I'm having issues evaluating the following integral using Cauchy's residue theorem.

$$\int_{-\infty}^{\infty} \frac{e^{ix}}{\sqrt{x^2 - 1}} dx $$

Here's what I have tried. We have to make a branch cut from $z = -a$ to $z = a$ and our contour runs along the real axis, with indents along $z = \pm 1$ and a semicircle in the upper half plane to close the contour. Our integrals are, $\oint_{C} \frac{e^{iz}}{\sqrt{z^2 - 1}} dx = \left[ \int_{C_R} + \int_{C_{\epsilon_1}} + \int_{C_{\epsilon_2}} + \int_{-\infty}^{-1-\epsilon_1} + \int_{-1 + \epsilon_1}^{1 - \epsilon_2} + \int_{1 + \epsilon_2}^{\infty} \right] \frac{e^{iz}}{\sqrt{z^2 - 1}} dx$

where $\epsilon_1$ is the radius of the indent around $z = -1$ and $\epsilon_2$ is the radius of the indent around $z = 1$.

We aren't enclosing any poles in our contour so $\oint_C$ must be zero. $\int_{C_R}$ must go to zero in the upper half plane because of Jordan's lemma. For the indent contours, I don't know how to deal with half poles, but I think we can bound their value using ML bound, $\begin{align} \left|\int_{C_{\epsilon_1}} \frac{e^{iz}}{\sqrt{z^2 - 1}} dx \right| \leq \int_{C_{\epsilon_1}} \frac{1}{\sqrt{|z| + 1)(|z| - 1)}} dx &\leq \frac{\pi \epsilon}{\sqrt{(\epsilon -1 )(\epsilon +1 )}} = 0 \end{align}$

In the last line we take $\epsilon \rightarrow 0$. By the same reasoning $\int_{C_{\epsilon_2}} \rightarrow 0$. This leaves everything zero so $\int_{-\infty}^{\infty} \frac{e^{ix}}{\sqrt{x^2 - 1}} dx = 0$. However just plugging the problem into Mathematica we get

$\pi (-i \text{BesselJ[0,1] - BesselY[0,1]})$.

Let me know why I'm wrong and how I can match Mathematica.

  • $\begingroup$ David, sorry that was my mistake. The integrand should be $e^{iz}/\sqrt{z^2 - 1}$. $\endgroup$ – cwin Aug 26 '14 at 1:16
  • $\begingroup$ You have two branch points. $\endgroup$ – Mhenni Benghorbal Aug 26 '14 at 5:33
  • $\begingroup$ Mathematica is not computing any contour integral. For $x < -1$, it treat $\sqrt{x^2-1}$ as positive. In contrast, when you evaluate it as a contour integral, $\sqrt{x^2-1}$ pick up a minus sign for $x < -1$. $\endgroup$ – achille hui Aug 26 '14 at 6:09
  • $\begingroup$ $\verb*Clear[x]; Integrate[Exp[I x]/Sqrt[x^2 - 1], {x, -Infinity, Infinity}]*$ yields $$ \pi\left[\,-{\rm Y}_{0}\left(\,1\,\right) - {\rm i\,J}_{0}\left(\,1\,\right)\,\right] $$ $\endgroup$ – Felix Marin Sep 18 '14 at 8:41
  • 1
    $\begingroup$ $\large x = \cosh\left(\,t\,\right)$. $\endgroup$ – Felix Marin Sep 18 '14 at 8:44

Your Answer

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

Browse other questions tagged or ask your own question.