May be this question has already been asked here. I’m looking for differents methods for handling this integral.

Edit: I am looking for a closed form. Any suggestion or method is welcome. Initially I wanted to find a closed form of $$\sum_{n=0}^{\infty}{\frac{(-1)^n}{ns+1}}$$ which leads me to this integral.


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Notation used in this answer:

$\ln(x)$ is the real-valued natural logarithm. It admits values $x\in (0,\infty)$

$\log(x)=\ln|x|+i\arg(x)$ is the complex-valued natural logarithm, which admits values $x\in\Bbb C\setminus \{0\}$

And $\arg(x)$ is the complex argument of $x$.

Answer: Note that $1+x^n$ may be factored as $$1+x^n=\prod_{k=0}^{n-1}(x-\lambda_{n,k})$$ Where $$\lambda_{n,k}=\exp\left[\frac{i\pi}n(2k+1)\right]$$ Hence $$H(x)=\frac1{1+x^n}=\prod_{k=0}^{n-1}\frac1{x-\lambda_{n,k}}$$ Since we have now factored $H(x)$, we may do partial fractions, i.e. we may write

$$\prod_{k=0}^{n-1}\frac1{x-\lambda_{n,k}}=\sum_{k=0}^{n-1}\frac{\Gamma_{n,k}}{x-\lambda_{n,k}}$$ for some coefficients $\Gamma_{n,k}$. We find $\Gamma_{n,k}$ by multiplying both sides by $\prod_{j=0}^{n-1}(x-\lambda_{n,j})$ to get $$1=\sum_{k=0}^{n-1}\left[\Gamma_{n,k}\prod_{j=0\\ j\neq k}^{n-1}(x-\lambda_{n,j})\right]$$ so for any $r\in\{0,1,\dots,n-1\}$ we plug in $x=\lambda_{n,r}$. The LHS stays the same, and every term on the RHS vanishes except for the term with $k=r$. Hence $$1=\Gamma_{n,r}\prod_{j=0\\ j\neq r}^{n-1}(\lambda_{n,r}-\lambda_{n,j})$$ $$\Rightarrow \Gamma_{n,r}=\prod_{j=0\\ j\neq r}^{n-1}\frac1{\lambda_{n,r}-\lambda_{n,j}}$$ See this answer for a proof that $$\Gamma_{n,k}=-\frac{\lambda_{n,k}}{n}$$ Hence we have $$\frac1{1+x^n}=-\frac1n\sum_{k=0}\frac{\lambda_{n,k}}{x-\lambda_{n,k}}$$ Anyway, we may now integrate with our summation formula: $$ \begin{align} I_n&=\int_0^1\frac{dx}{1+x^n}\\ &=-\frac1n\int_0^1\sum_{k=0}^{n-1}\frac{\lambda_{n,k}dx}{x-\lambda_{n,k}}\\ &=-\frac1n\sum_{k=0}^{n-1}\lambda_{n,k}j_{n,k}\\ \end{align}$$ Where $$j_{n,k}=\int_0^1\frac{dx}{x-\lambda_{n,k}}=\ln\sqrt{2-2\cos\frac{\pi(2k+1)}n}+i\arg(1-\lambda_{n,k})-\frac{i\pi}{n}(2k+n+1)$$ Which I can prove to you in full detail if you'd like.

The evaluation of $j_{n,k}$:

We have that $$\frac{d}{dz}\log z=\frac1z$$ So we immediately have that $$j_{n,k}=\log(1-\lambda_{n,k})-\log(-\lambda_{n,k})$$ Then from $\log(x)=\ln|x|+i\arg x$ we have that $$j_{n,k}=\ln|1-\lambda_{n,k}|+i\arg(1-\lambda_{n,k})-\ln|-\lambda_{n,k}|-i\arg(-\lambda_{n,k})$$ From Euler's formula we have $\lambda_{n,k}=\cos\frac{\pi(2k+1)}n+i\sin\frac{\pi(2k+1)}n$, so $$|-\lambda_{n,k}|=|\lambda_{n,k}|=\sqrt{\cos^2\frac{\pi(2k+1)}n+\sin^2\frac{\pi(2k+1)}n}=1$$ Thus $\ln|-\lambda_{n,k}|=\ln1=0$, giving $$j_{n,k}=\ln|1-\lambda_{n,k}|+i\arg(1-\lambda_{n,k})-i\arg(-\lambda_{n,k})$$ And since $\arg(xy)=\arg(x)+\arg(y)$ we have that $$\arg(-\lambda_{n,k})=\arg(-1)+\arg(\lambda_{n,k})=\pi+\frac\pi{n}(2k+1)=\frac{\pi}{n}(2k+n+1)$$ Then we see that $$\begin{align} |1-\lambda_{n,k}|&=\sqrt{\left(1-\cos\frac{\pi(2k+1)}n\right)^2+\sin^2\frac{\pi(2k+1)}n}\\ &=\sqrt{1-2\cos\frac{\pi(2k+1)}n+\cos^2\frac{\pi(2k+1)}n+\sin^2\frac{\pi(2k+1)}n}\\ &=\sqrt{2-2\cos\frac{\pi(2k+1)}n} \end{align}$$ So $$j_{n,k}=\int_0^1\frac{dx}{x-\lambda_{n,k}}=\ln\sqrt{2-2\cos\frac{\pi(2k+1)}n}+i\arg(1-\lambda_{n,k})-\frac{i\pi}{n}(2k+n+1)$$

  • $\begingroup$ Please provide that prove in detail $\endgroup$ – DINEDINE Apr 18 at 18:11
  • $\begingroup$ @HAMIDINESOUMARE see my edit $\endgroup$ – clathratus Apr 18 at 20:06
  • $\begingroup$ Thanks a lot.$\quad$ $\endgroup$ – DINEDINE Apr 18 at 23:10
  • $\begingroup$ you are very welcome :) $\endgroup$ – clathratus Apr 19 at 0:31

With $t=\tan^{2/n}u,\,x=\sin^2 u$ your integral becomes $$\frac{2}{n}\int_0^{\pi/4}\tan^{2/n-1}udu=\frac{1}{n}\int_0^{1/2}x^{1/n-1}(1-x)^{-1/n}dx=\frac{1}{n}\operatorname{B}\left(\frac{1}{2};\,\frac{1}{n},\,1-\frac{1}{n}\right).$$

  • $\begingroup$ Thanks for this $\endgroup$ – DINEDINE Apr 17 at 20:31

If n is an integer greater than 1.

Your contour is a section of a circle with an angle of $e^{\frac {2\pi i}{n}}$

$\oint_\gamma f(t) \ dt = \int_0^{\infty} f(t) \ dt + \lim_\limits{R\to\infty} \int_0^{\frac {2\pi}{n}} f(Re^{it}) iRe^{it} \ dt - \int_0^{\infty} f(e^{\frac {2\pi i}{n}} t)e^{\frac {2\pi i}{n}} \ dt$

$f(e^{\frac {2\pi i}{n}} t) = f(t)$

if $n > 1$

$\lim_\limits{R\to\infty} \int_0^{\frac {\pi}{n}} f(Re^{it}) iRe^{it} \ dt = 0$ (and does not converge otherwise)

$\int_0^{\infty} f(t) \ dt = \frac{1}{1-e^{\frac {2\pi i}{n}}}\oint_\gamma f(t) \ dt$

There is one pole inside the contour, and it evaluates to $-\frac{1}{n}e^{\frac{\pi i}{n}}$

$\int_0^{\infty} f(t) \ dt = \frac {\pi\csc \frac {\pi}{n}}{n}$


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