Compute $ \int_0^1\frac{{\rm d}x}{1+x^n} $

Question

$$I_n := \int_0^1\frac{{\rm d}x}{1+x^n} $$

It doesn't seem elementary to me. How do I solve this?

_{Related: Show that $\int_0^ \infty \frac{1}{1+x^n} dx= \frac{ \pi /n}{\sin(\pi /n)}$ , where $n$ is a positive integer.}

thanks.but can I change It from 0 to 1?will It be other methods to solve this?thanks for answering. — 1111XYZ, Oct 07 '20 at 07:15
Does this answer your question? Asymptotic expansion of $I_n=\int_0^1\frac{1}{1+x^n}dx$ — reyna, Oct 07 '20 at 07:28
Does this answer your question? Find the value of $\lim\limits_{n \to \infty}n\left(\left(\int_0^1\frac{1}{1+x^n},\mathrm{d}x\right)^n-\frac{1}{2}\right)$ (Note: solutions here include parts which might clear your doubt) — reyna, Oct 07 '20 at 07:33
Does this answer your question? Find the value of $\lim\limits_{n \to \infty}n\left(\left(\int_0^1\frac{1}{1+x^n},\mathrm{d}x\right)^n-\frac{1}{2}\right)$ — Toby Mak, Oct 07 '20 at 09:12

score 2 · Accepted Answer · answered Oct 07 '20 at 09:39

Depending on what you already know, it is a simple or difficult problem. One thing you could consider is that $$\frac 1 {1+x^n}=\prod_{k=1}^n \frac 1 {x-r_k}$$ where the $r_k$'s are the roots of unity.

Now, using partial fraction decomposition, you will have $$\frac 1 {1+x^n}=\sum_{k=1}^n \frac {a_k} {x-r_k}$$ which is easy to integrate, leading to $$\int_0^1 \frac {dx} {1+x^n}=\sum_{k=1}^n {a_k} \log \left(1-\frac{1}{r_k}\right)$$ This not really funny since the $r_k$ and the $a_k$ are complex numbers.

Now, this integral is in fact (do not worry, you will learn it soon) $$I_n=\frac{\psi\left(\frac{n+1}{2 n}\right)-\psi \left(\frac{1}{2 n}\right)}{2 n}$$ where appears the digamma function.

When $n$ is large $$I_n =1+\frac{\gamma +\psi \left(\frac{1}{2}\right)}{2 n}+\frac{\pi ^2}{12 n^2}+O\left(\frac{1}{n^3}\right)$$ with $\psi \left(\frac{1}{2}\right)=-1.96351$

Compute $ \int_0^1\frac{{\rm d}x}{1+x^n} $

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