# Laplace Transform f(t)=tcos(t)

## Homework Statement

I need to find the laplace transform of f(t)=tcos(t).

## Homework Equations

$$\int e^-^s^ttcos(t)dt$$

## The Attempt at a Solution

I just need help on how to integrate this. I can find the answer easily using the f(t)=tcos(kt) general formula but I wish to find it directly.

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Cyosis
Homework Helper
Use the complex form of the cosine.

Thank you :)

I get $$\mathcal{L} [tcost] = \frac {1}{2} ( \frac {1}{(i-s)^{2}} + \frac {1}{(i+s)^{2}})$$

Hopefully that simplifies to $$\frac {s^{2}-1}{(s^{2}+1)^{2}}$$

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Integration by parts also works but thanks :D

Integration by parts also works but thanks :D

Even if you use the complex form of cos(t), you still have to integrate by parts (albeit it's rather simple).

$$\frac {1}{2} \int^{ \infty}_{0} e^{-st}t \frac {1}{2}(e^{it} + e^{-it})dt$$

$$= \frac {1}{2} \int^{ \infty}_{0} te^{(i-s)t}dt + \frac {1}{2} \int^{ \infty}_{0} te^{-(i+s)t}dt$$

You don't need to integrate by parts at all. Just compute the Laplace transform of cos(t). Differentiating with respect to s will then bring down a factor of minus t in the Laplace integral.

You don't need to integrate by parts at all. Just compute the Laplace transform of cos(t). Differentiating with respect to s will then bring down a factor of minus t in the Laplace integral.

Cool shortcut. I've never seen it before.

$$\mathcal{L} [cos(t)] = \int^{ \infty}_{0} e^{-st}cos(t)dt$$

$$\frac {d}{ds} \mathcal{L} [cos(t)] = \frac {d}{ds} \int^{ \infty}_{0} e^{-st}cos(t)dt$$

Assuming it's okay to bring the derivative inside the integral,

$$= \int^{ \infty}_{0} \frac {d}{ds} e^{-st}tcos(t)dt$$

$$= -\int^{ \infty}_{0}te^{-st}cos(t)dt$$

$$= -\int^{ \infty}_{0}e^{-st}tcos(t)dt = - \mathcal{L} [tcos(t)]$$

then it would appear that $$\frac {d^{2}}{ds^{2}} \mathcal{L} [cos(t)] = \mathcal{L} [t^{2}cos(t)]$$

Another trick: To obtain the Laplace transform of, say, sin(t)/t you can compute the Laplace transform of sin(t) and then integrate w.r.t. s from p to infinity. The Laplace transform is then a function of the parameter p. If you put p = 0, you obtain the integral of sin(t)/t from zero to infinity. What's interesing about this is that the antiderivative of sin(t)/t cannot be evaluated in closed form.

Another trick: To obtain the Laplace transform of, say, sin(t)/t you can compute the Laplace transform of sin(t) and then integrate w.r.t. s from p to infinity. The Laplace transform is then a function of the parameter p. If you put p = 0, you obtain the integral of sin(t)/t from zero to infinity. What's interesing about this is that the antiderivative of sin(t)/t cannot be evaluated in closed form.

I have to see for myself.

$$\mathcal{L} [sin(t)] = \int^{ \infty}_{0} e^{-st}sin(t)dt$$

$$\int^{ \infty}_{p} \mathcal{L} [sin(t)] = \int^{ \infty}_{p} \int^{ \infty}_{0} e^{-st}sin(t)dtds$$

Changing the order of integration (which I assume is allowed),

$$= \int^{ \infty}_{0} \int^{ \infty}_{p} e^{-st}sin(t)dsdt$$

$$= \int^{ \infty}_{0} \frac {1}{t}e^{-pt}sin(t)dt$$

$$= \int^{ \infty}_{0} e^{-pt} \frac {sin(t)}{t}dt = \mathcal{L} [ \frac {sin(t)}{t}]$$

According to the table, $$\mathcal{L} [ \frac {sin(t)}{t}] = arctan( \frac {1}{p})$$

Should you then take the limit of $$arctan( \frac {1}{p})$$ as p goes to zero?

$$\int^{ \infty}_{0} \frac {sin(t)}{t}dt = \lim_{p \to 0} arctan( \frac {1}{p}) = \frac {\pi}{2}$$ ?

You can derive that the Laplace transform of sin(t)/t is arctan(1/p) by integrating the Laplace transform of sin(t). If you integrate
1/(s^2+1) from p to infinity, you get pi/2 - arctan(p).