I would like to compute this kind of Fourier transform $$I=\frac{1}{2\pi}\int_{-\infty}^{\infty}dw\; e^{-i w t}\frac{1}{\sqrt{1-w^2}(w^2+\epsilon^2)}$$ which has a branchcut and $\epsilon\in{\rm I\!R}$. I consider the argument of the complex numbers to be in the interval $\mbox{Arg}[z]\in[0,2\pi)$. Then, for $-\infty<\omega<-1$, $\sqrt{1-\omega^2}=i\sqrt{\omega^2-1}$ (similar for $1<\omega<\infty$)

Can analytic expressions be computed for this kind of integrals? I am more interested on learning the way to solve it than the result itself, so references are welcomed.

thanks in advance!

My try:

I have used the contour integration to compute this integral (the contour is represented in blue for $t>0$, see figure 1). The integral has 6 contributions $$I=I_1+I_2+I_3+I_4+I_5+I_6$$ where the contributions from the branchcut is given by $I_2$ and $I_4$. As usual, the contour is chosen to neglect the contribution from the contour closing $I_6\to 0$ when it is closed at infinity. Now, the contribution from the branchpoints can be written as $$I_2=\frac{1}{2\pi}\int_{\gamma_2}dw\,e^{-iwt} \frac{1}{\sqrt{1-w^2}(w^2+\epsilon^2)}$$ where I have taken the path in this part, $\gamma_2$, to be a half circumference of radius $\eta$. Then, I can change to polar variables at this point $z=w+1=\eta\, e^{i\theta}$ and this integral part can be written as $$I_2=\frac{1}{2\pi}\int_{\pi}^0 d\theta\,i\eta\,e^{i\theta}\frac{e^{-i(\eta\exp(i\theta)-1)t}}{\sqrt{\eta\,e^{i\theta}(\eta\,e^{i\theta}-2)}}\frac{1}{(\eta\,e^{i\theta}-1)^2+\epsilon^2}$$ Now, we can look for an upper bound for modulus of the integral $$||I_2||\leq \frac{1}{2\pi}\int_{\pi}^0 d\theta\,\frac{\eta}{\sqrt{\eta(\eta-2)}}\frac{1}{(\eta-1)^2+\epsilon^2}\to0,\;\mbox{when}\;\eta\to 0$$ Then, the contribution from the branchcut goes to zero (similar arguments can be done in the other branchpoint), and we can apply the usual residue theorem, finding that $$I=\sum_p 2\pi i\,res(p)=\frac{e^{-\epsilon t}}{2\sqrt{1+\epsilon^2}}$$ I don't know where my mistake is, but I found this result to be false. My feeling is that the reasoning on the integral surrounding the branchpoints is false, since these points have some finite contribution on the integration.

Contour for integrationFigure 1: Contour integration. The crosses in the real axis represent the branchcuts, located at $w=\pm1$ and the red points represent the poles of the function, locates at $w=\pm i\epsilon$. The blue line is the contour integration, which has been chosen to avoid the branchpoints, and closed at infinity. As usual, the x-axis represents the real part of $w$ ($\mbox{Re}(w)$) while the y-axis is the imaginary one ($\mbox{Im}(w)$).

  • $\begingroup$ I have posted my try on the contour integration. I think I made a mistake somewhere, that I am not able to find $\endgroup$
    – seoanes
    Mar 31 '16 at 9:42
  • $\begingroup$ Your integral path goes straight over two poles, so I don't see how it can be well defined. Aren't for the residue theorem the poles supposed to be inside the area and not on the integration path? $\endgroup$
    – Elsa
    Mar 31 '16 at 14:00
  • 1
    $\begingroup$ $$ \int_{-\infty}^{\infty}dw\; \frac{\cos (wt)}{\sqrt{1-w^2}(w^2+\epsilon^2)}=\left(\int_{-\infty}^{-1}+\int_{-1}^{1}+ \int_{-\infty}^{\infty}\right) dw\; \frac{\cos (wt)}{\sqrt{1-w^2}(w^2+\epsilon^2)} $$ To calculate these integrals you have to determine the value of $\sqrt{1-w^2}$ for $-\infty<w<-1$. $ \sqrt{1-w^2}=i\sqrt{w^2-1}$ or $ \sqrt{1-w^2}=-i\sqrt{w^2-1}$ ? Also for $1<w<\infty$, $ \sqrt{1-w^2}=i\sqrt{w^2-1}$ or $ \sqrt{1-w^2}=-i\sqrt{w^2-1}$ ? The difficult point is in it rather than in the residue theorem. $\endgroup$
    – ts375_zk26
    Apr 2 '16 at 22:16
  • 1
    $\begingroup$ I mistook the direction of the contour. So \begin{align} I&=\int_{-\infty}^{-1}dw\; \frac{e^{-iwt}}{-i\sqrt{w^2-1}(w^2+\epsilon^2)}+\int_{-1}^{1}dw\; \frac{e^{-iwt}}{\sqrt{1-w^2}(w^2+\epsilon^2)}+\int_{-\infty}^{\infty}dw\; \frac{e^{-iwt}}{i\sqrt{w^2-1}(w^2+\epsilon^2)}\\ &=\frac{e^{-\varepsilon t}}{2\varepsilon \sqrt{1+\varepsilon ^2}}. \end{align} $\endgroup$
    – ts375_zk26
    Apr 4 '16 at 6:39
  • 1
    $\begingroup$ If you want to get \begin{align} I&=\int_{-\infty}^{-1}dw\; \frac{e^{-iwt}}{i\sqrt{w^2-1}(w^2+\epsilon^2)}+\int_{-1}^{1}dw\; \frac{e^{-iwt}}{\sqrt{1-w^2}(w^2+\epsilon^2)}+\int_{-\infty}^{\infty}dw\; \frac{e^{-iwt}}{i\sqrt{w^2-1}(w^2+\epsilon^2)}, \end{align} the residue theorem would not work. $\endgroup$
    – ts375_zk26
    Apr 4 '16 at 6:42

