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Would everyone please help me on how to prove this value of Lebesgue integral of the function $f(x)=\frac{1}{x}$ in the interval [1,5] by using approximation by simple function $f_n$ step by step?

What if the function in this example defined on the interval of $[0,\infty]$? What will the value of integral be? (Still need help for the explanation by using simple function approximation)

Thank you.

math-learner
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1 Answers1

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For $k=0,\cdots,4n-1$, $x\in[1+\frac{k}n,1+\frac{k+1}n)$: $$f_n(x)=\frac1{1+(k+1)/n}=\frac{n}{k+n+1},$$ $$\int_1^5f_n=\sum_{k=0}^{4n-1}\frac{n}{k+n+1}\frac1{n}=\sum_{k=0}^{4n-1}\frac1{k+n+1}=\sum_{k=n+1}^{5n}\frac1k=H_{5n}-H_n\approx\log(5n)-\log n=\log 5.$$ In $[0,\infty]$ the harmonic series (divergent) is a lower bound of the integral. Try yourself, is much easier.

Martín-Blas Pérez Pinilla
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  • Also, the idea behind choosing $x$ in $[1+\frac{k}{n},1+\frac{k+1}{n})$? – math-learner Nov 20 '14 at 11:21
  • @learnie: First, take equal length intervals. Second, take an (lower in this case) approximation of $f$ in each interval. The $4n$ instead of $n$ simplifiles the calculations. – Martín-Blas Pérez Pinilla Nov 20 '14 at 11:30
  • I still do not understand. Please explain more of how you choose the interval of value of $x$. Can you explain me about your statement how to "take equal length intervals"? Thank you so much. – math-learner Nov 20 '14 at 11:42
  • Points for equal length subintervals of $[a,b]$: $a+k(b-a)/n=$ lower end + $k$(length of interval)/(num of intervals). If $f$ is monotonic (like in this case) what obvious approximations of $f$ can be taken in each subinterval? Answer: the values of $f$ in the interval ends. – Martín-Blas Pérez Pinilla Nov 20 '14 at 11:52
  • $k=0\rightarrow$ lower end. For the interval $[2,5]$ (length 3) is better $3n$. – Martín-Blas Pérez Pinilla Nov 20 '14 at 12:55
  • I still have no clue, in the interval $[1,5]$, we take equal length subinterval of $[1,5]$ : $1+k\frac{4}{n}$. It is the length of interval of each subinterval, right? So, how come we instead use $\frac{1}{n}$ in the length of each subinterval as the impact of using $x$ in $[1+\frac{k}{n},1+\frac{k+1}{n})$? I am a beginner on this – math-learner Nov 20 '14 at 13:17
  • The subinterval length is not 1+...! – Martín-Blas Pérez Pinilla Nov 20 '14 at 13:30