# Values of $\int_{0}^{1}{\frac{dt}{1+t^n}}$ [closed]

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.

## closed as off-topic by GNUSupporter 8964民主女神 地下教會, RRL, Winther, Clayton, rogerlApr 18 at 19:49

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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 $$j_{n,k}=\int_0^1\frac{dx}{x-\lambda_{n,k}}=\ln\sqrt{2-2\cos\frac{\pi(2k+1)}n}+i\arg(1-\lambda_{n,k})-\frac{i\pi}{n}(2k+n+1)$$

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

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

• Thanks for this – HAMIDINE SOUMARE 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 $$-\frac{1}{n}e^{\frac{\pi i}{n}}$$

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