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Converting a limit to integral form or vice-versa

  1. May 6, 2015 #1
    What is the proof for this
    $$ \int_a^b f(x) dx = 1/n\lim_{n\to\infty} (f(a) + f(a+h) + f(a+2h) +...+ f( a+ (n-1)h)) $$
    h = (b-a)/n

    Also I think there is some summation form which can be converted to integral form how?
    Last edited by a moderator: May 6, 2015
  2. jcsd
  3. May 6, 2015 #2
    What is your definition of the integral?
  4. May 6, 2015 #3


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  5. May 7, 2015 #4
    An integral is the area under a curve.
    For any curve we can find it's area by putting infinitely small rectangles, trapezoids etc. shapes.
    We can write $$ \int_a^b f(x) dx = \lim_{n\to\infty} \sum_{i=1}^n f(x_{i-1}) Δx $$
    Δx = (b-a)/n
    But how do we get this

    $$ \int_a^b f(x) dx = 1/n\lim_{n\to\infty} (f(a) + f(a+h) + f(a+2h) +...+ f( a+ (n-1)h)) $$
    h = (b-a)/n
    From above that I know?
  6. May 7, 2015 #5


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    By setting x0 = a, x1 = a + h, etc. (you already have ##\Delta x = h##) .

    I think there is a small error in the lower formula: 1/n can't be outside the ##\lim## .
    (It also doesn't fit dimensionwise)
    (b-a) should be the factor in front and 1/n should be at the end (within the ##\lim## scope)
  7. May 7, 2015 #6
    Okay, got that and yes 1/n should be inside lim scope.

    Applying this how $$ \lim_{n\to\infty}\frac{1}{n}\sum_{r=1}^n log(r/n) = \int_0^1 logx dx$$ ?
  8. May 7, 2015 #7


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    Ah ! A new turn to this thread. We're going to derive Stirling's formula !

    Your righthand side makes me feel uncomfortable. A primitive of ##\log x## is ##x\log x - x## and there is no ##\log 0##, so I suppose this is to be continued. And yes:

    If you take ##f(x) = \log x## and write out
    \int_a^b f(x) dx = (b-a) \lim_{n\to\infty} ( f(a+h) + f(a+2h) +...+ f( a+ nh)) 1/n
    $$you see your summation appearing (b = 1, a = 0, h = 1/n).

    Note that I sneakily shifted by h (starting from a+h instead of from a). So from lower sum to upper sum (which google) to avoid the ##\log 0##).​
  9. May 7, 2015 #8
    I saw that Stirling formula in Wikipedia and I am feeling uncomfortable with that. ( I think it is taught in university with a lot of basics first and I am in high school.)
    I also googled lower to upper sum and don't know that things trapezoidal rule in details.( I think university will be fun.:rolleyes:)

    So, according to wolframalpha
    Left hand side is 0 and Right hand side is coming -1
    Is that true?
    But someone was telling me that they are equal.
    Also 1/∞ is not strictly zero, it may be 0.00000....1
    So log 0 not comes in left hand side.
  10. May 7, 2015 #9
    ##1/\infty## is undefined. It doesn't exist. And if you do want to define it, then it will be exactly 0.
  11. May 7, 2015 #10
    Okay, so is this true?

    $$ \lim_{n\to\infty}\frac{1}{n}\sum_{r=1}^n log(r/n) = \int_0^1 logx dx$$

    [Edit - added ] is 0 * log0 = 0 ?
    Last edited: May 7, 2015
  12. May 7, 2015 #11


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    No, but ##\displaystyle \lim_{x\rightarrow 0} x\log x = 0##

    Where do you see a zero appearing on the left hand side ? Try it on your calculator (mine ?:) goes up to n = 170 and no way it's going to 0).

    So is Wolfram a dud ? No it's not. But you need to improve the incantation...

    By the way, my deepest respect for the JEE level. I find it hard to believe they can ask so much of people wanting to get into university ! All (or almost) all the stuff I have to mobilize in order to help you, I learned after grammar school.​
    Last edited: May 7, 2015
  13. May 8, 2015 #12
    ##\displaystyle \lim_{x\rightarrow 0} x\log x = 0##
    Now how is this true?

    Got the -1 in left hand side by improving incantation:)
  14. May 8, 2015 #13


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    As x goes to 0, log(x) along goes to -infinity so [itex]\frac{1}{log(x)}[/itex] goes to 0 and we can apply L'Hopital's rule to [itex]\frac{x}{\frac{1}{log(x)}}[/itex].

    The derivative of x is, or course, 1 and the derivative of [itex]\frac{1}{log(x)}= (log(x))^{-1}[/itex] is [itex]-(log(x))^{-2}\frac{1}{x}[/itex] so that the limit is the limit of [itex]\frac{1}{-\frac{1}{x}(log(x))^2}[/itex][itex]= -\frac{(log(x))^2}{x}= -\frac{xlog(x)}{x}[/itex].

    So if [itex]\lim_{x\to 0} x log(x)[/itex] is some finite number, A, then we must have [itex]A= -A(\lim_{x\to 0} x)= 0[/itex]
  15. May 8, 2015 #14
    Shouldn't that bold part have## (log(x))^{-2} ## in denominator ?
    and I am also not getting what you are doing after that
    [itex]\frac{1}{-\frac{1}{x}(log(x))^2}[/itex][itex]= -\frac{(log(x))^2}{x}[/itex]
    Last edited: May 8, 2015
  16. May 9, 2015 #15
  17. May 9, 2015 #16
    HaHa, no problem,
    Halls you took me in a complicated way,
    Here see this,
    $$ xlogx = \frac{logx}{\frac{1}{x}} $$
    Here limit x tends to zero and as we are getting num. and den -∞/∞ we can then also apply L Hopital rule.
    $$ \frac{-x^2}{x} = -x $$
    As here limit x tends to 0
    We get 0.

    Thanks BvU, Halls (although your way was very complicated but bringing L Hopital was crucial) and also thanks to μm .
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