Complex integration, possibly branch cut integral

Maurice7510

1. Homework Statement
The integral I want to solve is
$$D(x) = \frac{-i}{8\pi^2}\int dr\,d\theta \frac{e^{-irx\cos\theta}}{\sqrt{r^2+m^2}}r^2\sin\theta$$
which I've reduced to
$$D(x) = \frac{-i}{4\pi x}\int dr \frac{r\sin(rx)}{\sqrt{r^2+m^2}}$$
by integrating over $\theta$. However, I don't know how to go about solving this bad boy.
2. Homework Equations
I'm honestly not sure; maybe residue theorem or Cauchy's integral formula?

3. The Attempt at a Solution
I've tried to find a Laurent expansion but to no avail. I've also tried Feynman's trick of differentiating under the integral sign, but it complicates things even more. From the research I've done, it seems that the square root in the denominator indicates a branch cut integral, but I haven't been able to find a source that explains how to do this sufficiently well. Any help would be appreciated, thanks.

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Ray Vickson

Homework Helper
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1. Homework Statement
The integral I want to solve is
$$D(x) = \frac{-i}{8\pi^2}\int dr\,d\theta \frac{e^{-irx\cos\theta}}{\sqrt{r^2+m^2}}r^2\sin\theta$$
which I've reduced to
$$D(x) = \frac{-i}{4\pi x}\int dr \frac{r\sin(rx)}{\sqrt{r^2+m^2}}$$
by integrating over $\theta$. However, I don't know how to go about solving this bad boy.
2. Homework Equations
I'm honestly not sure; maybe residue theorem or Cauchy's integral formula?

3. The Attempt at a Solution
I've tried to find a Laurent expansion but to no avail. I've also tried Feynman's trick of differentiating under the integral sign, but it complicates things even more. From the research I've done, it seems that the square root in the denominator indicates a branch cut integral, but I haven't been able to find a source that explains how to do this sufficiently well. Any help would be appreciated, thanks.
What were the limits of integration in the $(\theta, r)$- integral? What are they now in the $r$-integral?

Maurice7510

Oh sorry, $\theta$ was from $0$ to $\pi$ and $r$ from $0$ to $\infty$

Ray Vickson

Homework Helper
Dearly Missed
Oh sorry, $\theta$ was from $0$ to $\pi$ and $r$ from $0$ to $\infty$
In your second integral the integrand $f(r) = r \sin(rx) /\sqrt{m^2+r^2}$ (with $x$ regarded as a fixed parameter) is an even function of $r$ when $x \neq 0$, so $\int_0^{\infty} f(r) \, dr = (1/2) \int_{-\infty}^{\infty} f(r) \, dr$. Now
$$\int_{-\infty}^{\infty} r \sin(rx)/\sqrt{r^2+m^2} \,dr = \text{Im} \int_{-\infty}^{\infty} r e^{irx}/\sqrt{m^2+r^2} \, dr$$
The second form allows completion of the integral in the upper or lower $r$-plane.

Maurice7510

Now I've seen an extremely similar argument before (the same integral with just $r^2+m^2$ in the denominator is commonly approached) but they never include the $Im$; perhaps this is just a result of physicists lacking proper mathematical rigour, but I've seen it in a few sources and was wondering why they don't include it? To be clear, they have exactly what you have, they just don't specify that it's the imaginary part. I'll see what I can do with this, but is there a specific method you would suggest?

Thanks

Ray Vickson

Homework Helper
Dearly Missed
Now I've seen an extremely similar argument before (the same integral with just $r^2+m^2$ in the denominator is commonly approached) but they never include the $Im$; perhaps this is just a result of physicists lacking proper mathematical rigour, but I've seen it in a few sources and was wondering why they don't include it? To be clear, they have exactly what you have, they just don't specify that it's the imaginary part. I'll see what I can do with this, but is there a specific method you would suggest?

