Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 7021 by FilupSmith last updated on 06/Aug/16

z=∫_0 ^( t) i(−1)^x dx  =∫_0 ^( t) ie^(iπx) dx    (1)  =i∫_0 ^( t) e^(iπx) dx  =i((1/(iπ))(e^(iπt) −1))  z=(1/π)((−1)^t −1)    (2)  =∫_0 ^( t) ie^(iπx) dx  u=e^(iπx)  ⇒ du=(1/(iπ))e^(iπx) dx  =∫_0 ^( t) ((iπ)/(iπ))ie^(iπx) dx  =∫_0 ^( t) iπidu  =−π∫_0 ^( t) du  =−π(e^(iπt) −1)    which is incorrect?

z=0ti(1)xdx=0tieiπxdx(1)=i0teiπxdx=i(1iπ(eiπt1))z=1π((1)t1)(2)=0tieiπxdxu=eiπxdu=1iπeiπxdx=0tiπiπieiπxdx=0tiπidu=π0tdu=π(eiπt1)whichisincorrect?

Commented by FilupSmith last updated on 07/Aug/16

Ah, i see my mistake, thank you!

Ah,iseemymistake,thankyou!

Commented by Yozzii last updated on 06/Aug/16

In (2), for u=e^(iπx) ⇒du=iπe^(iπx) dx  ⇒u=e^(πit)  at x=t and u=1 at x=0.  ⇒∫_0 ^t ie^(πix) dx=i×(1/(iπ))∫_1 ^e^(iπt)  du=(1/π)(e^(πit) −1)=(1/π)((−1)^t −1)

In(2),foru=eiπxdu=iπeiπxdxu=eπitatx=tandu=1atx=0.0tieπixdx=i×1iπ1eiπtdu=1π(eπit1)=1π((1)t1)

Answered by Yozzii last updated on 08/Aug/16

Check for a response in the comments.

Checkforaresponseinthecomments.

Terms of Service

Privacy Policy

Contact: info@tinkutara.com