Rewriting sin(x^(1/2)) for Laplace Transform

[twotaileddemon]

I haven't read through your results but the integral certainly isn't zero. It's the Gaussian integral:

$$\int_{-\infty}^{\infty} e^{-a v^2}dv=\sqrt{\frac{\pi}{a}}$$

isn't this some kind of error function though? from what I understand it's impossible to integrate exp(-x^2)?

 

[vela]

You can evaluate it since the limits are from -infinity to +infinity.

 

[twotaileddemon]

You can evaluate it since the limits are from -infinity to +infinity.

hmm ok. any other limits would fail though, right?

 

[vela]

Because of the symmetry of the Gaussian, you can also figure out the integral from 0 to infinity, but yeah, for arbitrary limits in general, you have to evaluate the integral numerically.

 

[Cyosis]

Any limit would work, but it just so happens that the integral has a neat solution with the limits you have.

 

[twotaileddemon]

oh ok!
so after combining the two integrals with (ia/2s) in that way, I ended up getting just a pi^1/2 for that. the other two with v*(exp(-v²) ended up canceling out, and so I was left with

1/(as)^1/2*exp(-a/4s)*(ia/2s)*pi^1/2
= (a*pi)^1/2 * exp(-a/4s) / s^3/2

It looks weird because I guess I'm used to laplace transforms with "neat" answers, but at least there are no imaginary numbers in the answer.

 

[vela]

I think you still have an overall factor of 1/2 missing from when you wrote the sine in terms of exponentials.

 

[twotaileddemon]

I think you still have an overall factor of 1/2 missing from when you wrote the sine in terms of exponentials.

I did bring that out front, but it ended up canceling with the 2 I had after making the u substitution dt = 2u/a du (at the very beginning)

 

[gabbagabbahey]

Your final answer is missing a factor of 1/2, but other than that it looks good.

As an aside, the trick to evaluating $\int_{-\infty}^{\infty}e^{-x^2}dx$ is to realize that it is identical to $\int_{-\infty}^{\infty}e^{-y^2}dy$ and so

$$\int_{-\infty}^{\infty}e^{-x^2}dx=\left(\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-(x^2+y^2)}dxdy\right)^{1/2}$$

which you can easily evaluate by swithcing to polar coordinates.

 

[vela]

I did bring that out front, but it ended up canceling with the 2 I had after making the u substitution dt = 2u/a du (at the very beginning)

Oh, yeah. Well, you still have a two missing somewhere then.

 

[twotaileddemon]

Your final answer is missing a factor of 1/2, but other than that it looks good.

As an aside, the trick to evaluating $\int_{-\infty}^{\infty}e^{-x^2}dx$ is to realize that it is identical to $\int_{-\infty}^{\infty}e^{-y^2}dy$ and so

$$\int_{-\infty}^{\infty}e^{-x^2}dx=\left(\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-(x^2+y^2)}dxdy\right)^{1/2}$$

which you can easily evaluate by swithcing to polar coordinates.

that's really quite interesting and good to know - thank you!

I checked my work over, and I don't -think- I'm missing a factor of 1/2. When I converted from sin[(at)^(1/2)] to sin u, I made the substitution u = (at)^(1/2), and so du = (a/2u)dt and dt = (2u/a)du. I brought this 2 out front, and when I rewrote sinu in terms of complex exponential, I brought out the 1/(2i) to the front and the 2's canceled.

EDIT:

Oh, yeah. Well, you still have a two missing somewhere then.

actually, yeah, I found it. I had it in my calculations, but when simplifying the terms I forgot to include it in the final answer. Thanks for pointing that out!

Thanks again for your patience and guidance (to everyone involved) =]

 

[gabbagabbahey]

The missing 1/2 came from the factor of ia/2s you got after the v-sub. You had it in your second last line of post#41, but then it inexplicably disappeared in your last line.