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We have to evaluate: $$\lim_{n\to\infty}nI_n$$ with $$I_n=\int_0^1\frac{x^n}{x^2+3x+2}\:dx.$$

There is an elegant way to solve this problem?

Here is all my steps:

  • My first ideea was to find a recurrence relation such that:

$$I_{n+2}+3I_{n+1}+2I_n=\frac{1}{n+1},\forall x\in\mathbb{N}$$

  • Next step I show that $\forall x\in[0,1]\Rightarrow I_{n}\ge I_{n+1}\ge I_{n+2}$

Therefore it involving that: $$6I_n\ge 4I_{n+1}+2I_n\ge\frac{1}{n+1},\forall x\in\mathbb{N}$$

As I said above $$6I_{n+2}\leq 4I_{n+2}+2I_n\leq\frac{1}{n+1}$$ $\Rightarrow \frac{n}{6(n+1)}\leq nI_n\leq\frac{n}{6(n-1)},\forall x\in\mathbb{N}$

Therefore by squeeze thereom:

$$nI_n\to\frac{1}{6}\:as\:n\to\infty$$

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    $\begingroup$ $$\implies I_{n+1}+3I_{n}+2I_{n-1}=\dfrac1n$$ $\endgroup$
    – lab bhattacharjee
    Commented Jun 1, 2015 at 6:15
  • $\begingroup$ I'd write the integrand as $\frac{x^n}{x+1}-\frac{x^n}{x+2}$ $\endgroup$
    – Someone
    Commented Jun 1, 2015 at 6:36
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    $\begingroup$ @Mann, I'm trying to show $$I_m=\dfrac1{6m}+O\left(\dfrac1{m^2}\right)$$ by setting $$I_n=\cdots+a_2n^2+a_1n+a_0+a_{-1}\dfrac1n+\cdots$$ $\endgroup$
    – lab bhattacharjee
    Commented Jun 1, 2015 at 6:39
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    $\begingroup$ Here is the general result :math.stackexchange.com/questions/128823/… $\endgroup$
    – user99914
    Commented Jun 1, 2015 at 7:03
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    $\begingroup$ @John: That was the idea I had in mind when I wrote my answer (+1) :-) $\endgroup$
    – robjohn ♦
    Commented Jun 1, 2015 at 7:28

2 Answers 2

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We may just integrate by parts, $$ \begin{align} I_n=\int_0^1\frac{x^n}{(x+1)(x+2)}dx&=\left. \frac{x^{n+1}}{(n+1)}\frac{1}{(x+1)(x+2)}\right|_0^1+\frac{1}{(n+1)}\int_0^1\frac{(2x+3)\:x^{n+1}}{(x+1)^2(x+2)^2}\:dx\\\\ &=\frac1{6(n+1)}+\frac{1}{n+1}\int_0^1\frac{(2x+3)}{(x+1)^2(x+2)^2}\:x^{n+1}dx\\\\ &=\frac1{6(n+1)}+\frac{1}{n+1}J_n \tag1 \end{align} $$ and one may observe that $$ 0\leq \int_0^1\frac{(2x+3)}{(x+1)^2(x+2)^2}\:x^{n+1}dx\leq \frac{(2\times1+3)}{(0+1)^2(0+2)^2}\int_0^1x^{n+1}dx $$ or $$ 0\leq J_n\leq \frac{5}{4}\frac{1}{(n+2)}. \tag2 $$ Then using $(1)$ and $(2)$ gives easily

