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

[twotaileddemon]

Homework Statement

I would like to know how to rewrite sin[x^(1/2)]? I have to do a laplace transform of a certain equation but I need to get it into a more usable form...

Homework Equations

The Attempt at a Solution

I was thinking of using the binomial theorem but 1/2 is neither a even or odd number so I can't use it?

 

[rock.freak667]

I don't think there is a way to get in a more usable form. Was your equation to solve directly involving sin(√x)?

 

[twotaileddemon]

I don't think there is a way to get in a more usable form. Was your equation to solve directly involving sin(√x)?

essentially, yes, where x = at. and

f(t) = sin(√at)

I didn't know how to solve it because there was a t^(1/2) term X_X. Do you have any ideas?

Can I just use the formula L(f) = F(s) = $$\int$$f(t)e^-st dt where the integration runs from 0 to infinity, even though I have a t^(1/2) term?

 

[gabbagabbahey]

Can I just use the formula L(f) = F(s) = $$\int$$f(t)e^-st dt where the integration runs from 0 to infinity, even though I have a t^(1/2) term?

Of course. That is the definition of the Laplace transform after all (you are missing your integration limits though)

 

[twotaileddemon]

Of course. That is the definition of the Laplace transform after all (you are missing your integration limits though)

can you have a Fourier transform with a complex number in it? The answer I got has a complex number in it...

 

[gabbagabbahey]

If $a$ is real, your answer should be too. When you take the Laplace transform of a real-valued function, you get a real-valued function.

Maybe you should show us your calculation...

 

[Hellabyte]

You could try 'u' substitution

 

[twotaileddemon]

If $a$ is real, your answer should be too. When you take the Laplace transform of a real-valued function, you get a real-valued function.

Maybe you should show us your calculation...

well I'm not sure how to do all the symbols and stuff very well, but basically, I did

$$\int$$sin[(at)^1/2]e^-stdt

substituted 1/(2i) * ( exp[i(at)^1/2] - exp[-i(at)^1/2] ) for the sine term, integrated with limits between 0 and infinity, and eventually got

-i / s

as my answer, where i is an imaginary number.

 

[gabbagabbahey]

well I'm not sure how to do all the symbols and stuff very well

See my sig

$$\int$$sin[(at)^1/2]e^-stdt

substituted 1/(2i) * ( exp[i(at)^1/2] - exp[-i(at)^1/2] ) for the sine term, integrated with limits between 0 and infinity, and eventually got

-i / s

as my answer, where i is an imaginary number.

Your method looks fine, but your result doesn't...show the rest of your calculations

 

[twotaileddemon]

See my sig



Your method looks fine, but your result doesn't...show the rest of your calculations

Okay!

1/(2i) * [ $$\int$$ exp[i(at)^1/2] - st] dt - $$\int$$ exp[-i(at)^1/2] - st] dt ] =

1/(2i) * [ 1 / [ i(a / 2(at)^1/2) - s] - 1 / [ -i(a / 2(at)^1/2) - s] ]

*note at this point I plugged in 0 for t for the lower limit and infinity for the upper limit of t from integration*
= 1/(2i) * [1/s + 1/s]
= 1/(2i) * [2/s]
= 1/is = -i/s

 

[gabbagabbahey]

You seem to be claiming that

$$\int e^{h(t)}dt=\frac{e^{h(t)}}{h'(t)}$$

Is that really true? (Use the product rule to differentiate $$\frac{e^{h(t)}}{h'(t)}$$ and see if you get [itex]e^{h(t)}[/tex])<br /> <br /> Instead, start with the substitution $u=\sqrt{at}$ and then complete the square on the exponent...[/itex]

 

[twotaileddemon]

You seem to be claiming that

$$\int e^{h(t)}dt=\frac{e^{h(t)}}{h'(t)}$$

Is that really true? (Use the product rule to differentiate $$\frac{e^{h(t)}}{h'(t)}$$ and see if you get [itex]e^{h(t)}[/tex])<br /> <br /> Instead, start with the substitution $u=\sqrt{at}$ and then complete the square on the exponent...[/itex]

$<br /> I did the substitution and now I have an exponent equal to<br /> <br /> exp(iu - su²/a)<br /> <br /> and another equal to <br /> <br /> exp(-iu - su²/a)<br /> <br /> can I complete the square with this and solve it?$

 

[gabbagabbahey]

Yes, start by completing the square...

