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Following expression was asked to be evaluated in TIFR GS 2015 exam,

$$G = \lim_{n\to\infty}(n+1)\int_{0}^{1} x^{n} f(x) dx$$

where $x \in [0, 1]$ and $f(x)$ be any real valued continuous function.

I have tried using Integration by parts technique but I am not able to solve the integral. I would like to know if any other approach should be used for solving this problem.

P.S. : As per answer key, $G = f(1)$.

Pranav
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    What are the hypotheses on $f$? – Olivier Oloa May 23 '16 at 09:24
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    if $f(x)$ is constant on $[a,1]$ then $(n+1) \int_a^1 x^n f(x) dx = f(1)(1-a^{n+1})$. since $|a| < 1$ it $\to 1$ when $n \to \infty$. in the same way, if $f(x) = C$ on $[b,a]$ then $(n+1) \int_b^a x^n f(x) dx = C(a^{n+1}-b^{n+1})$ which $\to 0$ since $|b| < 1$ and $|a| < 1$, etc.. this way we get that for any piecewise constant function, $G = f(1)$, and hence also for any (piecewise) continuous function. – reuns May 23 '16 at 09:30
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    After the transformation $x^n = t$ we have $\displaystyle \lim_{n\to \infty}(1+1/n)\int_0^1t^{1/n}f(t^{1/n}),dt = f(1). $ – JanG May 23 '16 at 21:30
  • An alternative to the splitting of interval proposed by @user1952009 is $[0,1-1/\sqrt{n}]$ and $[1-1/\sqrt{n},1]$. – mickep May 24 '16 at 18:30
  • @mickep : ?? it is the same idea, prove that $\int_0^a x^n f(x) dx \to 0$ whenever $a < 1$ and $f$ is bounded/ integrable, and show that the rest $\to f(1)$ when $ f$ is continuous – reuns May 24 '16 at 19:45
  • @user1952009 My wording was perhaps not optimal. It is indeed the same idea, but one let the point where the interval is divided depend on $n$. That was not clear (to me) from your comment. I'm sorry if you feel I stole your idea... – mickep May 24 '16 at 19:50

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Intuitively, no matter what happens with $f(x)$, only the region near $x=1$ will contribute to the integral.

This isn't very rigorous, but this gives you the idea:

Assume $f(x)$ is integrable. Divide the interval $[0,1]$ into $M$ intervals. On each interval $[a,b]$ take the average value of $f(x)$, so that: $$G_n = \sum_{m=1}^M \bar{f}(x)\int_{m-1/M}^{m/M}(n+1)x^n dx = \sum_{m=1}^M \bar{f}(x) \left((m/M)^{n+1}-((m-1)/M+1)^{n+1}\right)$$ In the limit $M,n\to\infty$, only the last element $m=M$ survives, so: $$G = f(1)$$

Nathaniel Bubis
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  • (when $f$ is continuous) taking the $\min$ and the $\max$ of $f$ on $[M/n,(M+1)/n]$ instead of the mean value – reuns May 24 '16 at 19:48