Question Number 149813 by mathdanisur last updated on 07/Aug/21

∫(−1)^([x]) dx = ?

Commented byDonQuichote last updated on 07/Aug/21

deal: you show me your solution idea first then I′ll show you mine

Answered by mathmax by abdo last updated on 08/Aug/21

f(x)=∫_0 ^x (−1)^([t]) dt ⇒f(x)=Σ_(k=0) ^([x]) ∫_k ^(k+1) (−1)^k dt +∫_([x]) ^x (−1)^([t]) dt =Σ_(k=0) ^([x]) (−1)^k +∫_([x]) ^x (−1)^([t]) dt =((1−(−1)^([x]+1) )/2) +∫_([x]) ^x (−1)^([t]) dt if x from N ∫_([x]) ^x (−1)^([t]) dt=0 ⇒∫_0 ^x (−1)^([t]) dt=((1−(−1)^(x+1) )/2)

Commented bymathdanisur last updated on 08/Aug/21

Thank you Ser but, (−1)=e^(iπ) ⇒ (−1)^(3iπ) = (−1)^(5iπ) = ... .?