Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

How to manipulate limits in this integral

  1. Oct 30, 2013 #1
    hi all,

    I have following integral and i was wondering if i can manipulate limits to simplify it.

    [itex]∫^{t}_{0} P(τ)[/itex] exp([itex] - a ∫^{t}_{τ} P(u) du[/itex]) dτ

    I know that the answer is [itex]\frac{1}{a}[/itex] - [itex]\frac{1}{a}[/itex] exp([itex] - a ∫^{t}_{0} P(t) dt[/itex])

    But dont know how to get there.

    Thanks in advance.
     
  2. jcsd
  3. Oct 30, 2013 #2
    If you let ##y(\tau)=-a\int_\tau^tP(u)\ du##, then ##y_0=y(0)=-a\int_0^tP(u)\ du##, ##y_t=y(t)= -a\int_t^tP(u)\ du=0##, and ##y'(\tau)=aP(\tau)##. Then you get

    ##\int_0^tP(\tau)\text{exp}(-a\int_\tau^tP(u)\ du)\ d\tau=\int_0^t\frac{1}{a}y'(\tau)\text{exp}(y(\tau))\ d\tau=\frac{1}{a}\int_{y_0}^{y_t}e^y\ dy##.
     
  4. Oct 31, 2013 #3
    Thanks a lot for your answer.
     
  5. Oct 31, 2013 #4
    Hi nitin7785. Welcome to Physics Forums!

    I have a slightly different way of doing it;
    [itex]∫^{t}_{0} P(τ) exp( - a ∫^{t}_{τ} P(u) du)dτ=\int^{t}_{0} {P(τ) exp (-a\int^{t}_{0} {P(u) du}+a\int^{τ}_{0} {P(u) du})}dτ=exp (-a\int^{t}_{0} {P(u) du})\int^{t}_{0} {P(τ) exp (a\int^{τ}_{0} {P(u) du})}dτ[/itex]

    This becomes:
    [itex]\int^{t}_{0} {P(τ) exp (-a\int^{t}_{0} {P(u) du}+a\int^{τ}_{0} {P(u) du})}dτ=exp (-a\int^{t}_{0} {P(u) du})\int^{t}_{0} {P(τ) exp (a\int^{τ}_{0} {P(u) du})}dτ[/itex]

    This becomes:

    [itex]exp (-a\int^{t}_{0} {P(u) du})\int^{t}_{0} {P(τ) exp (a\int^{τ}_{0} {P(u) du})}dτ=e^{-aI(t)}\int^{I(t)}_{0}{e^{aI'}dI'}[/itex]

    where [itex]I(t)=\int^{t}_{0} {P(u) du}[/itex]

    Chet
     
  6. Nov 1, 2013 #5
    Thank you Chestermiller for providing alternative way of solving my problem.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: How to manipulate limits in this integral
  1. Integration limits (Replies: 3)

Loading...