Improper Integral explicit function

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Ghost3k

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Hi, so I'm having a bit of trouble starting out part b of the following question.


(a) Calculate \(\displaystyle \, \int_0^{\infty}\, 2x^2\, \cdot\, e^{-2x}\, dx,\,\) if it exists.

Define a function of \(\displaystyle \, t,\, M,\, \) using the improper integral \(\displaystyle \, \int_0^{\infty}\, 2e^{tx}\, \cdot\, e^{-2x}\, dx,\, \) i.e., \(\displaystyle \, M(t)\, \) is the improper integral.

Determine

(b) the explicit function that is represented by the improper integral.


I was able to get part a (answer was 1/2). Can anybody help me in finding out how to get the explicit function that's represented by the improper integral?
 
Last edited by a moderator:
Ghost3k said:
Hi, so I'm having a bit of trouble starting out part b of the following question.


(a) Calculate \(\displaystyle \, \int_0^{\infty}\, 2x^2\, \cdot\, e^{-2x}\, dx,\,\) if it exists.

Define a function of \(\displaystyle \, t,\, M,\, \) using the improper integral \(\displaystyle \, \int_0^{\infty}\, 2e^{tx}\, \cdot\, e^{-2x}\, dx,\, \) i.e., \(\displaystyle \, M(t)\, \) is the improper integral.

Determine

(b) the explicit function that is represented by the improper integral.


I was able to get part a (answer was 1/2). Can anybody help me in finding out how to get the explicit function that's represented by the improper integral?
Click to expand...
What have you done so far? Please show us your work so we may try to help you.You might also read
http://www.freemathhelp.com/forum/threads/78006-Read-Before-Posting

You might want to start with the definition of an improper integral, i.e.
http://en.wikipedia.org/wiki/Improper_integral
and go from there.
 
Last edited by a moderator:
Ishuda said:
What have you done so far? Please show us your work so we may try to help you.You might also read
http://www.freemathhelp.com/forum/threads/78006-Read-Before-Posting

You might want to start with the definition of an improper integral, i.e.
http://en.wikipedia.org/wiki/Improper_integral
and go from there.
Click to expand...

Well for the first part I did the integration by parts table method to get my answer of 1/2. For the second question the only thing I did was rewrite the equation as integral from 0 to infinity 2e^(tx-2x) dx. Then I got 2* integral from 0 to infinity e^(t-2)x dx.
What I'm having trouble with is the explicit function part. Does this mean I just need to continue solving the integral?

Because if I would proceed I would make u = (t-2)x, du/(t-2) = dx and setting it up to be 2/(t-2)*integral e^u du which gives me, 2*e^((t-2)x)/(t-2) + C. But it is the improper integral so I would have to take that answer with the bounds of 0 to infinity. Not really sure if I'm in the right direction. Would appreciate any other guidance.

Edit: For clarity.
 
Last edited:
Ghost3k said:
Well for the first part I did the integration by parts table method to get my answer of 1/2. For the second question the only thing I did was rewrite the equation as integral from 0 to infinity 2e^(tx-2x) dx. Then I got 2* integral from 0 to infinity e^(t-2)x dx.
What I'm having trouble with is the explicit function part. Does this mean I just need to continue solving the integral?

Because if I would proceed I would make u = (t-2)x, du/(t-2) = dx and setting it up to be 2/(t-2) integral e^u du which gives me, 2*e^((t-2)x)/(t-2) + C. But it is the improper integral so I would have to take that answer with the bounds of 0 to infinity. Not really sure if I'm in the right direction. Would appreciate any other guidance.

Edit: For clarity.
Click to expand...
Did you read what the definition of an improper integral was? In this particular case it is given explicitly in the link where the function f(x) is given by 2 e(t-2)x and a is zero. If so then you should know the form of the explicit function takes [you have shown you know what the indefinite integral is] and the conditions it places on t.

As a hint,
\(\displaystyle \underset{x\, \to\, \infty}{lim}\, \underset{1}{\overset{x}{\int}}\, \frac{1}{u^2}\, du\, =\, \underset{x\, \to\, \infty}{lim} [-\frac{1}{u}]_{_1}^{^x}\, =\, \underset{x\, \to\, \infty}{lim} [1 -\frac{1}{x}]\)
 
Ishuda said:
Did you read what the definition of an improper integral was? In this particular case it is given explicitly in the link where the function f(x) is given by 2 e(t-2)x and a is zero. If so then you should know the form of the explicit function takes [you have shown you know what the indefinite integral is] and the conditions it places on t.

As a hint,
\(\displaystyle \underset{x\, \to\, \infty}{lim}\, \underset{1}{\overset{x}{\int}}\, \frac{1}{u^2}\, du\, =\, \underset{x\, \to\, \infty}{lim} [-\frac{1}{u}]_{_1}^{^x}\, =\, \underset{x\, \to\, \infty}{lim} [1 -\frac{1}{x}]\)
Click to expand...

Thank you for all your help. Unfortunately I can't seem to grasp the direction you're taking me. I've looked at the link and the hint you provided but I'm not sure how to proceed. Thank your for taking your time either way.
 
Ghost3k said:
Thank you for all your help. Unfortunately I can't seem to grasp the direction you're taking me. I've looked at the link and the hint you provided but I'm not sure how to proceed. Thank your for taking your time either way.
Click to expand...
What is
I(s, t) = \(\displaystyle 2 \underset{0}{\overset{s}{\int}}\, e^{(t-2)x}\, dx\)?

As s becomes very large what is the restriction on t to keep I(s, t) from becoming very large? What is the relationship between M(t) and I(s, t)?
 
Ishuda said:
What is
I(s, t) = \(\displaystyle 2 \underset{0}{\overset{s}{\int}}\, e^{(t-2)x}\, dx\)?

As s becomes very large what is the restriction on t to keep I(s, t) from becoming very large? What is the relationship between M(t) and I(s, t)?
Click to expand...

Okay so I gave it another shot and here's what I got.
Lim c -> infinity from 0 to c integral e^((t-2)x) dx

lim c -> infinity (2e^((t-2)x)/(t-2) from 0 to C

lim c -> infinity (2e^((t-2)c)/(t-2) - (2/(t-2))

So if the c> 2 it will approach infinity?
 
Ishuda said:
No! If t is greater than or equal to 2, it will approach infinity. What is I(s,t)?
Click to expand...
Sorry I meant to write t, although I still messed up and didn't include 2. So basically, while t<2 the limit is -2/(t-2)?
 
Ghost3k said:
Sorry I meant to write t, although I still messed up and didn't include 2. So basically, while t<2 the limit is -2/(t-2)?
Click to expand...
Yes. So now you have an explicit function for M(t) which is the limit as s goes to infinity of I(s,t) for t<2.
 
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