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01-13-2003, 04:45 PM
Hi. A basic question. I forgot how to and I know it's pathetic, but please somebody show me how to do (with a use of integration by part):

Improper integral (0 to +infinity) [x(e^x)]dx

Thanks a lot and don't laugh.

oedipus rex
01-13-2003, 07:50 PM
I think you mean &int; xe^(-x) dx from 0 to &infin; since the form you suggested is unbounded from above and the integral does not converge. Assuming this is what you mean, an integration by parts would proceed as follows: u = x, du = dx, dv = e^(-x) dx, v = -e^(-x).
Using the rule uv - &int; v du, we get [-xe^(-x) - e^(-x)] from 0 to &infin; = (0-0) - (0-1) = 1.
The problem gets interesting if you put a coefficient in the exponential (e^(-kx)) and raise x to integral powers, but that's for a rainy day...

01-13-2003, 08:15 PM
rex thanks! you are right. you should take the exam 3!!!

01-13-2003, 08:17 PM
oh wait. maybe you have already taken C3. ok thanks

oedipus rex
01-13-2003, 08:20 PM
hey, how did you know that I'm going to take Course 3? :o

01-13-2003, 08:51 PM
cuz you have mentioned some in C3 discussion sites.

oedipus rex
01-13-2003, 09:10 PM
Oh yeah. So you're taking Course 3 too?

Dr T Non-Fan
01-13-2003, 10:31 PM
Out with my anal-retentive journal, which tallies people who have passed an exam yet still post in that course's threads.

"Rex"..."13-Jan-2002"..."16:50"...(slam) ...got it.

Carry on.

oedipus rex
01-13-2003, 11:09 PM
you might want to add an entry for "DTNF" Jan 10 in the Course 3 thread named "From Course 1 to Course 3", and of course this thread as well... :roll:

Dr T Non-Fan
01-13-2003, 11:15 PM
As if I didn't already (rolly eyes)!

I only note it because it was one of the reasons given for not separating the exam threads. I contended that it wasn't true, and I wanted to make sure I was right (because there was a slight possibility that I might be wrong).

01-14-2003, 12:05 AM
yeah rex, i'm taking C3. I forgot a lot of algebra from Calculus, I need to refresh my memory. i'm so pathetic.

oedipus rex
01-14-2003, 12:48 AM
Aha. It looks like this thread then could be a contender for being moved to the Course 3 heading, but it's probably not worth it...

Dr T Non-Fan
01-14-2003, 01:07 AM
I was thinking the same thing...
...so I did it.

cubedbee
01-14-2003, 10:28 AM
Here's another stupid calc question.

How would you do the improper integral from 0 to infinity of something in the form cx^2/(k+x)^4, where c and k are constants. I'm thinking its gonna have to be done using partial fractions, but I don't really remember that method and wondered if there was an easier way.

Also, how do you do an integral sign in html? are there any good resources to learn html from?

Gandalf
01-14-2003, 10:43 AM
Here's another stupid calc question.

How would you do the improper integral from 0 to infinity of something in the form cx^2/(k+x)^4, where c and k are constants. I'm thinking its gonna have to be done using partial fractions, but I don't really remember that method and wondered if there was an easier way.

Just substitution: let t = k + x, then dt = dx and it becomes the integral from k to infinity of c (t - k)^2 / t^4. Then expand numerator to c (t ^ 2 - 2k + k^2); the resulting expression becomes a sum of (negative) powers of t.

Bama Gambler
01-14-2003, 10:57 AM
Calculus is RARELY needed for Course 3!! Go to Bob Batten's seminar.

Been There Done That
01-14-2003, 11:37 AM
You could use parts, but there is an easier way .....

The integral (after inserting the minus sign) is the expected value of an exponential random variable with parameter 1. The expected value is 1.

Avi
01-14-2003, 11:53 AM
Rex, how did you get the integral sign and infinity in your post?

Thanks.

Tommy Vercetti
01-14-2003, 12:16 PM
&int

Avi
01-14-2003, 12:36 PM
It's an HTML object? I know it's not in ASCII.

I see.

Thanks.

Is there a list of these math related HTML codes?

Theyre not in http://www.htmlgoodies.com/tutors/&command.html

Bama Gambler
01-14-2003, 12:37 PM
Testing...

&int

didn't work. what am I doing wrong?

Avi
01-14-2003, 12:39 PM
All ampersand commands must start with an & and close with a ;

Try &ampint;

Bama Gambler
01-14-2003, 12:40 PM
&int;

Avi
01-14-2003, 12:42 PM
There ya go :D

Bama Gambler
01-14-2003, 12:43 PM
thanks :D

Tommy Vercetti
01-14-2003, 01:56 PM
Complete expectation of life:

ė = &int; (tPx) dt

another way to incert a symbol from windows is
go Start -> Programs -> accessories -> system tools

and pick "Character Map"
copy the symbols and paste.

oedipus rex
01-14-2003, 02:38 PM
and &infin (with a semicolon, Avi, how do you just print the text and not the symbol?) for infinity and so on, i think avi posted a link in a seperate threads that gives more of these '&' codes

cubedbee
05-06-2003, 03:07 PM
I think you mean ∫ xe^(-x) dx from 0 to 8 since the form you suggested is unbounded from above and the integral does not converge. Assuming this is what you mean, an integration by parts would proceed as follows: u = x, du = dx, dv = e^(-x) dx, v = -e^(-x).
Using the rule uv - ∫ v du, we get [-xe^(-x) - e^(-x)] from 0 to 8 = (0-0) - (0-1) = 1.
The problem gets interesting if you put a coefficient in the exponential (e^(-kx)) and raise x to integral powers, but that's for a rainy day...

I have a little trick/technique I found when doing a past mixed distribution (poisson where lambda was uniform) problem.

∫ x<sup>k</sup> e<sup>-x</sup> dx can be done doing integration by parts k times, but it can be a real pain. I was looking for a way to memorize the first couple k's, and found an easy algorithm to generate the answer for any k.

∫ x<sup>k</sup> e<sup>-x</sup> dx = -e<sup>-x</sup> [ a<sub>1</sub>x<sup>k</sup>+ a<sub>2</sub> x<sup>k-1</sup>+ ...+ a<sub>k</sub>x<sup>1</sup> + a<sub>k+1</sub>x<sup>0</sup>]. The notation is a little poor, but a<sub>1</sub> = 1, a<sub>2</sub> = a<sub>1</sub> * k, a<sub>3</sub> = a<sub>2</sub> * (k-1) etc. You just basically start with the left most term, then multiply each terms coefficient by its power to get the next term.

An example: ∫ x<sup>4</sup> e<sup>-x</sup> dx

Start with knowing the answer is -e<sup>-x</sup> [ x<sup>k</sup>+ a<sub>2</sub> x<sup>3</sup>+ a<sub>3</sub>x<sup>2</sup> + a<sub>4</sub>x<sup>1</sup> + a<sub>5</sub>x<sup>0</sup>] (Since the first coefficient is always 1). Second coefficient is just first coefficient * first power = 1*4 = 4. Third coefficient = 4 * 3 = 12. Fourth coefficient = 12 * 2 = 24. Last coefficient = 24 * 1 = 24.

So the answer is -e<sup>-x</sup> [ x<sup>k</sup>+ 4x<sup>3</sup>+ 12x<sup>2</sup> + 24x<sup>1</sup> + 24x<sup>0</sup>]

VernSchil
05-06-2003, 03:25 PM
I think you gave a more specific example of Tabular integration by parts. A very time saving method for any integration by parts.