- #1

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

Could you please help me with it?

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- Thread starter spaghetti3451
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- #1

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

Could you please help me with it?

- #2

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

- #3

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How about this one:

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

- #4

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

- #5

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

- #6

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It's ##e^{i\phi}= \text{cos}\ \phi + i\text{sin}\ \phi##.The Euler formula is ##e^{i\phi}= \text{cos}\ \phi + \text{sin}\ \phi##.

Indeed, I too, do not see how that can be true.I don't see how ##\text{exp}(ikr\text{cos}\ \theta) = \frac{\text{sin}(kr)}{kr}##.

- #7

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

- #8

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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

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Thanks! I feel like a total idiot now!

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