Is this integral zero: $\int_{-\infty}^{\infty} e^{-ipr}p\sqrt{p^2 + m^2} dp$?

Question

I am calculating a propagator in Quantum Field Theory and I get the following integral:

$$\int_{-\infty}^{\infty} e^{-ipr}p\sqrt{p^2 + m^2} dp$$ where $r,m$ are constants. Now we see that the exponential can be split up into cosine and sine. Cosine is even and the remaining function is odd and so the $\cos(pr)p\sqrt{p^2 + m^2}$ is odd and so the integral with $\cos(pr)p\sqrt{p^2 + m^2}$ as the integrand is $0$. Now, sine is odd and the remaining function is odd and so the $\sin(pr)p\sqrt{p^2 + m^2}$ is even. Thus the integral is $-2i\int_{0}^{\infty} \sin(pr)p\sqrt{p^2 + m^2} dp$ . Now I plotted the integrand $\sin(pr)p\sqrt{p^2 + m^2}$ and I noticed rapid oscillations. Does this mean the integral is $0$? If not, how do I evaluate this integral?

Answer 1

Usually when one resorts to distribution theory to compute Fourier transforms, very little mechanical computation is used. The process asserts a few nice properties since, by the definition of the distributional Fourier transform, if they apply to a Schwarz function, they must apply to a tempered distribution as well. Using Fourier transform properties (with crucial sign changes due to the application on a test function), we can work backwards to show what we want is

$$\mathcal{F}\left\{p\sqrt{p^2+m^2}\right\} = i\frac{d}{dr}\mathcal{F}\left\{\sqrt{p^2+m^2}\right\} = \frac{d}{dr}\frac{1}{|r|}\mathcal{F}\left\{\frac{p}{\sqrt{p^2+m^2}}\right\}$$ $$ = i\frac{d}{dr}\frac{1}{|r|}\frac{d}{dr}\mathcal{F}\left\{\frac{1}{\sqrt{p^2+m^2}}\right\} = 2i\frac{d}{dr}\frac{1}{|r|}\frac{d}{dr}K_0(|mr|)$$

where the monomial product-to-derivative and derivative-to-monomial product properties were used, and $K_n$ is the $n$th modified Bessel function of the second kind. The absolute values are in place due to the property that the Fourier transform of a radial distribution is radial. Taking one derivative retrieves the formula Mathematica confirms

$$K_0'(x) = -K_1(x) \implies \mathcal{F}\left\{p\sqrt{p^2+m^2}\right\} = -2i\frac{d}{dr}\left|\frac{m}{r}\right|K_1(|mr|)$$

*Technically there is a missing Dirac mass in the second equality in the chain of properties, but it vanishes since the function at that step is odd.

Answer 2

An attempt at the computation with distributions that doesn't quite get to Ninad's formula. (I'll be replacing $p$ with $x$.) Even in the sense of distributions, $\mathcal F(xf)(\xi) = \frac{\partial_\xi}i\mathcal F f(\xi)$. So the $x$ can be dispensed with, and one should try to understand the Fourier transform of $\sqrt{x^2+m^2}$. Then note that $(m-\partial^2_\xi)e^{-ix\xi} = (m+x^2)e^{-ix\xi}$, so \begin{align}\newcommand{\ip}[1]{\left\langle#1\right\rangle} \ip{\mathcal F[\sqrt{x^2+m^2}](\xi),\phi(\xi)} &= \ip{\sqrt{x^2+m^2},\mathcal F\phi(x)} \\&= \iint\sqrt{x^2+m^2} \phi(\xi)e^{-ix\xi }d\xi dx \\&= \iint\frac1{\sqrt{x^2+m^2}} \phi(\xi)(m-\partial^2_\xi)e^{-ix\xi }d\xi dx \\&= \iint\frac1{\sqrt{x^2+m^2}} (m-\partial^2_\xi)\phi(\xi)e^{-ix\xi }d\xi dx \\&= \ip{\mathcal F(\frac1{\sqrt{x^2+m^2}}),(m-\partial^2_\xi)\phi(\xi)} \end{align} Putting it in this form makes it easier to search for (as $\frac1{\sqrt{x^2+m^2}}$ is in $L^2$). We find, for $m=1$ $$ \mathcal F(\frac1{\sqrt{x^2+1}})(\xi)=2K_0(|\xi|)$$ and hence $$\mathcal F[\sqrt{x^2+1}](\xi)=2(1-\partial_\xi^2)(K_0(|\xi|))$$ Generically, $\partial^2_\xi (f(|\xi|)) = f''(|\xi|) + 2f'(0)\delta_0$, but $K_0$ has a log divergence at the origin, so $K_0'=-K_1$ explodes at $0$...

Let $K_0(|\xi|)=K(\xi)$ for ease of notation. Since on changing variables $y=x/m$, $$\iint\sqrt{x^2+m^2} \phi(\xi)e^{-ix\xi }d\xi dx = |m|^2 \iint\sqrt{y^2+1} \phi(\xi)e^{-iym\xi }d\xi dy $$ We obtain for general $m$, $$ \mathcal F[\sqrt{x^2+m^2}](\xi)=2m^2（(1-\frac{\partial_\xi^2}{m^2})K）(m\xi) = 2((m^2 - \partial_\xi^2)K)(m\xi) $$ Putting back in the $x$ from the beginning we obtain $$\mathcal F[x\sqrt{x^2+m^2}](\xi)= -2i|m|(\partial_\xi(m^2 - \partial_\xi^2)K)(m\xi).$$

Answer 3

If you define it as: $$I=\lim_{x\to\infty^+}\int_{-x}^{x}\exp(ipr)p\sqrt{p^2+m^2}\mathrm dp$$ Trying to take a CPV you will notice that the problem is the $p\sqrt{p^2+m^2}$ as: $$\int_0^\infty\sin(pr)p\sqrt{p^2+m^2}$$ the negative will not "cancel out" the positive, since the function multiplied by it is still growing, and so it will not converge on a value. Think of it as if you put an arbitrarily large number in and look at the result, then put a larger one in they will likely not be close in magnitude and may have a different sign. An interesting but not comparable integral is: $$\int_0^\infty\sin(x)\,\mathrm dx$$ which has to be between $-1$ and $1$ but oscillates and a value cannot be defined.