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Question Number 7021 by FilupSmith last updated on 06/Aug/16
z=∫0ti(−1)xdx=∫0tieiπxdx(1)=i∫0teiπxdx=i(1iπ(eiπt−1))z=1π((−1)t−1)(2)=∫0tieiπxdxu=eiπx⇒du=1iπeiπxdx=∫0tiπiπieiπxdx=∫0tiπidu=−π∫0tdu=−π(eiπt−1)whichisincorrect?
Commented by FilupSmith last updated on 07/Aug/16
Ah,iseemymistake,thankyou!
Commented by Yozzii last updated on 06/Aug/16
In(2),foru=eiπx⇒du=iπeiπxdx⇒u=eπitatx=tandu=1atx=0.⇒∫0tieπixdx=i×1iπ∫1eiπtdu=1π(eπit−1)=1π((−1)t−1)
Answered by Yozzii last updated on 08/Aug/16
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