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$$\int_{0}^{1}\dfrac{x^{46}}{\sum_{n=0}^{100}x^n}\mathrm{d}x$$

How to integrate the above expression?
I am trying the above question by breaking the denominator.
I broke the denominator like $\dfrac{1-x^{101}}{1-x}$. Now $(1-x)$ will go into the numerator.
Therefore, the above integrand will become $\dfrac{x^{46}(1-x)}{1-x^{101}}$.
But I don't know how to approach further. I don't know how to factorize the denominator further.
Please help me out.

Cameron Williams
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DEEPANWITA ROY
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    Factoring the denominator takes you right back where you started. – Ted Shifrin Sep 19 '23 at 15:18
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    @Surb your comment has two errors. (1) is uses $n$ in two ways, and (2) it forgets the $n+1$ on the right side. – Thomas Andrews Sep 19 '23 at 15:20
  • But how to approach further @Surb. I have already approached and written what you are saying. – DEEPANWITA ROY Sep 19 '23 at 15:21
  • Where did this question come from? What makes you think there's an elementary integral? – Ted Shifrin Sep 19 '23 at 15:22
  • The roots of the denominator are $e^{i2\pi ki/101}$ for $k=1,\dots 100.$ – Thomas Andrews Sep 19 '23 at 15:23
  • @ThomasAndrews One $i$ too many :P – Ted Shifrin Sep 19 '23 at 15:24
  • @ Ted Shifrin I think there exists an elementary integral. – DEEPANWITA ROY Sep 19 '23 at 15:27
  • You think? With what justification? – Ted Shifrin Sep 19 '23 at 15:29
  • @Ted Shifrin, I think because the integrand is x^46(1-x)/(1-x^101) and the limit is from 0 to 1. So, I think the integration will exist. – DEEPANWITA ROY Sep 19 '23 at 15:33
  • Of course, the Riemann integral exists. That has nothing to do with why one ought to be able to calculate it explicitly. You haven't told us where you got this and you certainly have not answered the question I've asked several times. – Ted Shifrin Sep 19 '23 at 15:35
  • @ Ted Shifrin, Then how to factorize the above summation of the denominator without breaking it in the terms of (1-x^n)/(1-x). And what do you mean by elementary integral? – DEEPANWITA ROY Sep 19 '23 at 15:41
  • Is there a reason to expect an easy formula here? The partial fractions are a mess, either way you do them. – Thomas Andrews Sep 19 '23 at 15:56
  • Where did you get the question from? If you got it from a specific resource the chances are there is some trick to get a nice answer. If not and you just made the question up, the most likely scenario is that there is no nice form for the answer and it'll just end up being a big (yet finite) sum. Mind you, in comparison to most integrals, the fact it is algorithmically computable into a finite sum is nice because most integrals don't even have a closed form – Riemann'sPointyNose Sep 19 '23 at 15:59
  • Surprisingly, Wolfram will actually spit out a (tediously long) answer to the integral, which does indeed look like it ended up simply using partial fractions: see here. If you want to, you can simply then plug in ${x=1}$ and ${x=0}$ and subtract the two to get the definite integral, but the answer is not nice looking! – Riemann'sPointyNose Sep 19 '23 at 16:11
  • The answer is $$\frac{1}{101}\left(\psi(\tfrac{48}{101})-\psi(\tfrac{47}{101})\right)$$ I doubt this can be further simplified. – jjagmath Sep 19 '23 at 16:15
  • You can check the above question in desmos. It is giving a value. – DEEPANWITA ROY Sep 19 '23 at 16:19
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    Of course it's giving a numerical value. Computers/calculators do numerical evaluation of integrals. – Ted Shifrin Sep 19 '23 at 16:29
  • But how to get that value without using computers and calculators. I have made this question on my own. – DEEPANWITA ROY Sep 19 '23 at 16:37

2 Answers2

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$$I_{m,n}=\int\frac{x^{m}}{\sum_{k=0}^{n}x^k}\,dx=\int \frac{(1-x) x^m}{1-x^{n+1}}\,dx$$ The antiderivative is given in terms of the Gaussian hypergeometric function $$I_{m,n}=x^{m+1} \left(\frac{\, _2F_1\left(1,\frac{m+1}{n+1};\frac{m+n+2}{n+1};x^{n+1}\right)} {m+1}-\frac{x \, _2F_1\left(1,\frac{m+2}{n+1};\frac{m+n+3}{n+1};x^{n+1}\right)} {m+2}\right)$$

The definite integral is given in terms of the digamma function

$$J_{m,n}=\int_{0}^{1}\frac{x^{m}}{\sum_{k=0}^{n}x^k}\,dx=\frac{\psi \left(\frac{m+2}{n+1}\right)-\psi \left(\frac{m+1}{n+1}\right)}{n+1}$$

Then, for your specific case, the result already given by @jjagmath in comments.

