Iterated integrals involving exp(-t^2)

[JumboJam]

Hello all, I am new to these forums. Please let me know if I am off base a bit.

I am unsure of how to even start the following problem, so holding my hand would be appreciated.

Let \(\displaystyle A\, =\, \displaystyle{ \int_0^1 \, } e^{-t^2}\, dt\) and \(\displaystyle B\, =\, \displaystyle{ \int_0^{\frac{1}{2}} \,} e^{-t^2}\, dt.\) Evaluate the iterated integral:

. . . . .\(\displaystyle I\, =\, 2\, \displaystyle{ \int_{-\frac{1}{2}}^1 \, }\) \(\displaystyle \left[\displaystyle{ \int_0^x \,} \exp\left(-y^2\right)\, dy \right]\, dx\)

in terms of \(\displaystyle A\) and \(\displaystyle B.\) These are positive integers \(\displaystyle m\) and \(\displaystyle n\) such that:

. . . . .\(\displaystyle I\, =\, mA\, -\, nB\, +\, e^{-1}\, -\, e^{-\dfrac{1}{4}}\)

Use this fact to check your answer.

Thank you in advance.

 

[stapel]

Let \(\displaystyle A\, =\, \displaystyle{ \int_0^1 \, } e^{-t^2}\, dt\) and \(\displaystyle B\, =\, \displaystyle{ \int_0^{\frac{1}{2}} \,} e^{-t^2}\, dt.\) Evaluate the iterated integral:

. . . . .\(\displaystyle I\, =\, 2\, \displaystyle{ \int_{-\frac{1}{2}}^1 \, }\) \(\displaystyle \left[\displaystyle{ \int_0^x \,} \exp\left(-y^2\right)\, dy \right]\, dx\)

in terms of \(\displaystyle A\) and \(\displaystyle B.\) These are positive integers \(\displaystyle m\) and \(\displaystyle n\) such that:

. . . . .\(\displaystyle I\, =\, mA\, -\, nB\, +\, e^{-1}\, -\, e^{-\frac{1}{4}}\)

Use this fact to check your answer.

Is "These" supposed to be "There"? Thanks!

 

[JumboJam]

The problem statement says "these" right there, but maybe this is a typo?? Would this make the problem statement solvable??

Thanks for the help!

 

[doive]

Hello all, I am new to these forums. Please let me know if I am off base a bit.

I am unsure of how to even start the following problem, so holding my hand would be appreciated.

Let \(\displaystyle A\, =\, \displaystyle{ \int_0^1 \, } e^{-t^2}\, dt\) and \(\displaystyle B\, =\, \displaystyle{ \int_0^{\frac{1}{2}} \,} e^{-t^2}\, dt.\) Evaluate the iterated integral:

. . . . .\(\displaystyle I\, =\, 2\, \displaystyle{ \int_{-\frac{1}{2}}^1 \, }\) \(\displaystyle \left[\displaystyle{ \int_0^x \,} \exp\left(-y^2\right)\, dy \right]\, dx\)

in terms of \(\displaystyle A\) and \(\displaystyle B.\) These are positive integers \(\displaystyle m\) and \(\displaystyle n\) such that:

. . . . .\(\displaystyle I\, =\, mA\, -\, nB\, +\, e^{-1}\, -\, e^{-\dfrac{1}{4}}\)

Use this fact to check your answer.

Thank you in advance.

"These" and "There" look like a typo, but it's not really too pertinent to the question.
The integral \(\displaystyle { \int} e^{-t^2}dt \) does not have a general solution, I'm working on an almost forgotten memory here, but you may be able to get some joy from the form of integration by parts?

Since: \(\displaystyle \int u dv = uv - \int v du \)

let u equal the inside integral : \(\displaystyle u = \int^x_0 e^{-y^2} dy\) and \(\displaystyle du = e^{-x^2} -1 \)
trivially: dv = dx, v = x

NB I'm working on the assumption that \(\displaystyle u = \int^x_0 e^{-y^2} dy \implies du = e^{-y^2} |^x_0 = e^{-x^2} - 1\) but I'm not 100% convinced that is the case as it's been a while?
NB2 I'm also working with the assumption that \(\displaystyle \int v du \equiv v du \) which may also be dodgy!

This gives:
\(\displaystyle
I = 2 \Bigl[ x \int^x_0 e^{-y^2} dy - x (e^{-x^2} -1) \Bigr] ^t_{-\frac{1}{2}}
\)


That feels like a long way to solving it as we can now sub in those terms and get:
\(\displaystyle
I = 2 \Biggl[ \Bigl[ t \int^t_0 e^{-y^2} dy - t( e^{-t^2} - 1) \Bigr] - \Bigl[ -\frac{1}{2} {\int ^{-\frac{1}{2}} _0} e^{-y^2} dy + \frac{1}{2} ( e^{\frac{1}{4}} -1 ) \Bigr] \Biggr]
\)

Comparing to what we're given in the question and noticing a bit of symmetry the second bracket becomes quite neat:
\(\displaystyle ( B - e^{-\frac{1}{4}} + 1 ) \)
I leave the rest as an exercise to the reader because it's getting late

EDIT: if you can somehow set up t = 1 on the limit of your remaining integral you get the first bracket as:
\(\displaystyle 2(A - 2e^{-1} + 1)\)
Overall that gives you:
\(\displaystyle I = 2A + B - 2 e^{-1} - e^{-\frac{1}{4}} - \frac {1}{2} \)
which isn't quite right, but it pretty **** close for 2am, having not done any calculus since leaving college 3 yrs ago!
I'm sure you can do some tidy up and checking and get the proper solution - when you do, please post it up as I'd be interested to see how close I got!

 

[JumboJam]

Thanks for the help. Coincidentally, the integral actually is from -1/2 to 1, not -1/2 to t. It must have been accidentally altered when it was converted to LaTex from the image I posted.

With this in mind, I started with integration by parts, just as doive did. I got

\(\displaystyle I\, =\, 2\, (x\,\displaystyle{ \int_{0}^x \, }\) \(\displaystyle \left[ \exp\left(-y^2\right)\, dy \right]\, -x\,\exp\left(-x^2\right)\,-x\Bigr] ^1_{-\frac{1}{2}}\))

Plugging in terms, this gives

\(\displaystyle I\, =\, 2\, (1\,\displaystyle{ \int_{0}^1 \, }\) \(\displaystyle \left[ \exp\left(-y^2\right)\, dy \right]\, -1\,\exp\left(-1^2\right)\,-1\,-\frac{1}{2}\,\displaystyle{ \int_{0}^\frac{-1}{2} \, }\) \(\displaystyle \left[ \exp\left(-\frac{1}{2}\,^2\right)\, dy \right]\, -\frac{1}{2}\,\exp\left(-\frac{1}{2}^2\right)\,-\frac{1}{2}\))

Then, noting symmetry (\(\displaystyle \exp\left(-\frac{1}{2}^2\right)\, =\, exp\left(\frac{1}{2}^2\right)\))

I got (\(\displaystyle 2\,A - \,2\,exp\left(-1\right)\, - B - \,exp\left(\frac{-1}{4}\right)\,-1\))

This isn't QUITE what the problem statement said, but is pretty close? Anyone see what I did wrong?