Let us choose the branch cut of $\operatorname{Arg}$ function exactly as in OP so that $\operatorname{Arg}(z) \in [0, 2\pi)$. Correspondingly the square root is defined as

$$ \sqrt{z} = |z|^{1/2} e^{i\operatorname{Arg}(z)/2}. $$

The problem is that, with this definition the branch cut of the function $w \mapsto \sqrt{1-w^2}$ is $[-1, 1] \cup i\Bbb{R}$. This in particular crosses the complex plane vertically, rendering the Cauchy integration formula inapplicable.

This problem can be circumvented by replacing the function $\sqrt{1-w^2}$ by another function which agrees with $\sqrt{1-w^2}$ on $\Bbb{R}$ but is analytically more tamed on the lower half plane.

Our choice is the function $g(w) := i\sqrt{w-1}\sqrt{w+1}$. Then it is easy to check that the branch cut of $g$ is $[-1, 1]$ and we have

$$\sqrt{1-x^2} = \lim_{\delta\downarrow0} g(x-i\delta) \quad \text{for} \quad x \in \Bbb{R}.$$

(Remark. This is important because $g(x) = -\sqrt{1-x^2}$ for $|x|<1$ due to the choice of our branch cut. Only in the limit from below gives the correct sign.) Now consider the function

$$ f(w) = \frac{e^{-itw}}{g(w)(w^2 + \epsilon^2)}. $$

Using the contour considered by OP, we have

$$ \int_{-\infty}^{\infty} \frac{e^{-itw}}{\sqrt{1-w^2}(w^2 + \epsilon^2)} \, dw = -2\pi i \underset{w = -i\epsilon}{\operatorname{Res}} f(w). $$

(Here, the negative sign is introduced since the contour winds $-i\epsilon$ clockwise.) Since $f$ has simple pole at $w = -i\epsilon$, the corresponding residue can be computed as

$$ \underset{w = -i\epsilon}{\operatorname{Res}} f(w) = \lim_{w \to -i\epsilon} (w + i\epsilon)f(w) = \frac{e^{-\epsilon t}}{g(-i\epsilon)(-2i\epsilon)}. $$

The value $g(-i\epsilon)$ can be computed by noticing that

$$ \sqrt{-i\epsilon-1} = (1+\epsilon^2)^{1/4} e^{i(\pi + \arctan \epsilon)/2} \quad \text{and} \quad \sqrt{-i\epsilon+1} = (1+\epsilon^2)^{1/4} e^{i(2\pi - \arctan \epsilon)/2}.$$

Multiplying them gives $g(-i\epsilon) = i \sqrt{-i\epsilon-1}\sqrt{-i\epsilon+1} = \sqrt{1+\epsilon^2}$ and hence we have

$$ \frac{1}{2\pi} \int_{-\infty}^{\infty} \frac{e^{-itw}}{\sqrt{1-w^2}(w^2 + \epsilon^2)} \, dw = \frac{e^{-\epsilon t}}{2\epsilon \sqrt{1+\epsilon^2}}. $$

  • $\begingroup$ Thank you for your answer! I guess the problem on my reasoning was understanding the role of the branchcuts. I have two questions related to your answer: Does the final result of the integral depend on the choice of the branch? meaning that: Will I obtain the same result considering $Arg(z)\in[0,2\pi)$? I suppose this is not in general the case, so here it comes my second question: Is there any relation between your solution and the integral with the choice $Arg(z)\in[0,2\pi)$? $\endgroup$
    – seoanes
    Apr 3 '16 at 17:25
  • 1
    $\begingroup$ @seoanes, Of course the branch cut does not change the value of the integral unless it also changes the value along $\Bbb{R}$. Only the technique depends on the branch cut. You will get the same answer if you use your branch cut (since my choice is designed so), though your solution will differ from this. $\endgroup$ Apr 3 '16 at 20:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.