Laplace transform of $\frac{1-\cos (t)}{t}$

Question

In the documentation, it states that

The Laplace transform of a function $f(t)$ is defined to be $\int_0^{\infty } f(t) e^{-s t} \, \mathrm{d}t$.

But why can Mathematica not get the Laplace transform of $\frac{1-\cos (t)}{t}$?

$Assumptions=s>0;

LaplaceTransform[(1-Cos[t])/t,t,s]
(* EulerGamma+LaplaceTransform[1/t,t,s]+1/2 Log[1+s^2] *)

and the integral does converge

Integrate[(1-Cos[t])/t Exp[-s t],{t,0,∞}]
(* 1/2 Log[1+1/s^2] *)

?

Answer 1

You can use a trick to prevent Mathematica from taking your expression apart:

LaplaceTransform[Abs[1 - Cos[t]]/t, t, s]
(* 1/2 Log[1 + 1/s^2] *)

Answer 2

When Mathematica tries to pull the fraction apart, it gets

$$\mathcal{L}_t\left[\frac{1-\cos(t)}{t}\right](s)=\mathcal{L}_t\left[\frac{1}{t}\right](s) - \mathcal{L}_t\left[\frac{\cos(t)}{t}\right](s)$$

While the cosine term has a Laplace-transform, $1/t$ doesn't have a transform. That might be the reason why Mathematica cannot solve it. The problem is, that the $1/t$ term has a singularity at 0:

Limit[1/t, t -> 0, Direction -> -1]

(* ∞ *)

while the complete expression doesn't

Limit[(1 - Cos[t])/t, t -> 0, Direction -> -1]

(* 0 *)

On the other hand, calculating the back-transform works:

InverseLaplaceTransform[1/2 Log[1 + 1/s^2], s, t] // FullSimplify

(* (1 - Cos[t])/t *)

What you can do is the following. You expand your formula into a series

Series[(1 - Cos[t])/t, {t, 0, 10}] // Normal

(* t/2 - t^3/24 + t^5/720 - t^7/40320 + t^9/3628800 *)

then you use LaplaceTransform

LaplaceTransform[%, t, s]

(* 1/(10 s^10) - 1/(8 s^8) + 1/(6 s^6) - 1/(4 s^4) + 1/(2 s^2) *)

we see that this sum is pretty easy, so we write it down and let Mathematica calculate the value:

Sum[(-1)^(i/2 + 1)/(i*s^i), {i, 2, Infinity, 2}]

(* 1/2 Log[(1 + s^2)/s^2] *)

Answer 3

Ok, I think I know why the integral worked, but not the Laplace transform.

When using the integral, there is a pole at t=0 but this is a removable singularity.

Series[1 - Cos[t], {t, 0, 6}] // Normal

Now dividing by t

(#/t) & /@ r

So, the t in the denominator is gone. I do not know how Mathematica actually removed this pole at t=0 in the code, but it did it when calling Integrate. It might be because it is at start of the interval? or it have done something more advanced than the above, I do not know.

But when doing the Laplace transform, Mathematica must have first tried table lookup for each term. It must saw the 1/t term there. Using the lookup table, the Laplace transform for $t^n$ for negative $n$ is Gamma[n+1]/(s^(n+1)) reference and for $n=-1$ this gives Gamma[0] which is not defined. Hence LaplaceTransform gave up.

Basically what seems to have happened, is that LaplaceTransform does lookup first (for speed). It does not call Integrate to do the integration right away unless needed (else why have LaplaceTransform function in first place, no need to be calling a function which will just call integrate right away).

Answer 4

If you don't want to resort to tricks, you can differentiate the transform over s first, which would bring -t downstairs and cancel 1/t. You can then take the transform for:

LaplaceTransform[(1 - Cos[t]), t, s]

(* 1/s - s/(1 + s^2) *)

and then integrate this over s (with a negative sign, since differentiation produced an extra -1):

-Integrate[1/s - s/(1 + s^2), s]

(* -Log[s] + 1/2 Log[1 + s^2] *)

In principle, this was done up to an additive constant, but you can see that this constant is zero, by inspecting the behavior of the original transform, and also the result, at $s\to\infty$ - both tend to zero in this limit.

Answer 5

Another quick way is to use the following trig identity

In[13]:= TrigFactor[(1-Cos[2x])]
Out[13]= 2 Sin[x]^2

If t==2*x, then we have the integrand becomes:

LaplaceTransform[2 Sin[x]^2/x, x, 2*s]

(* 1/2 Log[1+1/s^2] *)