Thanks
Since $e^{irx} = \cos(rx) + i \sin(rx)$ we have
$$\int_{-\infty}^{\infty} \frac{r e^{irx}}{\sqrt{r^2+m^2}} dr = I_r + I_i,$$
where
$$I_r = \int_{-\infty}^{\infty} \frac{r \cos(rx)}{\sqrt{r^2+m^2}} dr = 0$$
because integrand is an odd function of $r$, and
$$I_i = \int_{-\infty}^{\infty} \frac{r i \sin (rx)}{\sqrt{r^2+m^2}} dr = i \int_{-\infty}^{\infty} \frac{r \sin (rx)}{\sqrt{r^2+m^2}} dr$$
So, if anybody leaves out the "Im" they will pick up an unwanted factor of $i$.

I don't believe that physicists mistakenly leave out the "Im" when it is not needed; they are much more educated than that.

Maurice7510

Yes, the factor of $i$ isincluded in the derivations I've seen, and I see now how it works. How might you recommend trying to solve this now? With Cauchy's integral formula, having $$\oint \frac{f(z)}{z-im}$$ would mean $f(z)$ would have a factor of $\sqrt{z-im}$ and would thus be zero at $z=im$. The expansion of the integrand only has non-negative powers of $r$ so I can't use residue theorem either, and that's about as far as my contour integration knowledge goes. I've come across some sources that suggest branch cut integration but I don't know how to do that and they don't explain it very well or generally.

Thanks

Ray Vickson

Homework Helper
Dearly Missed
Yes, the factor of $i$ isincluded in the derivations I've seen, and I see now how it works. How might you recommend trying to solve this now? With Cauchy's integral formula, having $$\oint \frac{f(z)}{z-im}$$ would mean $f(z)$ would have a factor of $\sqrt{z-im}$ and would thus be zero at $z=im$. The expansion of the integrand only has non-negative powers of $r$ so I can't use residue theorem either, and that's about as far as my contour integration knowledge goes. I've come across some sources that suggest branch cut integration but I don't know how to do that and they don't explain it very well or generally.

Thanks
I wish I could attach an appropriate diagram, but I cannot, so I will try to describe things in words. If $x > 0$ and $r = s + iy$ with $s$ real and $y \geq 0$, then $e^{ixr} = e^{-xy} e^{isx}$ falls off exponentially fast as $y \to +\infty$, so we can deform into the upper $r$-plane. Take the integral from $-R$ to $+R$, with large $R > 0$ that we let go to $\infty$ eventually. By definition, the integral we seek is the limit of that finite integral as $R \to \infty$. So, for the finite integral, deform the path from $-R+ i 0 \rightarrow R + i0$ along the real axis, to a part along the circular arc of radius $R$ from $-R + i0$ to $-\epsilon + iR$ (just a bit to the left of the imaginary axis), down from $-\epsilon + iR$ to $-\epsilon + im$, then from $-\epsilon + im$ to $+\epsilon + im$ around a semicircle of radius $\epsilon$, then up from $\epsilon + im$ to $\epsilon + iR$, then from $\epsilon + iR$ to $R + i0$ along a circular arc of radius $R$. The integral along the deformed path equals the original integral, since there are no singularities between the two paths.

Now, if we can show that the sum of the integrals on the circular arcs of radius $R$ go to zero as $R \to \infty$---and there is hope of that because they involve $\exp(-x R \sin \theta)$, with $\theta$ varying from $0$ to $\pi/2$. You would need to examine the details and perform a very careful limiting argument, but I would be willing to bet that the limit of 0 comes out, eventually. That leaves only the integral down one side of the imaginary axis to $im$ and up the other side. On each side you need to deal with an integral of the form
$$F(x) = \int_m^{\infty} \frac{ y e^{-xy}}{\sqrt{y^2-m^2}} ,$$
obtained by putting $r = iy$ in the integral.

After integrating by parts with $u = e^{-xy}$ and $dv = y/\sqrt{y^2-m^2}$ you will get an integration of the form
$$e^{-xy} \left. \sqrt{y^2-m^2} \,\right|_m^{\infty} + x \int_m^{\infty} e^{-xy} \sqrt{y^2-m^2} \, dy .$$
That last integral can be expressed in terms of a modified Bessel Function of the second kind.

