# How to manipulate limits in this integral

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 dont know how to get there.

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 dont know how to get there.

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##.

1 person

Chestermiller
Mentor
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 dont know how to get there.

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

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