\psi$ function, that by the reflection formula for the $\psi$ function simplifies to

$$ I_n = \color{red}{\frac{\pi}{n\sin\frac{2\pi}{n}}}.$$


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