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A simple integration [Solved!]

My question

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

Optimal government investment and public debt in an economic growth model - ScienceDirect

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 `

X

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  `

Re: A simple integration

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?

X

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-&gt;oo} { \int_{-T}^0 f(t) dt}`

In this case, we would have:

`\int_{-\infty}^0 e^{-\rho t} dt = lim_{T-&gt;oo} {\int_{-T}^0 e^{-\rho t} dt}`

Can you do the next step?

Re: A simple integration

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!

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!

Re: A simple integration

Hi again. OK, given your correction to the question, the integral you need to do is:

`\int_0^{\infty} e^{-\rho t} dt = lim_{T->oo} {\int_0^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`

X

Hi again. OK, given your correction to the question, the integral you need to do is:

`\int_0^{\infty} e^{-\rho t} dt = lim_{T-&gt;oo} {\int_0^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`

Re: A simple integration

Hi, I will do as below,

`\int e^{-\rho t} dt = -1/ \rho e^{-\rho t}`

X

Hi, I will do as below,

`\int e^{-\rho t} dt = -1/ \rho e^{-\rho t}`

Re: A simple integration

OK, good. The next step is to do it as a definite integral, and substitute in the upper and lower values:

`\int_0^T e^{-\rho t} dt = [-1/ \rho e^{-\rho t}]_0^T`

(That's an upper case `T` as an upper limit.)

X

OK, good. The next step is to do it as a definite integral, and substitute in the upper and lower values:

`\int_0^T e^{-\rho t} dt = [-1/ \rho e^{-\rho t}]_0^T`

(That's an upper case `T` as an upper limit.)

Re: A simple integration

Thank, I will do as below,
`\int_0^T e^{-\rho t} dt = [-1/ \rho e^{-\rho t}]_0^T = -1/ \rho (0 -1 ) = 1/ \rho `. Is it right?

X

Thank, I will do as below,
`\int_0^T e^{-\rho t} dt = [-1/ \rho e^{-\rho t}]_0^T = -1/ \rho (0 -1 ) = 1/ \rho `. Is it right?

Re: A simple integration

Where did the upper limit go?

After substituting `T`, it should be

`-1/ \rho (e^(-rhoT) -1 )`

Now can you do the limit of this as `T->oo`?

X

Where did the upper limit go?

After substituting `T`, it should be

`-1/ \rho (e^(-rhoT) -1 )`

Now can you do the limit of this as `T-&gt;oo`?

Re: A simple integration

Ok got it! Thank you for the explanation.

X

Ok got it! Thank you for the explanation.

Re: A simple integration

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`

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-&gt;oo} ( e^{-\rho T} - 1) = 0 - 1 = -1`

So the overall integral approaches the value:

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

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