Question

Fun integral question! (As long as my internet doesn't die!)

Answer 1

\[\Large \int\limits_1^\infty \lfloor x \rfloor ^{-2}dx\]

Answer 2

i remember this....

Answer 3

annoying not fun pls

Answer 4

Hahaha XD

Answer 5

convert to a \(\Sigma\)-sum?

Answer 6

1 + 1/4 + 1/9 + 1/16 + . . .

Answer 7

it is:
\[\int\limits_{0}^{\infty}[x]^{-2}dx=\sum_{0}^{\infty}\frac{ 1 }{ n ^{2} }=\frac{ \pi ^{2} }{ 6 }\]

Answer 8

Yeah, you got it. But make sure you check the lower bound, 1/0 =P

Answer 9

oops then it is:
\[\sum_{1}^{\infty} \frac{ 1 }{ n ^{2} }=\frac{ \pi ^{2} }{ 6 }\]

Answer 10

since x=0, has [x]=0

Answer 11

0+1/2^2+1/3^2+.....
nice

Answer 12

opps from 1 not zero

Answer 13

Also if you want to write the floor or ceiling function brackets, just type lfloor, rfloor, lceil, rceil for left and right each. =)

Answer 14

olk :P
next time

Answer 15

what math is that ?? :(

Answer 16

discrete

Answer 17

huh ?? what is that??
*nnesha out of this post*
thanks
sorry for disturbed u

Answer 18

nice question! :)

Answer 19

Yeah if anyone has any experience with integrating floor or ceiling functions or changing between integrals and series, I'd like to hear about it since I pretty much know nothing about it.

Answer 20

hmm infact we should imply it like this
sum (m-(n-1) ) 1/(n-1)^2 from n=1 to
the only confusing part is that m is strictly less than n-1 :P

Answer 21

but since its an area of rectangles , then (m-(n-1) is an interval of length 1