Question Number 149854 by mathdanisur last updated on 07/Aug/21

Commented bymr W last updated on 08/Aug/21

i gave an answer. you didn′t agree, but gave an other answer which is in fact the same as my answer.

Commented bymr W last updated on 08/Aug/21

you said it is zero, not i.

Commented bymr W last updated on 07/Aug/21

= { ((x+C if [x] is even)),((−x+C if [x] is odd)) :}

Commented bymathdanisur last updated on 08/Aug/21

Thanks Ser, but answer periodik

Commented bymr W last updated on 08/Aug/21

please tell what concretely is not correct. when i say if [x] is even, it is the same as to say if 0≤x<1, 2≤x<3, 4≤x<5, ... and when i say if [x] is odd, it is the same as to say if 1≤x<2, 3≤x<4, 5≤x<6, ...

Commented bymathdanisur last updated on 08/Aug/21

x+C [x] even −x+C[x] odd Ser but your answer here is zero.?

Commented bymathdanisur last updated on 08/Aug/21

Der Ser, the answer is given periodically −1=e^(iπ) ⇒−1=e^(3iπ) =e^(5iπ) =...

Commented bymr W last updated on 08/Aug/21

i don′t understand that, too high to me. you even didn′t tell what the concrete answer is.

Commented bymathdanisur last updated on 08/Aug/21

Ser, answer: ∫(−1)^([x]) dx = (1/π) arccos(cosπx)+C

Commented bymr W last updated on 08/Aug/21

don′t you see that this is absolutely the same as my answer? ((cos^(−1) (πx))/π)=x if 2k≤x<2k+1 ((cos^(−1) (πx))/π)=−x if 2k+1≤x<2k+2

Commented bymathdanisur last updated on 08/Aug/21

Very nice Ser thankyou