7

How to integrate $$\int_0^1 \frac{x^{49}}{1+x+x^2+\cdots+ x^{100}}\;dx$$

I am trying by breaking the summation in the denominator.$$\sum_{i=0}^{100} x^i=\frac{1-x^{101}}{1-x}$$Therefore the above integrand will become $$\frac{x^{49}(1-x)}{1-x^{101}}$$ But how to approach further? As far as I can guess the only way to proceed after this is by applying the Maclaurian series expansion of $1/(1-x)$; which is $$\sum_{i=0}^n x^i$$. But the Series expansion of $1/(1-x)$ is applicable only when $-1 \le x \lt 1$. In the integration, the limit is from $0$ to $1$, i.e. we have to find the area in the interval $[0,1]$. So, how to proceed further? Please help me out.

Chinnapparaj R
  • 11,989
Shivam Kumar
  • 381
  • 6
    The series expansion works for $-1<x<1$, so it also works for $0\leq x <1$, the point $x=1$ has measure zero so it doesn't matter. You can use the series expansion here – Yuriy S Oct 15 '23 at 13:42
  • I haven't understood what are you trying to say@Yuriy S. What do you mean by x=1 has measure 0? We have to find the area of the above function from [0,1]. – Shivam Kumar Oct 15 '23 at 14:31
  • Though you are correct @Yuriy S. If the series expansion works for -1<x<1, so it also works for 0$\le$x<1 – Shivam Kumar Oct 15 '23 at 14:32
  • Also after applying the Maclaurian series expansion of 1/(1-x), I am unable to find the summation of $$\sum_{n=0}^{\infty} 1/(50+101n)-1/(51+101n)$$ – Shivam Kumar Oct 15 '23 at 14:36
  • @ShivamKumar why do you belive this sum has a closed form? Where is this problem from? – Sine of the Time Oct 15 '23 at 15:19
  • I don't know that whether the summation has a closed form or not. But the fact is that finally I am getting this summation after solving this integration. – Shivam Kumar Oct 15 '23 at 15:35
  • With Mathematica: $\int_0^1 \frac{(-1+x) x^{49}}{-1+x^{101}} , dx=\frac{1}{101} \left(\psi ^{(0)}\left(\frac{51}{101}\right)-\psi ^{(0)}\left(\frac{50}{101}\right)\right)=\frac{1}{101} \pi \tan \left(\frac{\pi }{202}\right)$ – Mariusz Iwaniuk Oct 15 '23 at 16:10
  • Using Substitution we arrive at: $$\frac{1}{101}\int_{0}^{1}\left(\frac{\ 1-t^{-\frac{50}{101}}}{1-t}-\frac{1-t^{-\frac{51}{101}}}{1-t}\right)dt$$ Then using Digamma Reflection Formula we arrive at the result mentioned by Mariusz Iwaniuk. – Miracle Invoker Oct 15 '23 at 16:34
  • Ok Thanks to everyone for helping me out with this question. – Shivam Kumar Oct 16 '23 at 03:56
  • @SineoftheTime it has as we know that $\psi(1+n) = H_n - \gamma$ –  Oct 25 '23 at 11:30

2 Answers2

4

Note that the Taylor expansion

$$ \frac{1}{1-x} = 1+x+x^2+x^3+\ldots $$

converges only when $|x|<1$. In the integral, $x$ goes from $0$ to $1$, as you've acknowledged. Thankfully, the point $x=1$ has zero measure. This doesn't affect anything and you can use the expansion safely.

Here's the hard part — proving the fascinating equality down below.

$$ \bbox[15px,border:4px groove red]{\int_{0}^{1}\frac{x^{49}\left(1-x\right)}{1-x^{101}}dx = -\frac{\pi}{101}\cot\left(\frac{51\pi}{101}\right)} $$

