http://faculty.missouristate.edu/l/lesreid/potw.html

Missouri State University’s Advanced Problem of February 2010 Vincent Pantaloni (Orléans, France) Problem : Evaluate the following integral : Z

1

0

  1 dx x

where ((x)) denotes the fractional part of a number x. Solution : . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ∀x ∈ R, ((x)) = x − [x]. Where [x] is the integer part of x. The above integral is undefined in 0 so we have to evaluate : Z 1   1 dx lim n→+∞ 1 x n

R1 Let n be an integer greater than 2 and consider : In = 1 x1 dx. Then : n

  1 1 1 dx − In = 1 x x n Z 1  Z 1 1 1 dx − dx = 1 1 x x n n Z 1  1 = ln(n) − dx 1 x n n Z 1   X k−1 1 dx = ln(n) − 1 x k=2 k n Z 1 X k−1 k − 1 dx = ln(n) − Z

1

k=2 k n  X

 1 1 = ln(n) − × (k − 1) − k−1 k k=2  n  X k−1 1− = ln(n) − k = ln(n) −

k=2 n X

k=2

= 1 + ln(n) −

1 k n X 1 k

k=1

We know that lim

n→+∞

Pn


1

k=1 k




− ln(n) = γ. Where γ ≈ 0.5772156649 is the Euler constant. Hence : Z

0

http://prof.pantaloni.free.fr

1

  1 dx = 1 − γ x

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