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When calculating the Green's Function to the Wave Equation in two spatial dimensions, I came across the integral

$$\int_0^{\infty}e^{ickt}J_0(kr)dk$$

The only idea I had was to put in the integral representation of the Bessel Function and try to interchange integrals. This lead me to

\begin{align} \int_0^{\infty}e^{ickt}J_0(kr)dk &= \frac{1}{2\pi}\int_{-\pi}^{\pi}\int_0^{\infty}e^{i(ct + r\cos\phi)k}dk d\phi\\ &= \int_{-\pi}^{\pi}\delta(ct+r\cos\phi)d\phi \end{align}

Then, I used the fact that $\delta(f(x)) =\sum\frac{\delta(x-x_i)}{|f'(x_i)|}$, where $x_i$ are the roots of the function $f(x)$.

Thus, we have

$$ct+r\cos\phi = 0\Rightarrow\cos\phi = -\frac{ct}{r}\Rightarrow\phi=\arccos\left(-\frac{ct}{r}\right)$$

and

\begin{align} f'\left(\arccos\left(-\frac{ct}{r}\right)\right) &= -r\sin\left(\arccos\left(-\frac{ct}{r}\right)\right) \\ &=-r\sqrt{1-\cos^2\left(\arccos\left(-\frac{ct}{r}\right)\right)}\\ &=-r\sqrt{1-\frac{c^2 t^2}{r^2}}\\ &=-\sqrt{r^2 - c^2 t^2} \end{align}

Putting this into our integral, we arrive at

\begin{align} \int_{-\pi}^{\pi}\delta(ct+r\cos\phi)d\phi &=\int_{-\pi}^{\pi}\frac{\delta\left(\phi - \arccos\left(-\frac{ct}{r}\right)\right)}{\sqrt{r^2 - c^2 t^2}}d\phi\\ &=\frac{1}{\sqrt{r^2 - c^2 t^2}}\int_{-\pi}^{\pi}\delta\left(\phi - \arccos\left(-\frac{ct}{r}\right)\right)d\phi \end{align}

Now, this last step should be straightforward, but I'm not sure how to proceed with the limits of integration here. My first thought was to argue that because the range of $\arccos(x)$ is $0\leq x\leq \pi$, for any value that $\arccos\left(-\frac{ct}{r}\right)$ can take, it is in the interval of integration and therefore, the integral is simply $1$.

However, for negative values of $\phi$, the expression inside the Delta function is never zero, so do I have to exclude those values from consideration somehow, and if so, what do I do with negative values of $\phi$? Or does it simply mean that the result would be the same if I were to only integrate from $0$ to $\pi$?

I've entered the integral into Wolfram Alpha, and I've gotten the result

$$\int_{-\pi}^{\pi}\delta\left(\phi - \arccos\left(-\frac{ct}{r}\right)\right)d\phi = \theta\left(\pi - \arccos\left(-\frac{ct}{r}\right)\right)\theta\left(\pi + \arccos\left(-\frac{ct}{r}\right)\right)$$

which seems to just be a complicated way to write $1$ (as $\arccos(x) \leq\pi$ for any $x\in[-1,1]$, the argument of either step function is always positive) - but why is written like that? Am I missing something fundamental here?

P.S.: Apologies if I've provided lots of unnecessary details, but I've seen how poorly questions are received that don't show any work done, so I thought better be safe than sorry.

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  • $\begingroup$ A lot of details are fine, but you should make an effort to draw attention to what exactly your question is. Also, I'm changing the tags to increase visibility. $\endgroup$ – Omnomnomnom Dec 16 '16 at 23:15
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    $\begingroup$ The question is focused on how to treat the last integral containing the Delta function, especially how the limits relate to the range and domain of $\arccos(x)$ function within the delta function. Apologies if that isn't clear, should I edit the question? EDIT: I see you already did that to highlight my question. Thanks! $\endgroup$ – Tom Dec 16 '16 at 23:18
  • $\begingroup$ Wouldn't this $$\frac{1}{2\pi}\int_{-\pi}^{\pi}\int_0^{\infty}e^{i(ct + r\cos\phi)k}dk d\phi = \int_{-\pi}^{\pi}\delta(ct+r\cos\phi)d\phi$$ be true only if the limits were $(-\infty,\infty)$? That is, I thought the complex integral definition of the delta function needed to be integrated across the entire real line. $\endgroup$ – zahbaz Dec 16 '16 at 23:34
  • $\begingroup$ Damn, you're right. What a stupid mistake. :( Thanks for pointing that out, I guess I'll have to find another way of computing that integral. $\endgroup$ – Tom Dec 17 '16 at 21:50
  • $\begingroup$ the physicist's way to compute this integral is 1) regularize it by adding a small +ve imaginary part to "ct" and integrate over $k$, 2) evaluate the $\phi$ integral by turning it to a contour integral over unit circle. 3) remove the regularization, The small +ve imaginary part of $ct$ help you fix the sign of square root in the final result. $\endgroup$ – achille hui Dec 18 '16 at 2:06
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Disregarding my comment above, and starting from where you have a delta integral, I think there may be an issue from relying on the principal branch of the inverse cosine to give you all the roots. Let $a=ct/r$. $$ \cos\phi = -a\Rightarrow\phi=\arccos(-a) \text{ for } \phi\in[0,\pi)$$

But if you want to account for all $\phi\in(-\pi,\pi)$, you need more than the principal branch. If my physics sense tells me anything here, your parameters are some radial distance, the speed of light, and time such that it's safe to assume $a>0$. Thus $\cos \phi < 0$ and $\phi$ is a quadrant II or III angle. In the domain of integration, it suffices to say

$$ \cos\phi = -a\Rightarrow\phi=\pm\arccos(-a) \text{ for } \phi\in(-\pi,\pi)$$

Then we get two terms via the $\delta(f(x))$ identity.

\begin{align} \int_{-\pi}^{\pi}\delta(ct+r\cos\phi)d\phi &=\int_{-\pi}^{\pi} \frac{\delta\left(\phi - \arccos(-a)\right) + \delta\left(\phi + \arccos(-a)\right) } {\sqrt{r^2 - c^2 t^2}}d\phi =\frac{2 } {\sqrt{r^2 - c^2 t^2}} \end{align}

This would again evaluate to step function via Wolfram, but as you say, $ 0\le\arccos(-a) \le \pi$. So the delta functions would both activate.

If you have Mathematica, you may like to explore this integral via:

Integrate[ DiracDelta[x- x0] + DiracDelta[x + x0], {x, -Pi, Pi}, Assumptions -> {0 < x0 < Pi}]
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  • $\begingroup$ Thanks for your help understanding Delta Functions! It sadly won't help me much with my actual problem due to the fact that I made a mistake in deriving the integral your post treats, but I'm sure that it'll come in handy some time in the future anyway. $\endgroup$ – Tom Dec 17 '16 at 21:51

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