Let $\mathcal{I}$ be the integral in question. We use the substitution $x = u^{1/101}$ and some non-intuitive algebra manipulations to get

$$ \begin{align} \mathcal{I} &= \int_{0}^{1}\frac{u^{49/101}\left(1-u^{1/101}\right)}{1-u}\cdot\frac{1}{101}u^{-100/101}du \\ &= \frac{1}{101}\int_{0}^{1}\left(\frac{u^{-51/101}}{1-u}-\frac{u^{-50/101}}{1-u}\right)du \\ &= \frac{1}{101}\int_{0}^{1}\left(\frac{u^{-51/101}}{1-u}-\frac{u^{-50/101}+1-1}{1-u}\right)du \\ &= \frac{1}{101}\int_{0}^{1}\left(\frac{1-u^{-50/101}}{1-u}-\frac{1-u^{-51/101}}{1-u}\right)du \\ &= \frac{1}{101}\int_{0}^{1}\frac{1-u^{\frac{51}{101}-1}}{1-u}du-\frac{1}{101}\int_{0}^{1}\frac{1-u^{\frac{50}{101}-1}}{1-u}du\,. \\ \end{align} $$

Next, we use one of the integral representations for the Digamma Function, which is

$$ \psi(z) = -\gamma + \int_0^1 \frac{1-x^{z-1}}{1-x}dx\,, $$

where $\Re(z) > 0$, I think, and $\gamma$ is the Euler-Mascheroni Constant. The functional relation is proven in this YouTube video by the YouTuber named Flammable Maths. We proceed further by writing

$$ \begin{align} \mathcal{I} &= -\frac{\gamma}{101} + \frac{1}{101}\int_{0}^{1}\frac{1-u^{\frac{51}{101}-1}}{1-u}du + \frac{\gamma}{101} - \frac{1}{101}\int_{0}^{1}\frac{1-u^{\frac{50}{101}-1}}{1-u}du \\ &= \frac{1}{101}\left(\psi\left(\frac{51}{101}\right) - \psi\left(\frac{50}{101}\right)\right) \\ &= \frac{1}{101}\left(\psi\left(\frac{51}{101}\right) - \psi\left(1-\frac{51}{101}\right)\right)\,. \end{align} $$

We end with the Digamma Reflection identity,

$$ \psi(1-z) - \psi(z) = \pi \cot(\pi z)\,, $$

which is derived in this YouTube video (timestamp 8:57 to 18:37) by the YouTuber named Gamma Digamma.

Finally, the answer is

$$ \bbox[15px,border:4px groove green]{\int_0^1 \frac{x^{49}}{1+x+x^2+\cdots+ x^{100}}\;dx = -\frac{\pi}{101}\cot\left(\frac{51\pi}{101}\right)} $$

and we're done!

Cheers ❤️

Accelerator
  • 6,203
1

In general, $$\boxed{\int_0^1 \frac{x^{n-1}}{1+x+\ldots+x^{2 n}} d x=\frac{\pi}{2 n+1} \cot \frac{n \pi}{2 n+1}}$$ Proof: $$ \begin{aligned} &\int_0^1 \frac{x^{n-1}}{1+x+x^2+\cdots+x^{2n}} d x \\= & \int_0^1 \frac{x^{n-1}-x^{n}}{1-x^{2n+1}} d x \\ = & \int_0^1 \frac{\left(1-x^{n}\right)-\left(1-x^{n-1}\right)}{1-x^{2n+1}} d x \\ = & \int_0^1 \frac{1-x^{n}}{1-x^{2n+1}} d x-\int_0^1 \frac{1-x^{n-1}}{1-x^{2n+1}} d x \\ = & \frac{1}{2n+1}\left[\psi\left(\frac{n+1}{2n+1}\right)-\psi\left(\frac{1}{2n+1}\right)-\psi\left(\frac{n}{2n+1}\right)+\psi\left(\frac{1}{2n+1}\right)\right] \cdots (1)\\ = & \frac{1}{2n+1}\left[\psi\left(\frac{n+1}{2n+1}\right)-\psi\left(\frac{n}{2n+1}\right)\right] \\ = & \frac{\pi}{2n+1} \cot \frac{n \pi}{2n+1} \cdots (2) \end{aligned} $$ where $(1)$ comes from my post, and $(2)$ from the reflection property of digamma function: $$\psi(1-x)-\psi(x) =\pi \cot (\pi x) \quad \forall x\notin \mathbb{Z}.$$

In particular, $$ \int_0^1 \frac{x^{49}}{1+x+x^2+\cdots+x^{100}} d x= \frac{\pi}{101} \cot \left(\frac{50 \pi}{101}\right) $$

Lai
  • 31,615