# 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!