How do I use the Laplace Transform to solve for the integral of sin ax?

[misogynisticfeminist]

I can't seem to integrate this properly and can't find the proper algebraic substituition for it. There's a table of laplace transforms and sin ax is included but I'll really like to do it myself.

$${\cal L} (sin ax)$$

 

[dextercioby]

$$L (\sin ax)=:\int_{0}^{+\infty} e^{-sx}\sin ax dx$$
Use Euler's formula to get
$$L (\sin ax)=Im[\int_{0}^{+\infty} e^{-x(s-ia)} dx]$$
Make the substitution
$$-x(s-ia)\rightarrow -z$$,with $$|z|=x^{2}(a^{2}+s^{2})$$
and solve the integral to get:
$$L (\sin ax)=Im(\frac{1}{s-ia})=\frac{a}{a^{2}+s^{2}}$$
,s>0.

Daniel.

 

[misogynisticfeminist]

hey, haven't thought about Euler's formula, thanks a lot...

 

[da_willem]

Partial integration would also have worked, but maybe takes a little longer.

 

[dextercioby]

Partial integration does not work.U try it and convince yourself I'm right.You expressed an opinion without checking the computations...

Daniel.

 

[da_willem]

Partial integration does not work.U try it and convince yourself I'm right.You expressed an opinion without checking the computations...

Daniel.

Check again, I did, got the same answer...

 

[da_willem]

$$\int_0^\infty e^{-sx}sin(ax)dx=-\frac{1}{a} cos(ax)e^{-sx}|_0^\infty - \frac{s}{a} \int_0^\infty cos(ax)e^{-sx}dx$$
$$= \frac{1}{a}-\frac{s}{a} [\frac{1}{a}sin(ax)e^{-sx}|_0^\infty + \frac{s}{a} \int_0^\infty sin(ax)e^{-sx}dx ]$$
$$= \frac{1}{a}-\frac{s^2}{a^2} \int_0^\infty sin(ax)e^{-sx}$$

so
$$(1+\frac{s^2}{a^2}) \int_0^\infty e^{-sx}sin(ax)dx = \frac{1}{a}$$
$$\int_0^\infty e^{-sx}sin(ax)dx = \frac{a}{a^2+s^2}$$

 

[dextercioby]

I'm sorry,u were right. There are 2 ways to part integrate.I chose the one that didn't go anywhere.

$$\int_{0}^{+\infty} e^{-sx}\sin ax dx=-\frac{1}{a}(\cos ax) e^{-sx}|_{0}^{+\infty}-\frac{s}{a}\int_{0}^{+\infty} (\cos ax) e^{-sx} dx$$
$$=\frac{1}{a}-\frac{s}{a}[\frac{1}{a}(\sin ax e)^{-sx}|_{0}^{+\infty}+\frac{s}{a}\int_{0}^{+\infty} (\sin ax) e^{-sx} dx]$$

Denote the first integral by 'I' and u'll get:
$$I=\frac{1}{a}-\frac{s^{2}}{a^{2}}I\Rightarrow I=\frac{a}{s^{2}+a^{2}}$$

I don't know why,but i like the complex integral version much more...

Daniel.

EDIT:At the same time...U missed a 'dx' in your antepenultimate integral...