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#1 2006-04-13 16:05:28

coolwind
Member
Registered: 2005-10-30
Posts: 30

A integral problem

∫(t+1)e^(-jwkt)dt


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#2 2006-06-01 08:56:02

Prakash Panneer
Member
Registered: 2006-06-01
Posts: 110

Re: A integral problem

You didn't give any information about j,k,and w. So, i took these are constants.

Problem: Integral ( t+1 ) e ^ ( -jwkt ) dt

  To solve this problem, we have to  use "Integration by parts method,"

  Take u = t+1 , dv = e^ (-jwkt)dt

       du = dt  and v = e^(-jwkt)/(-jwk)       (Oops)

Integrantion by parts formula  Integral ( u dv) = uv- integral(vdu)

So, we will get,     Integral ( t+1 ) e ^ ( -jwkt ) dt = {(t+1)e^(-jwkt) /(-jwk)} - Integral ( -e^(-jwkt)/(jwk)) dt
                                                                           
                                                                           = {-(t+1)e^(-jwkt)/(jwk)} - e^ (-jwkt)/ (jwk)^2.

                                                                           = {e^(-jwkt)/(jwk)} {-(t+1)- (1/jkw)}
   
      This is our required answer.

Last edited by Prakash Panneer (2006-06-04 01:05:40)


Letter, number, arts and science
of living kinds, both are the eyes.

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#3 2006-06-01 11:11:28

John E. Franklin
Member
Registered: 2005-08-29
Posts: 3,562

Re: A integral problem

I'm rusty on calculus.
Why in above is the integral of  dv = e^(-jwkt)dt
become:
        v = (e^(-jwkt)) / (-jwkt)
Does this mean that if y = e^(kx) then y'=kx e^(kx) ?


igloo myrtilles fourmis

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#4 2006-06-03 21:22:56

George,Y
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Registered: 2006-03-12
Posts: 1,306

Re: A integral problem

is there an "i' instead of a "j"?

Last edited by George,Y (2006-06-03 21:23:28)


X'(y-Xβ)=0

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#5 2006-06-04 01:09:22

Prakash Panneer
Member
Registered: 2006-06-01
Posts: 110

Re: A integral problem

Oops....................

I made a mistake.

  That means  dv = e^(kx) dx

Integrating with respect to x on both sides, we get,

                         v = e^(kx)/k.

 
If   y = e^(kx) then y' = k e^kx.

Last edited by Prakash Panneer (2006-06-04 01:12:21)


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