Solving Integration Problem: I = \int_0^\infty e^{-ax} \sin{bx} dx

[Logarythmic]

Can someone give me a hint on how to solve

I = \int_0^\infty e^{-ax} \sin{bx} dx

?

 

[courtrigrad]

Use integration by parts, and set the upper limit as t. Then let t--> infinity (i.e. improper integral)

 

[Logarythmic]

But integration by parts will only give me a similar integral to solve...?

 

[courtrigrad]

you have to use it multiple times, so that you can solve the integral

 

[jpr0]

You can write the trig function as a complex exponential:

<br /> \sin(bx) = \frac{1}{2i}\left[e^{ibx}-e^{-ibx}\right]\,.<br />

Then you will have two similar integrals to solve, which are straight forward.

 

[Logarythmic]

jprO: Of course. That will do it.

courtrigad: Partial integration multiple times will give me a loop between e^{-ax} \sin{bx} and e^{-ax} \cos{bx}.

 

[HallsofIvy]

Yes, it does:
I = \int_0^\infty e^{-ax} \sin{bx} dx
Let u= e^{-ax}, dv= sin(bx)dx
Then du= -ae^{-ax}dx, v= -\frac{1}{b}cos(bx)
so the integral becomes
I= -\frac{1}{b}e^{-ax}cos(bx)\left|_0^\inftjy- \frac{a}{b}\int_0^\inty}e^{-ax}cos(bx)dx
Since e^-ax goes to 0 as x goes to infinity, while cos(bx) is bounded, that is
I= \frac{1}{b}- \frac{a}{b}\int_0^\inty}e^{-ax}cos(bx)dx
Now let u= e^{-ax}, dv= cos(bx). Again we have du= -a e^{-ax}dx, v= \frac{1}{b}sin(bx)[/tex]<br /> The integral is now:<br /> I= \frac{1}{b}- \frac{a^2}{b^2}\int_0^\infty e^{-ax}sin(bx)dx<br /> But remember that I= \int_0^\infty e^{-ax}sin(bx)dx<br /> so that equation says<br /> \int_0^\infty e^{-ax}sin(bx)dx= \frac{1}{b}- \frac{a^2}{b^2}\int_0^\infty e^{-ax}sin(bx)dx[/tex]&lt;br /&gt; so &lt;br /&gt; \left(1+frac{a^2}{b^2}\right)\int_0^\infty e^{-ax}sin(bx)dx= \frac{1}{b}&lt;br /&gt; \int_0^\infty e^{-ax}sin(bx)dx= \frac{b}{a^2+b^2}&lt;br /&gt; (Modulo any silly litte errors from working too fast!)

 

[Logarythmic]

I don't understand why you do the substitutions. Can't you just solve the integral by parts without doing the substitutions?

 

[courtrigrad]

do you agree that integration by parts if of the form:

\int udv = uv-\int vdu? You need substitutions to do integration by parts.

 

[Logarythmic]

I see what you mean. I'm just not used to do it your way. Actually, I'm doing it your way. I just don't think about what I'm really doing. ;)