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

[gabbagabbahey]

Do the signs work out for that? The function $ve^{-v^2}$ is odd, and though the function $e^{-v^2}$ is even, the integrals with it as the integrand have the wrong overall signs in the beginning, right?

(I'll just mention I just scribbled some stuff on paper, not being particularly careful, so I could very well have messed up the signs myself.)

Works out fine for me (I even verified with Mathematica, and I get the correct results)...you end up subtracting the two odd integrals (which leaves you an integral from -something to +something of the odd function, yielding zero) and adding the two even integrals.

 

[gabbagabbahey]

I don't even know, to be honest.

I've been working on this problem for a long, long time today and it's really confusing me =|. I got to a point where I had to integrate exp(-x²) which is apparently not integrable, and wolfram suggested to use the error function and gave me some weird calculation. My answer now not only involves complex numbers but the error function as well.

I'm sure I did the substitution, completing the square, and distributing the integrals while keeping the limits of integration correct, but I'm truly at a loss at this point! I think I may just go for extra help on tuesday unless I can figure out something tomorrow.

I'll consider everything that has been said though and think about it some more. Thanks for your help!

Are you alright getting to the point

$$\mathcal{L}\left[\sin(\sqrt{at}\right]=\frac{{\rm e}^{-\frac{a}{4s}}}{{\rm i}\sqrt{as}}\left[\int_{-\alpha}^{\infty}\left(\sqrt{\frac{a}{s}}v+\frac{{\rm i}a}{2s}\right){\rm e}^{-v^2}dv-\int_{\alpha}^{\infty}\left(\sqrt{\frac{a}{s}}v-\frac{{\rm i}a}{2s}\right){\rm e}^{-v^2}dv\right]$$

where $\alpha=\frac{{\rm i}a}{2s}\sqrt{\frac{a}{s}}$?

If so, why not post your work from then on...

 

[vela]

Works out fine for me (I even verified with Mathematica, and I get the correct results)...you end up subtracting the two odd integrals (which leaves you an integral from -something to +something of the odd function, yielding zero) and adding the two even integrals.

Oh, OK, I was looking at just one integral, assuming the problem would work out differently, and wasn't getting what I thought I would need.

 

[twotaileddemon]

Are you alright getting to the point

$$\mathcal{L}\left[\sin(\sqrt{at}\right]=\frac{{\rm e}^{-\frac{a}{4s}}}{{\rm i}\sqrt{as}}\left[\int_{-\alpha}^{\infty}\left(\sqrt{\frac{a}{s}}v+\frac{{\rm i}a}{2s}\right){\rm e}^{-v^2}dv-\int_{\alpha}^{\infty}\left(\sqrt{\frac{a}{s}}v-\frac{{\rm i}a}{2s}\right){\rm e}^{-v^2}dv\right]$$

where $\alpha=\frac{{\rm i}a}{2s}\sqrt{\frac{a}{s}}$?

If so, why not post your work from then on...

I have some work, but I cannot understand how to do the latex for it (i did read your signature, but I easily get lost in all the latex text) But basically, you distribute the integrals, then for the one that has -(ia/2s)*exp(-v^2), change the limits of the integral from -alpha to -infinity because it is even and add a plus sign in front. then switch the limits from (-alpha to -infinity) to (-infinity to -alpha) by adding a minus sign in front again. so then you have

(integral of -alpha to infinity) of (ia/2s)*exp(-v^2) + (integral of -infinity to -alpha) of (-ia/2s)*exp(-v^2),

which is equal to integral from - infinity to infinity of (ia/2s)*exp(-v^2) ... which is zero.
is this right?

 

[Cyosis]

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

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

 

[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.