- #1
- 2,570
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I have two related questions. First of all, we have the identity:
[tex] \int_{-\infty}^{\infty} e^{ikx} dk = 2 \pi \delta(x) [/tex]
I'm wondering if it's possible to get this by contour integration. It's not hard to show that the function is zero for x non-zero, but the behavior at x=0 is bugging me.
Next, you can similarly derive something like:
[tex] \int_{0}^{\infty} e^{ikx} dk = \frac{i}{x} + \pi \delta(x) [/tex]
Now if I integrate both sides with respect to z, I get:
[tex] \int_{0}^{\infty} e^{ikx} \frac{dk}{k} = - \ln(x) + i \pi \theta(x) [/tex]
But the integral on the LHS can be rewritten with a change of variables so that it doesn't depend on x. How can this be?
[tex] \int_{-\infty}^{\infty} e^{ikx} dk = 2 \pi \delta(x) [/tex]
I'm wondering if it's possible to get this by contour integration. It's not hard to show that the function is zero for x non-zero, but the behavior at x=0 is bugging me.
Next, you can similarly derive something like:
[tex] \int_{0}^{\infty} e^{ikx} dk = \frac{i}{x} + \pi \delta(x) [/tex]
Now if I integrate both sides with respect to z, I get:
[tex] \int_{0}^{\infty} e^{ikx} \frac{dk}{k} = - \ln(x) + i \pi \theta(x) [/tex]
But the integral on the LHS can be rewritten with a change of variables so that it doesn't depend on x. How can this be?