How do I use integration by parts to find the Laplace transformation of tsin(t)?

[bmb2009]

Homework Statement

Find the Laplace transformation of the following function by using iterations of integration by parts:

f(t) = tsin(t)

Homework Equations

The Attempt at a Solution

I know how to do integration by parts (as learned in calculus) but have never seen a funtion that has dependence on t in three places ie. definition of laplace transform is ∫f(t)e^(-st)dt which for this problem would go to... ∫tsin(t)e^(-st)dt...how would I go about starting this? thanks

 

[LCKurtz]

Homework Statement

Find the Laplace transformation of the following function by using iterations of integration by parts:

f(t) = tsin(t)

Homework Equations

The Attempt at a Solution

I know how to do integration by parts (as learned in calculus) but have never seen a funtion that has dependence on t in three places ie. definition of laplace transform is ∫f(t)e^(-st)dt which for this problem would go to... ∫tsin(t)e^(-st)dt...how would I go about starting this? thanks

You could undoubtedly do it if it were just $t$ times an exponential. Since $\sin t$ is the imaginary part of $e^{it}$, try$$
\int_0^\infty te^{(-s+i)t}~dt$$ and take the imaginary part of the answer.

 

[Jufro]

The complex integral approach is completely correct but you would want to use the fact that
sin(t)= (e^it- e^-it) / 2i


Just to clarify on my notation ∫UdV = UV + ∫Vdu
So, if we let U= t and dV= sint(t)e^st all we have to do is figure out the integral of dV.

This is another common problem done in integral calculus on its own. The approach is to do IBP twice and return with the original integral so that you can solve for it in a less than kosher way. Just remember to keep the limits 0,∞ throughout the problem in the appropriate places.

For your convenience if dV= sin(t) e^st, then V= 1/(1+s²). This is only valid for the limits of integration 0, ∞.

 

[vanhees71]

Another trick is to use
\int_0^{\infty} \mathrm{d t} t \sin t \exp(-s t)=-\frac{\mathrm{d}}{\mathrm{d} s} \int_0^{\infty} \sin t \exp(-s t),
which is a trivial integral, using the exponential form of the sine function.

 

[Curious3141]

Another trick is to use
\int_0^{\infty} \mathrm{d t} t \sin t \exp(-s t)=-\frac{\mathrm{d}}{\mathrm{d} s} \int_0^{\infty} \sin t \exp(-s t)\mathrm{d t},
which is a trivial integral, using the exponential form of the sine function.

Fixed (missing a dt in the last integral).

 

[vela]

Or you could just try breaking it up as u=t sin t and dv=e^-st dt.