If $n$ is large, we can expand the result as series $$J_{m,n}=\frac{1}{(m+1) (m+2)}+\frac {\pi^2}{6(n+1)^2}-\frac{ (2 m+3) \zeta (3)}{(n+1)^3}+\frac{\pi^4 (3 m^2+9 m+7 ) }{90(n+1)^4}-\frac{(2m+3)(2 m^2+6 m+5) \zeta (5)}{(n+1)^5}+O\left(\frac{1}{n^6}\right)$$

For your case, this truncated series would give $0.000521782$ corresponding to a relative error of $3$%.

Claude Leibovici
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Partial answer.

The roots of $p(x)=\sum_{j=0}^{100} x^j$ are $x_k=e^{2i\pi k/101}$ for $k=1,\dots,100.$

Note that $x_k=x_1^k,$ and that $\overline{x_k}=x_{101-k}=x_k^{-1}.$

Letting $$p_k(x)=(x-x_k)(x-x_{101-k})=x^2-2\cos(2\pi k/101)x+1,$$ we can factorize $p(x)$ as:

$$p(x)=\prod_{k=1}^{50} p_k(x)$$

Let $q_k(x)=p(x)/p_k(x).$ Then $q_k(x_k)=p''(x_k)$ and $q_k(x_{101-k})=p''(x_{101-k}).$

Writing $$\frac{x^{46}}{p(x)}=\sum_{k=1}^{50}\frac{a_kx+b_k}{p_k(x)}$$ you get

$$\frac{x_k^{46}}{p''(x_k)}=a_kx_k+b_k\\ \frac{x_{k}^{-46}}{p''(x_{k}^{-1})}=a_kx_{k}^{-1}+b_k.$$

You can solve for $a_k$ and $b_k.$ They will be real numbers.

I don't see an easier way to get the answer. But I don't see an obvious way to get closed formula for $a_k,b_k.$ The hard part seems to me to be evaluating $p''.$

Then you get the integral is $$\sum_{k=1}^{50}\int_{0}^1\frac{a_kx+b_k}{p_k(x)}dx, $$

These will involve $\log$ and $\tan^{-1}.$ In general, for $\theta\in(0,\pi),$ $$\int_{0}^{1}\frac{Ax+B}{(x-\cos\theta)^2+\sin^2\theta}\,dx=\frac A2\log(2-2\cos\theta)+\frac{A\cos\theta+B}{\sin \theta}\left(\pi-\frac32\theta\right)$$

Thomas Andrews
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  • But how to find the exact summation. I think we can write 1/(1-x^101) as summation from n=0 to infinity (x^(101n)) – DEEPANWITA ROY Sep 19 '23 at 17:16
  • You can certainly take that approach. But you won't get a close formula, I suspect. @DEEPANWITAROY I've told you how to get the exact value, but it requires a lot of work. You have to find $a_k$ and $b_k.$ You can actually find just $a_1$ and $b_1$ and then find the images of these values under various automorphism,s, but it is a lot of work. Why do you think there will be an approach which doesn't? – Thomas Andrews Sep 19 '23 at 17:23
  • Your approach gives you $$\sum_{n=0}^{\infty}\left(\frac1{46+101n}-\frac1{47+101n}\right)=\sum_{n=0}^{\infty}\frac1{(46+101n)(47+101n)}.$$ But that doesn't give you a closed formula. – Thomas Andrews Sep 19 '23 at 17:26
  • Is there any formula to find this summation? – DEEPANWITA ROY Sep 19 '23 at 17:33
  • "Is there any formula to find this summation?" @DEEPANWITAROY I literally just answered that question. Can you read my comments before asking further questions? There might be a formula in terms of special function, but those functions are essentially defined as sums like these. – Thomas Andrews Sep 19 '23 at 17:35
  • Sorry. Kindly don't mind anything. – DEEPANWITA ROY Sep 20 '23 at 02:13