can ANYONE solve this integral?

[flood]

integral of e ot the power of t spuared times e to the power of t
it looks like:
S(e^(t^2)) (e^t) dt
S representing the integral sign

I need to get it down to something with no integrals.

 

[Napalm381]

Did you try integration by parts?
Sorry, don't have time to do it now myself.

 

[yakko]

42

 

[Gatsby]

Have you tried the integrator?

Integrator

Gatsby

 

[Wangel]

21

 

[flood]

Integration by parts is the idela way to do this, but I've done it several times and it still hasnt worked.

The method that I tried to use is:
S(e garbage) = stuff - S(e garbage)

2(S(e garbage) ) = stuff

S(e garbage) = (1/2) stuff

but what happens is you have to do partial integration twice, and so the "S(e garbage)" comes out to be positive on the right side.

 

[DAM]

u = e^t

du = e^t dt


replace your substitution


new problem is S u^2 du


i hope you know how to do that problem




dam()

 

[flood]

DAM-

that would work if it was
[e^t]^2

but its e^(t^2)

which can be rewritten as [e^t]^t
hmm... havent look at it that way before

 

[DAM]

dohh


thats right





dam()

 

[flood]

Gatsby-
the integrator gave me a real messy answer with "Erfi" in it
I have no idea what "Erfi" means

 

[Passions]

integrate by parts

 

[flood]

BOBBY RIBS-
integration by parts barfs ^^^

 

[flood]

^
did the forums just go down for a minute?

 

[DAM]

u = e^t^2
du = 2te^t^2

v = e^t
dv = e^tdt

the integral becomes: (e^t^2)(e^t) - S (e^t)(2e^t^2)dt

which can be written as: (e^t^2)(e^t) - 2S (e^t^3)dt


then use u subs

u = t
du = dt

integral becomes: (e^u^2)(e^u) - 2S (e^u^3)dt

and just take the derivative of (e^u^2)




dam()




(e^u^2)(e^u) - 2S (e^u)(e^u^

 

[flood]

DAM-
how'd you get from u = e^t^2 to du = 2t3^t^2 ?

derivitive of u = (e^t^2) 2t dt

 

[DAM]

typo dude 3 = e



dam(_)

 

[flood]

DAM-
you selected a u and a v
what youre supposed to do is select a u and a dv

 

[DAM]

you selected a u and a v
what youre supposed to do is select a u and a dv


i posted this:

u = e^t^2
du = 2te^t^2

v = e^t
dv = e^tdt

original problem:

S(e^(t^2)) (e^t) dt


now what do you mind asking again your question cause i dont understand?




dam()

 

[perry]

Integration by parts is:

S(u dv) = u * v - S(v du)

 

[perry]

And it'll probably require multiple integration by parts or integration by parts and a substitution if you can figure it out.

First integration by parts:
dv = e^t u = e^[t^2]
v = e^t du = 2*t*e[t^2]

So:
e^[t^2] * e^t - S(e^t * 2 * t * e^[t^2] dt) =
e^[t^2] * e^t - 2 * S(t * e^[t^2 + t] dt)

Second starts to get really messy, which is why I think you need substitution. I worked the second out, but it won't help too much. Gots ta get back to my own work..

Edit: That's probably wrong, but who knows. I haven't take a math course in a while.

 

[Thanatopsis]

Yakko,

The answer is 42, but what is the question? If only we had 5 more minutes to know...

 

[flood]

when you do integration by parts, you take the integral you start with
and break it into two pieces: a u and a dv
you take the derivitive of u to get your du
you take the integral of dv to get v
you do no select v or du
you get those from the other parts. you select only the u and dv.

when you did:
u = e^t^2
du = 2te^t^2

v = e^t
dv = e^tdt

you choose the parts of the original problem to be u and v
(or u and dv)

 

[hendon]

The integral cannot be done by hand, in particular, the integral of e^(t^2) dt cannot be formed from any of the elementary functions that we know.

Using the integrator that was mentioned earlier in the thread, I think you get something involving an 'erfi'...
check out this website for the definition... some complex stuff

http://mathworld.wolfram.com/Erfi.html