Hi, I am a junior and met a simple problem. How can I solve the integration as below, `\int^{-\infty}0 e^{-\rho t} dt`
Thanks in advance!

Relevant page
<a href="http://www.sciencedirect.com/science/article/pii/S1043951X16300906">Optimal government investment and public debt in an economic growth model - ScienceDirect</a>
What I've done so far
`\int^{-\infty}_0 e^{-\rho t} dt = 1/ \rho \int^{-\infty}_0 e^{-\rho t} d {\rhot} = 1/ \rho `

Hello Zhanghui
As your integral starts from a small number (`-oo`) to a larger number (`0`), we would normally write it as:
`\int{-\infty}^0 e^{-\rho t} dt`
An infinite integral is calculated by finding a general integral and then finding the limit, as follows:
`\int{-\infty}^0 f(t) dt = lim{T->oo} { \int{-T}^0 f(t) dt}`
In this case, we would have:
`\int{-\infty}^0 e^{-\rho t} dt = lim{T->oo} {\int{-T}^0 e^{-\rho t} dt}`
Can you do the next step?

Thanks for your advice?
Sorry I made a mistake, the infinite bound is rewritten as positive infinite limit,
`\int^{\infty}0 e^{-\rho t} dt = 1/ \rho \int^{0\infty}0 e^{-\rho t} d {\rhot} = 1/ \rho `

Dou you think that is right? Thanks!

X

Dear Murray,
Thanks for your advice?
Sorry I made a mistake, the infinite bound is rewritten as positive infinite limit,
`\int^{\infty}0 e^{-\rho t} dt = 1/ \rho \int^{0\infty}0 e^{-\rho t} d {\rhot} = 1/ \rho `
Dou you think that is right? Thanks!

I'm not sure why you are doing `d {\rhot}` Where has the exponential part gone?

Try to integrate this indefinite integral first only:

`\int e^{-\rho t} dt`

X

Hi again. OK, given your correction to the question, the integral you need to do is:
`\int0^{\infty} e^{-\rho t} dt = lim{T->oo} {\int0^T e^{-\rho t} dt}`
I'm not sure why you are doing `d {\rhot}` Where has the exponential part gone?
Try to integrate this indefinite integral first only:
`\int e^{-\rho t} dt`

OK, good. The next step is to do it as a definite integral, and substitute in the upper and lower values:
`\int0^T e^{-\rho t} dt = [-1/ \rho e^{-\rho t}]0^T`
(That's an upper case `T` as an upper limit.)

Now for the limit as `T` approaches infinity for the expression in brackets, we have:

`lim{T->oo} ( e^{-\rho T} - 1) = 0 - 1 = -1`

So the overall integral approaches the value:

` - 1/ \rho(-1) = 1/rho`

X

For those wondering, zhanghui's earlier answer was correct, but it didn't show all the steps, which is what I was getting at. They are as follows.
`[-1/ \rho e^{-\rho t}]0^T = -1/ \rho[ e^{-\rho T} - e^{-\rho xx 0}]`
`= -1/ \rho (e^(-rhoT) -1 )`
Now for the limit as `T` approaches infinity for the expression in brackets, we have:
`lim{T->oo} ( e^{-\rho T} - 1) = 0 - 1 = -1`
So the overall integral approaches the value:
` - 1/ \rho(-1) = 1/rho`

Reply

You need to be logged in to reply.

Related Applications of Integration questions

Shell Method[Solved!] URL:Shell Method | Brilliant Math & Science Wiki
Under the section, "When to use the shell...phinah 04 Jun 2020, 05:35

Applications of Integrations #11[Solved!] I haven't had calculus in many years but I like to work problems every now...Kabookiep 01 Apr 2020, 08:04

Finding volume using shells[Solved!] Exercise 2:
The area between the curve y = 1/x, the y-axis and the lines y...phinah 09 Mar 2020, 11:53