Laplace transform of sin(ωt)/[1+cos^2(ωt)]

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Homework Statement


L{sin(ωt)/[1+cos^2(ωt)]} =

Homework Equations


d {arctan[cos(ωt)]} /dt =
- ω•sin(ωt)/[1+cos^2(ωt)]

The Attempt at a Solution


∫e^(-st)•[sin(ωt)/(1+cos²(ωt)] dt =

-(1/ω)•∫e^(-st)•{arctan[cos(ωt)]}' dt =

= (integrating by parts and taking Re(s) > 0) =

= π/(4ω) -(s/ω)•∫ e^(-st)•arctan[cos(ωt)]

= π/(4ω) -(s/ω)•L{arctan[cos( ωt)]}.

But at this point I don't know how to compute the last laplace transform.
So I don't know if this is the best way or even if my question has a (simple) solution at all.

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lightarrow
 
Last edited:

Answers and Replies

  • #2
Ray Vickson
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Homework Statement


L{sin(ωt)/[1+cos^2(ωt)]} =

Homework Equations


d {arctan[cos(ωt)]} /dt =
- ω•sin(ωt)/[1+cos^2(ωt)]

The Attempt at a Solution


∫e^(-st)•[sin(ωt)/(1+cos²(ωt)] dt =

-(1/ω)•∫e^(-st)•{arctan[cos(ωt)]}' dt =

= (integrating by parts and taking Re(s) > 0) =

= π/(4ω) -(s/ω)•∫ e^(-st)•arctan[cos(ωt)]

= π/(4ω) -(s/ω)•L{arctan[cos( ωt)]}.

But at this point I don't know how to compute the last laplace transform.
So I don't know if this is the best way or even if my question has a (simple) solution at all.

--
lightarrow
The problem has a solution in terms of the non-elementary "LerchPhi" functions, but it is nasty. Here is how it goes:

Using the identity ##\cos^2 u = (1/2) ( \cos 2u + 1),## we have
$$f(t) = \frac{1}{1 + \cos^2 (\omega t)} = \frac{2}{3 + \cos(2 \omega t)}.$$ Let ##L_0(S) = \int_0^\infty e^{-S t} f(t)\, dt,## and use
$$\sin(\omega t) = \frac{1}{2i} \left( e^{i \omega t} - e^{-i \omega t} \right),$$
so $$e^{-st} \sin(\omega t) = \frac{1}{2i} \left( e^{-(s-i \omega) t} - e^{-(s + i \omega )t} \right). $$
Therefore, your desired Laplace transform ##L(s)## is given by
$$L(s) = \frac{1}{2i} \left( L_0(s+i \omega) - L_0(s + i \omega) \right).$$ If you have access to software that can compute the LerchPhi function at complex arguments, then you can evaluate the formula for ##L(s)## at various numerical values of ##s##. The details are given in the attachment, which shows the output from a Maple session.

Note added in edit: if you have access to Maple, you can evaluate and/or plot ##L(s)##, because Maple has built-in routines for all the functions it employs. Despite the complex-looking nature of the answer, ##L(s)## comes out as real for real ##\omega## and ##s > 0.## (In Maple, ##I## is the reserved symbol for ##i = \sqrt{-1}.##) I suppose you can do similar things in Mathematica, but I do not have access to it.

If you do not have access to Maple but still want to evaluate ##L(s)## for various ##s## but fixed ##\omega## you could avoid all the LerchPhi business and proceed instead to a pure numerical approach. All you need do is evaluate a fairly simple integral over a fixed, finite interval. The point is that
$$g(t) = \frac{\sin(\omega t)}{1 + \cos^2 (\omega t)} $$ is periodic, with half-period ##\tau = \pi/ \omega##, and alternates in sign as we traverse various ##\tau##-intervals. That is:
$$g(t + \tau ) = - g(t), \:g(t + 2 \tau) = g(t), \: g(t + 3 \tau) = - g(t), \ldots.$$ This means that if we set
$$F(s) = \int_0^{\tau} e^{-st} g(t) \, dt$$ then we have
$$L(s) = F(s) \left[ 1 - e^{-s \tau} + e^{- 2 s \tau } - e^{-3 s \tau} + \cdots \right] = \frac{F(s)}{1 + e^{s \tau}}. $$
Since ##F(s)## is given by an integral over a finite interval, when ##\omega## is specified we can easily evaluate ##F(s)## numerically using one of the many, well-studied numerical integration schemes.
 

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The problem has a solution in terms of the non-elementary "LerchPhi" functions, but it is nasty. Here is how it goes:

Using the identity ##\cos^2 u = (1/2) ( \cos 2u + 1),## we have
$$f(t) = \frac{1}{1 + \cos^2 (\omega t)} = \frac{2}{3 + \cos(2 \omega t)}.$$ Let ##L_0(S) = \int_0^\infty e^{-S t} f(t)\, dt,## and use
$$\sin(\omega t) = \frac{1}{2i} \left( e^{i \omega t} - e^{-i \omega t} \right),$$
so $$e^{-st} \sin(\omega t) = \frac{1}{2i} \left( e^{-(s-i \omega) t} - e^{-(s + i \omega )t} \right). $$
Therefore, your desired Laplace transform ##L(s)## is given by
$$L(s) = \frac{1}{2i} \left( L_0(s+i \omega) - L_0(s + i \omega) \right).$$
Obviously the first term at right of the last equation should have been
$$L_0(s-i \omega)$$
but it was clear.
If you have access to software that can compute the LerchPhi function at complex arguments, then you can evaluate the formula for ##L(s)## at various numerical values of ##s##. The details are given in the attachment, which shows the output from a Maple session.

Note added in edit: if you have access to Maple, you can evaluate and/or plot ##L(s)##, because Maple has built-in routines for all the functions it employs. Despite the complex-looking nature of the answer, ##L(s)## comes out as real for real ##\omega## and ##s > 0.## (In Maple, ##I## is the reserved symbol for ##i = \sqrt{-1}.##) I suppose you can do similar things in Mathematica, but I do not have access to it.

If you do not have access to Maple but still want to evaluate ##L(s)## for various ##s## but fixed ##\omega## you could avoid all the LerchPhi business and proceed instead to a pure numerical approach. All you need do is evaluate a fairly simple integral over a fixed, finite interval. The point is that
$$g(t) = \frac{\sin(\omega t)}{1 + \cos^2 (\omega t)} $$ is periodic, with half-period ##\tau = \pi/ \omega##, and alternates in sign as we traverse various ##\tau##-intervals. That is:
$$g(t + \tau ) = - g(t), \:g(t + 2 \tau) = g(t), \: g(t + 3 \tau) = - g(t), \ldots.$$ This means that if we set
$$F(s) = \int_0^{\tau} e^{-st} g(t) \, dt$$ then we have
$$L(s) = F(s) \left[ 1 - e^{-s \tau} + e^{- 2 s \tau } - e^{-3 s \tau} + \cdots \right] = \frac{F(s)}{1 + e^{s \tau}}. $$
Since ##F(s)## is given by an integral over a finite interval, when ##\omega## is specified we can easily evaluate ##F(s)## numerically using one of the many, well-studied numerical integration schemes.
Thank you very much Ray!

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lightarrow
 

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