# Thread: Difficult Integration

1. ## Difficult Integration

(sigh) I know I've seen how to do this before, but I can't find the notes!

The problem is this:
$\displaystyle \int_0^{\pi} d \theta \, sin \theta \, e^{ia \, sin \theta}$ where a is real.

It is simple enough to show that the integral is equal to a real number, so
$\displaystyle \int_0^{\pi} d \theta \, sin \theta \, e^{ia \, sin \theta} = \int_0^{\pi} d \theta \, sin \theta \, cos(a \, sin \theta)$
as an alternate formula.

Comparing two lines in the book I got this from the integral should be equal to
$\displaystyle 2 \frac{sin \, a}{a}$
(For the case a = 0, the integral is simple and comes out to be equal to 2. Obviously I'm not worried about this case.)

Any takers? Thanks in advance.

-Dan

2. You should know that the answer is:
$\displaystyle \int_0^{\pi} d \theta \, sin \theta \, e^{ia \, sin \theta}=2$

You should also know never to listen to me when you get past algebra

3. Originally Posted by Quick
You should know that the answer is:
$\displaystyle \int_0^{\pi} d \theta \, sin \theta \, e^{ia \, sin \theta}=2$

You should also know never to listen to me when you get past algebra
No no no! When it's Calculus the answer is always 8.

-Dan

4. Originally Posted by topsquark
No no no! When it's Calculus the answer is always 8.

-Dan
Oh, sorry you're right (I really need to review my notes)

5. Originally Posted by topsquark
(sigh) I know I've seen how to do this before, but I can't find the notes!

The problem is this:
$\displaystyle \int_0^{\pi} d \theta \, sin \theta \, e^{ia \, sin \theta}$ where a is real.

It is simple enough to show that the integral is equal to a real number, so
$\displaystyle \int_0^{\pi} d \theta \, sin \theta \, e^{ia \, sin \theta} = \int_0^{\pi} d \theta \, sin \theta \, cos(a \, sin \theta)$
as an alternate formula.
Why do you think the integral is real?

See attachment.

RonL

6. Originally Posted by CaptainBlack
Why do you think the integral is real?

See attachment.

RonL
Heavens. I had transformed the integral by substituting $\displaystyle \theta' = \theta - \pi/2$ and took a look at $\displaystyle sin(acos(\theta'))$ and decided it was an odd function over $\displaystyle [-\pi/2, \pi/2]$. It isn't, of course.

Now I've got a "real" problem (pun intended) because my book clearly indicates the real.

Thanks! Back to the books to find out what's going on.

-Dan

7. Okay, I fixed my boo-boo. I wish I could say I used the wrong coordinate system or something face-saving, but I simply used the wrong trig function.

The correct integral is:
$\displaystyle \int_0^{\pi} d \theta \, sin \theta \, e^{ia \, cos \theta} = \int_0^{ \pi} d \theta \, sin \theta \, cos(a \, cos \theta) = 2 \frac{sin(a)}{a}$ where $\displaystyle a > 0$.

Anyone know how to do this one? Thanks!

-Dan

8. Originally Posted by topsquark
Okay, I fixed my boo-boo. I wish I could say I used the wrong coordinate system or something face-saving, but I simply used the wrong trig function.

The correct integral is:
$\displaystyle \int_0^{\pi} d \theta \, sin \theta \, e^{ia \, cos \theta} = \int_0^{ \pi} d \theta \, sin \theta \, cos(a \, cos \theta) = 2 \frac{sin(a)}{a}$ where $\displaystyle a > 0$.

Anyone know how to do this one? Thanks!

-Dan
Perhaps the following might help:

$\displaystyle \sin (\theta)\ e^{ia \cos( \theta)} =-\frac{1}{ia}\ \frac{d}{d\theta} e^{ia \cos( \theta)}$

RonL

9. Originally Posted by CaptainBlack
Perhaps the following might help:

$\displaystyle \sin (\theta)\ e^{ia \cos( \theta)} =-\frac{1}{ia}\ \frac{d}{d\theta} e^{ia \cos( \theta)}$

RonL
I don't mind that you pointed out something I should have seen almost immediately. I don't mind that I figured this out late last night and didn't get to the post early enough to "save face" by mentioning I figured it out.

What I mind is that I spent hours working on a problem that I should have been able to answer in minutes.

Ah well. The learning process! Thanks CaptainBlack.

-Dan