 

[twotaileddemon]

Yes, start by completing the square...

ok well..

exp(iu - su²/a)

= exp[-(su²/a - iu)]
= exp[-(s/a)(u² - i(a/s)u)]
= exp[-(s/a)(u-(ia/2s)² - (i²/(4(s/a))]
= exp[-(s/a)(u-(ia/2s)² + a /(4s)]
...?

 

[vela]

ok well..

exp(iu - su²/a)

= exp[-(su²/a - iu)]
= exp[-(s/a)(u² - i(a/s)u)]

Up to here it looks okay

= exp[-(s/a)(u-(ia/2s)² - (i²/(4(s/a))]
= exp[-(s/a)(u-(ia/2s)² + a /(4s)]
...?

Your parentheses don't match up here, but what you actually have is probably right.

Note that e^a/(4s) is a constant, so you can pull it out of the integral.

 

[gabbagabbahey]

Right, now $$e^{-\frac{s}{a}\left(u-\frac{ia}{2s}\right)^2+\frac{a}{4s}}=e^{\frac{a}{4s}}e^{-\frac{s}{a}\left(u-\frac{ia}{2s}\right)^2}$$ and $$e^{\frac{a}{4s}}$$ is a constant and can be pulled out of the integral.

Now, use the substitution [itex]v=\sqrt{\frac{s}{a}}\left(u-\frac{ia}{2s}\right)[/tex]...[/itex]

 

[twotaileddemon]

Right, now $$e^{-\frac{s}{a}\left(u-\frac{ia}{2s}\right)^2+\frac{a}{4s}}=e^{\frac{a}{4s}}e^{-\frac{s}{a}\left(u-\frac{ia}{2s}\right)^2}$$ and $$e^{\frac{a}{4s}}$$ is a constant and can be pulled out of the integral.

Now, use the substitution [itex]v=\sqrt{\frac{s}{a}}\left(u-\frac{ia}{2s}\right)[/tex]...[/itex]

$<br /> also.. I have a "u term" from what I did<br /> <br /> u = (at)^1/2<br /> t = u²/a<br /> du = (a/2u)*dt<br /> dt = (2u/a)*du<br /> <br /> so I actually have a 1/ai in front of the integral and another u in it.<br /> That means I have to do integration by parts in addition to what you said, right?$

 

[gabbagabbahey]

You won't need to use IBP...what does your integral become after the v-sub?

 

[twotaileddemon]

You won't need to use IBP...what does your integral become after the v-sub?

a big mess!
the limits of integration are from -(i/2)(a/s)^1/2 to infinity now, and the equation under the integral is
(a/s)^1/2 * [(a/s)^1/2v + (ia/2s)]exp(-v²) dv

note: there was a u term before the whole exponential term, so I accounted for that. and then made a substitution for du as well

 

[gabbagabbahey]

Well, I'm sure you know how to integrate $\int ve^{-v^2}dv$ right? And if you use the substitution $v=\sqrt{\frac{s}{a}}\left(u+\frac{ia}{2s}\right)$ on the $$e^{-\frac{s}{a}\left(u+\frac{ia}{2s}\right)^2$$...I think you will get some fortuitous cancellation

 

[twotaileddemon]

Well, I'm sure you know how to integrate $\int ve^{-v^2}dv$ right? And if you use the substitution $v=\sqrt{\frac{s}{a}}\left(u+\frac{ia}{2s}\right)$ on the $$e^{-\frac{s}{a}\left(u+\frac{ia}{2s}\right)^2$$...I think you will get some fortuitous cancellation

I do know how to integrate that, but how can you make a substitution twice using the same variable? you set v equal to two different things in two different equations?

 

[gabbagabbahey]

Separate into two integrals first...your substitution variable is then a dummy variable in each integral

 

[twotaileddemon]

Separate into two integrals first...your substitution variable is then a dummy variable in each integral

I did separate into two integrals, using different substitutions for v, namely the one with the +(ia/2s) in the equation and the other with -(ia/2s) in the equation.

but then the limits of integration are different for each integral and you can't combine and cancel.

hence.. I was thinking of splitting up each integral further, and having to end up using integration by parts at least once.

