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Why are these 2 expressions equal?

  1. Oct 5, 2014 #1
    Why is this true (at least, according to my calculator)? Is there something obvious that I've missed?
    \int_{0}^{1}\frac{1}{x^x}dx=\sum_{x=1}^{\infty }\frac{1}{x^x}[/tex]
    Last edited: Oct 5, 2014
  2. jcsd
  3. Oct 8, 2014 #2


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    What do you mean by "equal, at least according to my calculator"? Does your calculator do such integrations and infinite sums exactly (a remarkable calculator but computer algebraic system can be remarkable!) or are you doing an approximation taking very large x but not to infinity? Certainly the two will be approximately the same since the sum on the right is a Riemann sum approximation to the integral on the left.
  4. Oct 8, 2014 #3
    My graphing calculator managed to calculate the integral to 10 decimal places, and I calculated the summation to x=20 using the graphing calculator as well (to 10 DP). And they're the same.

    I recently just learnt Riemann sums in school and I don't fully understand why the 2 are equal (approximately?). Could you explain a little?
  5. Oct 9, 2014 #4


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    The whole point of Riemann sums is that they are used to show that the integral gives "area under a curve". If we have the curve y= f(x) such that y> 0 for a< x< b, we can divide the area into many thin rectangles, with width [itex]\Delta x[/itex] and height [itex]f(x^*)[/itex] where [itex]x^*[/itex] is a point on the x-axis inside that rectangle. The area of each rectangle is, of course, [itex]f(x^*)\Delta x[/itex] so the entire area is approximated by [itex]\sum f(x^*)\Delta x[/itex], the sum of the areas of all the rectangles. Taking the limit, taking more and more rectangles with [itex]\Delta x[/itex] smaller and smaller, we get, by definition, the "Riemann integral", [itex]\int_a^b f(x)dx[/itex].

    So for large n, [itex]\int_a^b f(x)dx[/itex] is approximated by the sum [itex]\sum_{i= 0}^n f(x_n^*)\Delta x[/itex].
  6. Oct 9, 2014 #5


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    Either I am missing something or HallsofIvy is on the wrong track. The limits of the integral and the limits of the sum are dissimilar. There is no immediately obvious way in which to associate the terms of the sum with rectangles in the Riemann sum.

    However, I was able to find a reference that indicates that the equality in question is exact. Check out the remark from Antonio Vargas:

  7. Oct 9, 2014 #6


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  8. Oct 10, 2014 #7
  9. Oct 10, 2014 #8
    Oh yes, my favourite: the gamma function! I've always been fascinated by the minimum in the gamma function....as if it had mysterious properties. X! = 1/2 (pi).5. Solve for x.
  10. Oct 10, 2014 #9
    Also, I've always found it an incredible feat of mathematical ingenuity to create meaningful solutions to non-integer factorials.
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