The remaining issue is whether the integrations down and up opposite sides of the imaginary axis reinforce one another or cancel one another. That has to do with the choice of square root in the two integrations, and I will leave it to you to puzzle out.

Note that for $x < 0$ you could repeat the analysis, but deforming the path into the lower $r$-plane. However, your original integral is an even function of $x$ (involving an odd denominator and an odd numerator), so $D(x) = D(|x|)$ when $x < 0$, so recognizing that makes life a lot easier.

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Maurice7510

Thanks a lot! The thing I'm trying to show in the end is that the integral is proportional to $e^{-mx}$, which clearly this is.

Ray Vickson

Homework Helper
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Thanks a lot! The thing I'm trying to show in the end is that the integral is proportional to $e^{-mx}$, which clearly this is.
No, I don't think so: I get the answer
$$F(x) \equiv \int_0^{\infty} \frac{r \sin(rx)}{\sqrt{r^2+m^2}} \, dr = m K_1(mx),$$
where $K_a(y)$ is the modified Bessel function of the second kind. It is the solution of the modified Bessel DE
$$x^2 \frac{d^2 y}{dx^2} + x \frac{dy}{dx} - (a^2+x^2) y = 0,$$
with an appropriate boundary condition, specified in such a way that its asymptotic expansion for large $y$ is
$$K_a(y) \sim \frac{e^y}{\sqrt{2 \pi y}} \left( 1 + \frac{4 a^2-1}{8y} + \cdots \right)$$
See, eg.,
http://mathworld.wolfram.com/ModifiedBesselFunctionoftheSecondKind.html
and
https://en.wikipedia.org/wiki/Bessel_function.

The function $K_a$ is non-elementary, but is tabulated and is available in several algebraic computer packages: Maple calls it BesselK(a,y), while Mathematica uses BesselK[a,y].

The basic problem you must face is that
$$D(x) = -\frac{i}{4 \pi^2 x} F(x)$$
(with $\pi^2$ in the denominator, not $\pi$!)
involves a divergent integral. The integrand of $F(x)$ is of the form $\sin(rx) g(r)$, where $g(r) = r/ \sqrt{r^2+m^2} \to 1$ as $r \to +\infty$, so the integral is not well-defined. As is often done in Physics, we can try to re-interpret the integration as some type of "generalized function", similar to a Dirac delta-function or the like; in this case, it turns out that the generalized function we get is actually an ordinary function, taking the form of a Bessel function.

Basically (as in several intuitive treatments of generalized functions) we can regard the integral as the $p \to 0+$ limit of
$$F_p(x) \equiv \int_0^{\infty} e^{-px} \sin(rx) g(r) \, dr$$
We can further isolate the troublesome part by writing $g(r) = [g(r)-1] + 1 \equiv h(r) + 1$, to get
$$F_p(x) = \int_0^{\infty} e^{-px} \sin(rx) \, dr + \int_0^{\infty} e^{-px} \sin(rx) h(r) \, dr$$
The first integral is do-able, and the second one can be simplified by putting $p = 0$ right away, since $h(r) \sim - \frac{m^2}{2 r^2} + O(1/r^4)$ for large $r > 0$, so the integral converges nicely without the exponential damping factor. We end up with
$$\lim_{p \to 0+} F_p(x) = \frac{1}{x} + \int_0^{\infty} \sin(rx) h(r) \, dr,$$
You can use integration by parts in the second integral, with $u = h(r)$ and $dv = \sin(rx)\, dr$; eventually, this gives the second integral as
$$\int_0^{\infty} \sin(rx) h(r) \, dr =- \frac{1}{x} + m K_1(mx).$$
This uses integral expressions for $K_a$ as found in the cited links.

A basic question here is whether or not to trust any particular interpretation of a divergent integral. If the problem was part of a design plan for a life-critical system, would you really be willing to trust your survival to such an interpretation? It is a pretty common approach in physics, at least, but could you defend it in court?

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"Complex integration, possibly branch cut integral"

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