How to manipulate limits in this integral

  • Thread starter nitin7785
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  • #1
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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.
 

Answers and Replies

  • #2
575
76
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.
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##.
 
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  • #3
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Thanks a lot for your answer.
 
  • #4
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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)dτ[/itex]

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.
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
 
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  • #5
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Thank you Chestermiller for providing alternative way of solving my problem.
 

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