Given: $$\ I(n) = \int_{0}^1 \frac {nx^{n-1}}{1+x}dx$$ Find $\lim_{n \to\infty} I(n)$.
Please help me with this. I cannot find any way to do this.
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2Please include your work, up till which point you could go, and where you got stuck. – Mar 18 '18 at 15:49
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i cannot think of any way to this .. – Satyam Kumar Mar 18 '18 at 15:50
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i used sandwich but am not sure of whether its right to do or not and by that way I am getting zero as answer – Satyam Kumar Mar 18 '18 at 15:52
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1What is $nx^{n-1}$? Have you seen this before? – Arnaud Mortier Mar 18 '18 at 15:53
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$I(n)$ is a definite integral and $x$ is a dummy variable. You don't need $0<x<1$ – CY Aries Mar 18 '18 at 15:53
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its given in the question – Satyam Kumar Mar 18 '18 at 15:54
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https://math.stackexchange.com/questions/292251/limit-of-s-n-int-limits-01-fracnxn-11x-dx-as-n-to-infty – StubbornAtom Jan 27 '20 at 18:38
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Hint Use integration by parts. Indeed, $$ \int_{0}^1\frac{nx^{n-1}}{1+x}\,dx=\left[\frac{x^n}{1+x}\right]_{0}^1+\int_{0}^1\frac{x^n}{(1+x)^2}\,dx $$ Let $n\to\infty$ and apply the dominated convergence theorem to the second integral on the RHS.
robjohn
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Sri-Amirthan Theivendran
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can you elaborate it .. I dont know dominated convergence theorem – Satyam Kumar Mar 18 '18 at 15:59
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2@SatyamKumar this is why it is so important to include context and your efforts. Without it, you will get answers, (that users work hard to write!) which do not help you – operatorerror Mar 18 '18 at 17:18
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@satyamkumar One doesn't need to apply the DCT. The remaining integral, call it $f_n$,.satisfies the inequalities $\frac1{4(n+1)}\le f_n\le \frac1{n+1}$. Do you understand why that is true? – Mark Viola Mar 18 '18 at 17:57
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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ $\ds{\Large\left. a\right)}$
With Laplace's Method:
\begin{align} \lim_{n \to \infty}\int_{0}^{1}{nx^{n - 1} \over 1 + x}\,\dd x & = \lim_{n \to \infty}\bracks{n\int_{0}^{1}{\pars{1 - x}^{n - 1} \over 2 - x} \,\dd x} = \lim_{n \to \infty}\bracks{n\int_{0}^{1} {\exp\pars{\bracks{n - 1}\ln\pars{1 - x}} \over 2 - x}\,\dd x} \\[5mm] & = \lim_{n \to \infty}\bracks{n\int_{0}^{\infty} {\exp\pars{-\bracks{n - 1}x} \over 2 - 0}\,\dd x} = {1 \over 2}\lim_{n \to \infty}\pars{n\,{1 \over n - 1}} = \bbx{1 \over 2} \end{align}
$\ds{\Large\left. b\right)}$ \begin{align} \lim_{n \to \infty}\int_{0}^{1}{nx^{n - 1} \over 1 + x}\,\dd x & = \lim_{n \to \infty}\bracks{n\int_{0}^{1}{x^{n - 1} - x^{n} \over 1 - x^{2}} \,\dd x} = \lim_{n \to \infty}\bracks{n\int_{0}^{1}{x^{n/2 - 1} - x^{n/2 - 1/2} \over 1 - x}\,{1 \over 2}\dd x} \\[5mm] & = {1 \over 2}\lim_{n \to \infty}\bracks{n\pars{% \int_{0}^{1}{1 - x^{n/2 - 1/2} \over 1 - x}\,\dd x - \int_{0}^{1}{1 - x^{n/2 - 1} \over 1 - x}\,\dd x}} \\[5mm] & = {1 \over 2}\lim_{n \to \infty}\bracks{n\pars{H_{n/2 - 1/2} - H_{n/2 - 1}}} \qquad\pars{~H_{z}:\ Harmonic\ Number~} \\[5mm] & = \bbx{1 \over 2} \end{align}
Note that $\ds{H_{z} \sim \ln\pars{z} + \gamma + {1 \over 2z} - {1 \over 12z^{2}}}$ as $\ds{\verts{z} \to \infty.\quad}$ $\ds{\gamma}$ is the Euler-Mascheroni Constant.
Felix Marin
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