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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

### 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,

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,

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