# Integration problem

1. Oct 14, 2006

### Logarythmic

Can someone give me a hint on how to solve

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

?

2. Oct 14, 2006

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

Last edited: Oct 14, 2006
3. Oct 14, 2006

### Logarythmic

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

4. Oct 14, 2006

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

5. Oct 14, 2006

### jpr0

You can write the trig function as a complex exponential:

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

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

6. Oct 16, 2006

### Logarythmic

jprO: Of course. That will do it.

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

7. Oct 16, 2006

### HallsofIvy

Staff Emeritus
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] The integral is now: $$I= \frac{1}{b}- \frac{a^2}{b^2}\int_0^\infty e^{-ax}sin(bx)dx$$ But remember that [itex]I= \int_0^\infty e^{-ax}sin(bx)dx$
so that equation says
[itex]\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]
so
$$\left(1+frac{a^2}{b^2}\right)\int_0^\infty e^{-ax}sin(bx)dx= \frac{1}{b}$$
$$\int_0^\infty e^{-ax}sin(bx)dx= \frac{b}{a^2+b^2}$$
(Modulo any silly litte errors from working too fast!)

8. Oct 16, 2006

### Logarythmic

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

9. Oct 16, 2006

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

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

Last edited: Oct 16, 2006
10. Oct 17, 2006

### 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. ;)