Evaluate the following definite integral :
0 ∫ 1 ⎩ ⎪ ⎨ ⎪ ⎧ ( − 1 ) ⌊ x 1 ⌋ x 1 ⎭ ⎪ ⎬ ⎪ ⎫ d x
where { x } denotes the fractional part of x and ⌊ x ⌋ denotes the greatest integer function ( floor function ).
Note: Use the following definition to solve the problem.
{ x } = x − ⌊ x ⌋ ∀ x ∈ R
where R denotes the set of all reals.
Put your answer to 3 decimal places.
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@Refaat M. Sayed Nice work! But when I computed the integral numerically, I was getting an answer of − 0 . 1 4 4 7 ??
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Could you tell me How you did that? .
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@Refaat M. Sayed How do I add an image?? in comments
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@Kunal Gupta – @Hasan Kassim say that :first upload any pic u want to insert, on www.postimage.org, they provide a link for that image.
while writing solutions , write the following:
![](link
You should replace the word "link" by the link provided by that website, then close the paranthesis... and I think Master. @Calvin Lin will help us
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@Refaat M. Sayed – @Refaat M. Sayed thanks!! here you
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@Kunal Gupta – But the correct graph like this http://s11.postimg.org/lscirf7ur/Screenshot.png
Really nice way of computing that logarithm sum! Innovative! I really liked it. I have always used polylogarithm or zeta function derivative for such sums but this wake me up. Thanks!
How is that pi/2?
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this product is well known as Wallis product and here is the proof
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According to the given identity we can rewrite the internation in this form I = ∫ 0 1 { ( − 1 ) ⌊ x 1 ⌋ x 1 } d x = ∫ 0 1 ( − 1 ) ⌊ x 1 ⌋ x 1 d x − 0 ∫ 1 ⌊ ( − 1 ) ⌊ x 1 ⌋ x 1 ⌋ d x Now let x 1 = y . We get the internation above as I = 1 ∫ ∞ ( − 1 ) ⌊ y ⌋ y 1 d y − 1 ∫ ∞ ⌊ ( − 1 ) ⌊ y ⌋ y ⌋ y 2 1 d y Now I 1 = 1 ∫ ∞ ( − 1 ) ⌊ y ⌋ y 1 d y = m = 1 ∑ ∞ ( − 1 ) m m ∫ m + 1 y d y = m = 1 ∑ ∞ ( − 1 ) m l n ( m m + 1 ) so I 1 = − l n [ m = 1 ∏ ∞ ( 2 m − 1 2 m × 2 m + 1 2 m ) ] = − l n ( 2 π ) I 2 = 1 ∫ ∞ ⌊ ( − 1 ) ⌊ y ⌋ y ⌋ y 2 1 d y = m = 1 ∑ ∞ m ∫ m + 1 ⌊ ( − 1 ) m y ⌋ y 2 1 d y then I 2 = m = 1 ∑ ∞ 2 m ⎝ ⎜ ⎛ 2 m ∫ 2 m + 1 y 2 1 d y − 2 m − 1 ∫ 2 m y 2 1 d y ⎠ ⎟ ⎞ so I 2 = m = 1 ∑ ∞ 2 m + 1 1 − m = 1 ∑ ∞ 2 m − 1 1 = − 1 And : I = I 1 − I 2 = 1 − l n ( 2 π ) = 0 . 5 4 8