$$ \lim_{n \to +\infty}nI_n=\frac16.$$

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    $\begingroup$ excellent answer. deserving an "upvote".... $\endgroup$
    – DeepSea
    Commented Jun 1, 2015 at 7:11
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    $\begingroup$ Evaluating $\frac{2x+3}{(x^2+3x+2)^2}$ at $0$ and $1$ for the bounds assumes that it is monotone on $[0,1]$, which is not obvious to me. Using the min of the numerator and max of the denominator and vice versa, gives $$\frac3{36}\frac1{n+1}\le J_n\le\frac54\frac1{n+1}$$ which is not as tight an estimate, but tight enough. $\endgroup$
    – robjohn ♦
    Commented Jun 1, 2015 at 7:20
  • $\begingroup$ @OlivierOloa one observation that $\int_0^1 x^{n+1}dx=\frac{1}{n+2}$ $\endgroup$
    – Lucas
    Commented Jun 1, 2015 at 21:30
  • $\begingroup$ @Lucas Yes, I will edit my answer. Thank you very much. $\endgroup$
    – Olivier Oloa
    Commented Jun 1, 2015 at 23:20
  • $\begingroup$ @OlivierOloa can you explain why you multiply at numerator with different number than denominator ? can you give some details about this method ? $\endgroup$
    – Lucas
    Commented Jun 2, 2015 at 14:42
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Substitute $x\mapsto x^{1/(n+1)}$ and use Dominated Convergence: $$ \begin{align} n\int_0^1\frac{x^n}{x^2+3x+2}\,\mathrm{d}x &=\frac{n}{n+1}\int_0^1\frac1{x^2+3x+2}\,\mathrm{d}x^{n+1}\\ &=\frac{n}{n+1}\int_0^1\frac1{x^{2/(n+1)}+3x^{1/(n+1)}+2}\,\mathrm{d}x\\ &\to1\int_0^1\frac1{1+3\cdot1+2}\,\mathrm{d}x\\ &=\frac16 \end{align} $$

A More Basic Approach $$ \begin{align} \frac16-\left(n\int_0^1\frac{x^n}{x^2+3x+2}\,\mathrm{d}x\right) &=\int_0^1\left(\frac16-\frac{x^{1/n}}{x^{2/n}+3x^{1/n}+2}\right)\,\mathrm{d}x\tag1\\ &=\frac16\int_0^1\left(\frac{x^{2/n}-3x^{1/n}+2}{x^{2/n}+3x^{1/n}+2}\right)\,\mathrm{d}x\tag2\\ &\le\frac1{12}\int_0^1\left(x^{2/n}-3x^{1/n}+2\right)\,\mathrm{d}x\tag3\\[3pt] &=\frac1{12}\left(\frac{n}{n+2}-3\frac{n}{n+1}+2\right)\tag4\\[6pt] &=\frac{7n-2}{12(n+1)(n+2)}\tag5 \end{align} $$ Explanation:
$(1)$: substitute $x\mapsto x^{1/n}$ then bring the $\frac16$ inside the integral
$(2)$: algebra; $x^{2/n}-3x^{1/n}+2=\left(x^{1/n}-1\right)\left(x^{1/n}-2\right)\ge0$ on $[0,1]$
$(3)$: since the integrand is positive, bound it by replacing the denominator with its minimum
$(4)$: integrate
$(5)$: simpllfy

Therefore, $$ \frac16-\frac{7n-2}{12(n+1)(n+2)}\le\left(n\int_0^1\frac{x^n}{x^2+3x+2}\,\mathrm{d}x\right)\le\frac16\tag6 $$ Apply the Squeeze Theorem to get $$ \lim_{n\to\infty}n\int_0^1\frac{x^n}{x^2+3x+2}\,\mathrm{d}x=\frac16\tag7 $$

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  • $\begingroup$ Can you justify taking the limit inside the integral? $\endgroup$
    – Rory Daulton
    Commented Jun 1, 2015 at 7:00
  • $\begingroup$ @RoryDaulton Is it not enough that limits commute with Riemann integration? $\endgroup$
    – Anthony Peter
    Commented Jun 1, 2015 at 7:02
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    $\begingroup$ @RoryDaulton: yes, using Dominated Convergence $\endgroup$
    – robjohn ♦
    Commented Jun 1, 2015 at 7:03
  • $\begingroup$ @robjohn I don't think the OP will be familiar with DCT or Lebesgue integration, so it may not ultimately be helpful $\endgroup$
    – Anthony Peter
    Commented Jun 1, 2015 at 7:03
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    $\begingroup$ @AnthonyPeter: they tagged the question real-analysis, so I used a basic tool of real analysis. If they don't know Dominated Convergence, they can look it up at the link provided and learn something new. $\endgroup$
    – robjohn ♦
    Commented Jun 1, 2015 at 7:06

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