I Imaginary sinusoidal exponential

1. May 1, 2016

spaghetti3451

I need to convince myself that

$e^{ikr\text{cos}\ \theta}=\frac{\text{sin}\ \theta}{kr}$.

2. May 1, 2016

blue_leaf77

Cannot be true, the left side is complex whereas the right side is real.

3. May 1, 2016

spaghetti3451

Sorry, there was a typo.

$\text{exp}({ikr\text{cos}\ \theta}) = \frac{\text{exp}(ikr)-\text{exp}(-ikr)}{2ikr}$?

4. May 1, 2016

blue_leaf77

What do you know about the Euler formula for a complex exponential $e^{i\phi}$?

5. May 1, 2016

spaghetti3451

The Euler formula is $e^{i\phi}= \text{cos}\ \phi + \text{sin}\ \phi$.

I don't see how $\text{exp}(ikr\text{cos}\ \theta) = \frac{\text{sin}(kr)}{kr}$.

6. May 1, 2016

blue_leaf77

It's $e^{i\phi}= \text{cos}\ \phi + i\text{sin}\ \phi$.
Indeed, I too, do not see how that can be true.

7. May 1, 2016

spaghetti3451

The following is taken from page 39 of Matthew Schwartz's 'Introduction to Quantum Field Theory.' My above posts refer to the third and fourth lines of the following calculation:

a

Would you please have a look and let me know what I'm missing in the previous posts?

8. May 1, 2016

blue_leaf77

$$\int_{-1}^1 d(\cos \theta) e^{ikr\cos \theta} = \frac{1}{ikr}e^{ikr\cos \theta}\Big|_{-1}^1 = \frac{e^{ikr}-e^{-ikr}}{ikr}$$
Then in the third line you have $\frac{\sin kr}{k}$ as the integrand to be integrated from 0 to infiinity. The function $\frac{\sin kr}{k}$ is an even function, thus
$$\int_0^\infty \frac{\sin kr}{k} \, dk = \frac{1}{2} \int_{-\infty}^\infty \frac{\sin kr}{k} \, dk$$

9. May 1, 2016

spaghetti3451

Thanks! I feel like a total idiot now!