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I Riemann sum problem

  1. Jan 16, 2017 #1
    Hi.
    I try to solve the integral $$\int_{0}^{1} x^{x} dx$$
    Through sums of riemann But I came to the conclusion that the result is 0 that is wrong
    $$\int_{0}^{1} x^{x} dx = \lim_{n\rightarrow \infty }\frac{1}{n}\sum_{k=1}^{n} \left ( \frac{k}{n} \right )^{\frac{k}{n}}$$
    $$= \lim_{n\rightarrow \infty }\frac{\frac{1}{n}^{\frac{1}{n}}}{n} + \frac{\frac{2}{n}^{\frac{2}{n}}}{n} + ... +\frac{\frac{n-1}{n}^{\frac{n-1}{n}}}{n} + \frac{1}{n}=0$$
    $$\int_{0}^{1} x^{x} dx = 0 $$

    I'm sure the mistake is expand the ##\frac{1}{n}\sum_{k=1}^{n} \left ( \frac{k}{n} \right )^{\frac{k}{n}}##

    because in some easy integrals like ##\int_{0}^{1} x dx## if you expand the ##\frac{1}{n}\sum_{k=1}^{n} \left ( \frac{k}{n} \right )## you get
    $$= \lim_{n\rightarrow \infty } \frac{1}{n^{2}}+\frac{2}{n^{2}}+...+\frac{n-1}{n^{2}}+\frac{1}{n} = 0$$
    Instead of finding a "closed" formula like:
    $$= \lim_{n\rightarrow \infty } \frac{n(n+1)}{2\, n^{2}}=\frac{1}{2}$$
    so If I can not find a formula like ## \frac{n(n+1)}{2}## for ## \sum_{k=1}^{n} \left ( \frac{k}{n} \right )^{\frac{k}{n}}##
    the limit of the riemman sum is wrong?
     
  2. jcsd
  3. Jan 16, 2017 #2

    mfb

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    2016 Award

    Staff: Mentor

    Where does the first "=0" come from? The limit of each fraction is 0, but you add more terms in each step, you cannot simply look at the limits of each fraction.

    The limit is wrong, but the problem is not the lack of a general formula for the sum.
     
  4. Jan 16, 2017 #3
    Ok then what is the correct way to solve the problem?
    Because in simple cases as in polynomials it is enough Arrive at a formula to resolve the limit.
    in more complex function is "easy" to solve the problem Through riemann sums or is very difficult?
    Maybe I have to change the method?
     
  5. Jan 16, 2017 #4

    mfb

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    Staff: Mentor

    What makes you think there is a nice closed form for the solution?
    In general Riemann sums are not easy to evaluate. For polynomials they work, but even there they are not the easiest approach.
     
  6. Jan 16, 2017 #5

    ShayanJ

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    Gold Member

    According to wolframalpha, the series diverges!
     
  7. Jan 16, 2017 #6

    Stephen Tashi

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    Science Advisor

  8. Jan 16, 2017 #7
    well I suspected it but as the sum of riemann is the definition of integral I thought they might be useful for something other than polynomials
    so Then I'll try another approach. thanks
    I believe you but wolframalpha hates me
    https://www.wolframalpha.com/input/...rmassumption={"C",+"limit"}+->+{"Calculator"}
    I already knew the result but wanted to get it myself And as it is a sum I thought that the sums of riemann or the series of taylor would be a good approximation
     
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