Certain type of integrals $ \int_0^{\pi}d\theta\sin\theta \frac{1}{x+i\epsilon - \sqrt{y+z\cos\theta}}, $

Question

I would like to do the following integral $$ \int_0^{\pi}d\theta\sin\theta \frac{1}{x+i\epsilon - \sqrt{y+z\cos\theta}}, $$ where the $i\epsilon$ has been added to avoid some possible divergencies. One should take $\epsilon\rightarrow0$ at the end of the calculation. Mathematica gives an answer to this integral for $z<y$ in terms of roots, logarithms and tangents.

Is the integral zero for $z>y$? And does anybody know how to do it analytically? I guess integration in the complex plane has to be invoked?

NOTE: $x,y,z \in \mathbb{R}$

Answer 1

Alright, I'm getting rid of your $i \epsilon$, because there is no need for it at all at this point (the analytical integration).

Notice, that we are not thinking about any values for $x,y,z$ for now. We don't need to. We just take the integral, and then we should worry about the parameters.

$$ \int_0^{\pi}d\theta\sin\theta \frac{1}{x - \sqrt{y+z\cos\theta}} $$

First, we do the obvious substitution $u=\cos \theta$:

$$ \int_{-1}^{1} \frac{du}{x - \sqrt{y+z u}} $$

Then we do another obvious substitution:

$$v=\sqrt{y+z u}$$

And this is where your problem lies. For $u=-1$ (or $\theta=\pi$) you get $v=\sqrt{y-z}$, for $u=1$ (or $\theta=0$) you get $v=\sqrt{y+z}$. For different values of $y,z$ any or both of these roots can become imaginary. Notice, you did not say anywhere, that $x,y,z$ are positive, so they can all be negative as well.

Now on for the integral:

$$u=\frac{v^2-y}{z}$$

$$du=\frac{2}{z} v dv$$

$$ \frac{2}{z} \int_{\sqrt{y-z}}^{\sqrt{y+z}} \frac{vdv}{x - v}= $$

$$=\frac{2}{z} \int_{\sqrt{y-z}}^{\sqrt{y+z}} \frac{(v-x)dv}{x - v}+\frac{2x}{z} \int_{\sqrt{y-z}}^{\sqrt{y+z}} \frac{dv}{x - v}= $$

$$ =-\frac{2}{z} (\sqrt{y+z}-\sqrt{y-z})-\frac{2x}{z} \int_{x-\sqrt{y-z}}^{x-\sqrt{y+z}} \frac{d(x - v)}{x - v}= $$

$$=-\frac{2}{z} (\sqrt{y+z}-\sqrt{y-z})-\frac{2x}{z}(\ln (x-\sqrt{y+z})- \ln (x-\sqrt{y-z})) $$

$$\int_0^{\pi}d\theta\sin\theta \frac{1}{x - \sqrt{y+z\cos\theta}}=\frac{2}{z} \left(x \ln \left( \frac{x-\sqrt{y-z}}{x-\sqrt{y+z}} \right)+\sqrt{y-z}-\sqrt{y+z} \right)$$

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This is the only correct analytical solution for this integral, independent of the values of real parameters $x,y,z$.

If you introduce your $i \epsilon$ (for example take $x \to x+i \epsilon$ in this solution), and then take the limit $i \epsilon \to 0$ you will get the exact same answer.

It is not zero for $y<z$.

Now, if you want this integral to be real, you have certain restricitions on the parameters.

First, let's deal with roots.

$\sqrt{y-z} \in \mathbb{R}$ if $y-z \geq 0$.

$\sqrt{y+z} \in \mathbb{R}$ if $y+z \geq 0$.

So you have the following two conditions which need to be satisfied simultaneously for your integral to be real:

$$y \geq 0, ~~~~ y \geq |z|$$

If one or both of them are not satisfied, then you will get an imaginary unit in your answer. But what to do about the logarithm? Well, the complex logarithm (just like a complex root) is multivalued. But if you take the main value (branch) of the logarithm, everything becomes easy.

For example, let's have $x=4, y=2, z=3$:

$$\sqrt{y-z}=\sqrt{2-3}=i \sqrt{3-2}=i$$

$$\sqrt{y+z}=\sqrt{5}$$

Now for the logarithm:

$$\ln ( x-\sqrt{y-z})=\ln (4-i)$$

How to deal with this logarithm? We just convert the complex number to the exponential form:

$$a+ ib=\sqrt{a^2+b^2} e^{i \arctan (b/a)}$$

$$4-i=\sqrt{17} e^{i \arctan (1/4)}$$

$$\ln (4-i)=\frac{\ln 17}{2}-\ln e^{i \arctan (1/4)}=\frac{\ln 17}{2}-i \arctan \frac{1}{4}$$

You can check with Mathematica for these values of the parameters, it will give you the same answer.

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Now there is much worse case, if we get the logarithm of a negative number:

$x-\sqrt{y-z}<0$ and/or $x-\sqrt{y+z}<0$

But there is actually a very clear answer here.

If one of these two cases is true, then the integral does not converge, because if one of these cases is true, then the function in the denominator:

$$f(u)=x-\sqrt{y+zu}$$

goes to zero at least once between $u=-1$ and $u=1$. Which means we have an infinity under the integral, and it doesn't converge.

And here is where your $i \epsilon$ should come in.

But I don't if we actually can 'get rid' of the divergency with it. It's probably related to contour integration, and I'm not good at it. So you are welcome to try.

You question wasn't about this though, so I hope it's answered.