How to manipulate limits in this integral

[nitin7785]

hi all,

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

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

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

But don't know how to get there.

Thanks in advance.

 

[gopher_p]

hi all,

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

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

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

But don't 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

$\int_0^tP(\tau)\text{exp}(-a\int_\tau^tP(u)\ du)\ d\tau=\int_0^t\frac{1}{a}y'(\tau)\text{exp}(y(\tau))\ d\tau=\frac{1}{a}\int_{y_0}^{y_t}e^y\ dy$.

 

[nitin7785]

Thanks a lot for your answer.

 

[Chestermiller]

hi all,

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

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

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

But don't know how to get there.

Thanks in advance.

Hi nitin7785. Welcome to Physics Forums!

I have a slightly different way of doing it;
∫^{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

This becomes:
\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

This becomes:

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'}

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

Chet

 

[nitin7785]

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