Integration

[Axioms]

I am struggling to remember the rules regarding e within integration.

 

I know that the integral of e^x is e^x. This means that the integral of e^2x must be equal to (1/2)e^2x. (derivative of (1/2)e^2x = e^2x)

 

Ok now I am having trouble with two integrals. I am getting the right answer for the one in the book but I am not sure if my method is right.

 

Can anyone please confirm if my integration of (xe^-x^-2) is correct? (I get (1/2)(e)^(-x)^(-2)

 

My next question is how do you integrate (e^-x^2). Do you have to manipulate it algebratically? I know you can determine if it is divergent or convergent with a comparison test. I basically just want to know how you would start the evaluation of the indefinite integral of (e^-x^-2)dx.

 

Any help would be much appreciated.

[Axioms]

Does anyone know where I can find worked out examples of comparison theory for indeterminate integrals? I cant find any good explinations in any of my text books. I need to remember how to tell if a graph is converging or diverging if the integral is indeterminate. What I know is that you compare it to a similar equation but i cannot remember what the conditions are in doing so. Any help would be appreciated >.<

[Axioms]

Any ideas? I still dont know how to integrate (e^-x^2). I assume you must integrate by parts. let u = (e^-x^2), du = (-2xe^-x^2), dv = dx, x = v.

 

[integral of (e^-x^2)dx] = (e^-x^2)(x) - [integral of (-2xe^-x^2)(x)dx] ....1

 

let J = [integral of (-2xe^-x^2)(x)dx]

 

Thus let s = x, ds = dx, dt =(-2xe^-x^2) t= (e^-x^2)

 

Therefore J = x(e^-x^2) - [integral of (e^-x^2)dx]

 

Substitute J into 1

 

[integral of (e^-x^2)dx] = (e^-x^2)(x) - [x(e^-x^2) - [integral of (e^-x^2)dx]]

 

And this proves that zero = zero ........ -.- lol any1 know if this is the right track though? If I change s and t around and then sub from there maybe? I just cannot seem to simplify it. Probably missing something stupid or differentiation it wrong.

[Zorgoth]

Don't waste your time on integral(e^-(x^2))

 

It's indefinite integral is called erf (the error function) and doesn't have a closed form (i.e. it cannot be expressed algebraically using standard functions). To prove it converges, just use a test rather than evaluating it.

 

You *can* integrate it from -infinity to infinity using a trick from multivariable calculus, and the answer is sqrt(pi).Don't say that in your homework though, because you presumably can't prove it using calculus 2 methods.

Edited November 5, 2012 by Zorgoth

[Axioms]

Thanks a lot for the help. I was stuck on this question for a while. I've never used erf in calculus before so I am not sure how you can identify it. Would it then be: erf(z)= 2/sqrt(pi)* integral(e^-y^2)dy from 0 to z? Not sure where you can get the erf values from though? I dont need to do it so its ok. I do this in other subjects though so it would be interesting to know how they combine it with pure culculus.