How to find this limit;
$$\lim_{n\to \infty}\left[(n+1)\int_{0}^{1}x^n\ln(x+1)dx\right]$$
What techniques can I use here? Thanks for reading and help.
This question was asked in GATE 2008.
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1Maybe write $\ln(x+1)$ as a power series? – D_S Jan 22 '19 at 02:29
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@D_S Thanks for the help. I will try it out. – Shweta Aggrawal Jan 22 '19 at 02:30
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2Please give more context. Providing context not only assures that this is not simply copied from a homework assignment, but also allows answers to be better directed at where the problem lies and to be within the proper scope. Please avoid "I have no clue" questions. Defining keywords and trying a simpler, similar problem often helps. – robjohn Jan 22 '19 at 03:10
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This was asked in PhD entrance examination GATE 2008, @robjohn – Shweta Aggrawal Jan 22 '19 at 03:11
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1The idea is to put that information into the question. Often, questions lacking context are closed because of that. – robjohn Jan 22 '19 at 03:13
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1@robjohn I will edit the question. Thanks for pointing it out. – Shweta Aggrawal Jan 22 '19 at 03:14
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It seems that question about this limit has been asked before: Evaluate the limit of $(n+1)\int_0^1x^n\ln(1+x),dx$ when $n\to\infty$. I found the linked question using Approach0. More tips on searching: How to search on this site? – Martin Sleziak Jan 22 '19 at 04:26
2 Answers
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Hint:
You can use IBP to evaluate the integral as: $$\bigg[\frac{x^{n+1}\ln(x+1)}{n+1}\bigg]^1_0-\frac{1}{n+1}\int_0^1{\frac{x^{n+1}}{x+1}dx}$$
$$=\frac{1}{n+1}\bigg(\ln 2 -\int_0^1{\frac{x^{n+1}}{x+1}dx}\bigg)$$
This means you are looking for $$\lim_{n\to\infty}\bigg(\ln(2)-\int_0^1{\frac{x^{n+1}}{x+1}dx}\bigg)$$
Can you prove this integral goes to $0$?
Rhys Hughes
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Sorry I spent some time but can't see why this integral goes to zero. Can you give some hint? – Shweta Aggrawal Jan 22 '19 at 03:25
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1Remember that integrals calculate the area under the curve. If $f(x)=\frac{x^{n+1}}{x+1}$, can you see that $f(0)=0, f(1)=\frac12$ and $f(x)$ is increasing? Then you just need to show the values between these two get smaller as you increase $n$. This follows from $x^m<x^n$ for $x\in(0,1)$ if $m>n$ – Rhys Hughes Jan 22 '19 at 03:35
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1or Dominated Convergence since $\frac{x^{n+1}}{x+1}\le1$ or comparison with the integral of $x^{n+1}\le\frac{x^{n+1}}{x+1}$. – robjohn Jan 22 '19 at 11:05
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Too long for a comment but added for your curiosity (even if too complex).
You have received good answers already so I shall focus on the asymptotics of $$I_n=(n+1)\int_{0}^{1}x^n\ln(x+1)\,dx$$
First of all, if you know about the Gaussian hypergeometric function, there is an antiderivative $$\int x^n\ln(x+1)\,dx=\frac{x^{n+1} (\, _2F_1(1,n+1;n+2;-x)+(n+1) \log (x+1)-1)}{(n+1)^2}$$ and, assuming $n>0$, using the bounds $$\int_0^1 x^n\ln(x+1)\,dx=\frac{(n+1)\left( H_{\frac{n}{2}}- H_{\frac{n-1}{2}}\right)+n \log (4)+2\log (2)-2}{2 (n+1)^2}$$ Using the asymptotics of harmonic numbers $$H_p=\gamma +\log(p) +\frac{1}{2 p}-\frac{1}{12 p^2}+O\left(\frac{1}{p^4}\right)$$ and continuing with Taylor series, we should end with $$I_n=\log (2)-\frac{1}{2 n}+\frac{3}{4 n^2}+O\left(\frac{1}{n^3}\right)$$
For sure, the same result would be obtained using Rhys Hughes's answer since $$\int \frac{x^{n+1}}{x+1}\,dx=-\frac{x^{n+1} (\, _2F_1(1,n+1;n+2;-x)-1)}{n+1}$$ $$\int_0^1 \frac{x^{n+1}}{x+1}\,dx=\frac{1}{2} \left(H_{\frac{n+1}{2}}-H_{\frac{n}{2}}\right)$$
Claude Leibovici
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