 

[vela]

The integrands don't have any poles, so you can shift the contours without changing the values of the integral. In particular, you can just drop the imaginary part in the limits, so the integrals both go from 0 to ∞.

 

[gabbagabbahey]

You'll want to use the fact that if $g(v)$ is even,

$$\begin{aligned}\int_{-\alpha}^{\infty} g(v)dv+\int_{\alpha}^{\infty} g(v)dv &= -\int_{\alpha}^{-\infty} g(v)dv+\int_{\alpha}^{\infty} g(v)dv \\ &= \int^{\alpha}_{-\infty} g(v)dv+\int_{\alpha}^{\infty} g(v)dv \\ &= \int_{-\infty}^{\infty} g(v)dv\end{aligned}$$

 

[twotaileddemon]

The integrands don't have any poles, so you can shift the contours without changing the values of the integral. In particular, you can just drop the imaginary part in the limits, so the integrals both go from 0 to ∞.

so if the limits of the first integral were -i(1/2)(a/s)^1/2 to infinity and the second were +i(1/2)(a/s)^1/2 to infinity, I can just make it 0 to infinity in both cases by ignoring the imaginary parts?

I do see how there are no poles, as we covered this a while ago. I never knew this could affect limits of integration though.. thanks for letting me know!

You'll want to use the fact that if $g(v)$ is even,

$$\begin{aligned}\int_{-\alpha}^{\infty} g(v)dv+\int_{\alpha}^{\infty} g(v)dv &= -\int_{\alpha}^{-\infty} g(v)dv+\int_{\alpha}^{\infty} g(v)dv \\ &= \int^{\alpha}_{-\infty} g(v)dv+\int_{\alpha}^{\infty} g(v)dv \\ &= \int_{-\infty}^{\infty} g(v)dv\end{aligned}$$

we don't have the same function under both integrals though,

under one integral I have (a/s)^1/2 * [(a/s)^1/2v + ia/(2s)]exp(-v²) and the other (a/s)^1/2 * [(a/s)^1/2v - ia/(2s)]exp(-v²)

where the first has limits from -i(1/2)(a/s)^1/2 to infinity and the other has +i(1/2)(a/s)^1/2 to infinity.

 

[gabbagabbahey]

Break up the integrals first...$$\int\left[\sqrt{\frac{a}{s}}v + \frac{ia}{2s}\right]e^{-v^2}dv=\sqrt{\frac{a}{s}}\int ve^{-v^2}dv +\frac{ia}{2s}\int e^{-v^2}dv$$

 

[twotaileddemon]

Break up the integrals first...$$\int\left[\sqrt{\frac{a}{s}}v + \frac{ia}{2s}\right]e^{-v^2}dv=\sqrt{\frac{a}{s}}\int ve^{-v^2}dv +\frac{ia}{2s}\int e^{-v^2}dv$$

$$\int\left[\sqrt{\frac{a}{s}}v + \frac{ia}{2s}\right]e^{-v^2}dv=\sqrt{\frac{a}{s}}\int ve^{-v^2}dv +\frac{ia}{2s}\int e^{-v^2}dv$$

---

$$\int\left[\sqrt{\frac{a}{s}}v - \frac{ia}{2s}\right]e^{-v^2}dv=\sqrt{\frac{a}{s}}\int ve^{-v^2}dv -\frac{ia}{2s}\int e^{-v^2}dv$$

then I can combine the two integrals with exp(-v²) from this using the equation you gave me? okay, let me try that...

BTW - thank you both for all of your help so far :) it has been invaluable

so , I ended up getting:

$$\begin{aligned}\int_{-\alpha}^{\infty} g(v)dv-\int_{\alpha}^{\infty} g(v)dv &= -\int_{\alpha}^{-\infty} g(v)dv-\int_{\alpha}^{\infty} g(v)dv \\ &= \int^{\alpha}_{-\infty} g(v)dv-\int_{\alpha}^{\infty} g(v)dv \\ &= 0\end{aligned}$$

because it's even... right?

 

[vela]

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.)

 

[twotaileddemon]

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.)

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!