1

Using contour methods, evaluate the following integral:

$$\int_{-\infty}^{\infty} e^{i(k+iV)x^2}\,\mathrm{d}x$$

as $V$ approaches positive zero.

As far as I'm concerned I don't see any singularity so I thought that I should use a semi circle contour and set the contour integral to zero. That way the integral on the real line is the negative of the integral of the semi circle. However I feel that that approach is wrong because I get an integral with an exponential to an exponential. Also as $r$ goes to infinity it looks like it goes to zero which cannot be right.

Any help will be appreciated.

Dan Uznanski
  • 11,025
Daniel Berkowitz
  • 143
  • Formatting tips here. – Em. Nov 05 '16 at 04:16
  • anything unclear in my answer ?.. – reuns Nov 05 '16 at 04:36
  • @user1952009 your answer is clear but the professor only wants us to use contour integration. If you can give me an hint on what contour to use that would be great. As of now I'm trying to use a rectangular contour but I'm having trouble figuring out what height I should use. If another contour would be easier to use I would gladly use it. – Daniel Berkowitz Nov 06 '16 at 03:28
  • use the Cauchy integral theorem – reuns Nov 06 '16 at 16:24
  • @user1952009 I get that part. I know that when I Integrate this around a contour I get zero in accordance to Cauchy integral theorem. The problem is that I have no idea what contour use. I was trying to use a rectangle but I'm running into trouble. Is there a better contour then a rectangle to use? – Daniel Berkowitz Nov 06 '16 at 16:38
  • $\int_0^\infty e^{-z s^2}ds = z^{-1/2} \int_0^{z^{1/2}\infty} e^{-s^2}ds$ and $\int_0^{z^{1/2}\infty}+\int_{z^{1/2}\infty}^\infty+\int_{\infty}^0 e^{-s^2}ds = 0$ by the Cauchy integral theorem – reuns Nov 06 '16 at 17:06
  • @user1952009 ah I see, is this with a rectangular contour these integrals? I know that two of them will to go to zero leaving one integral equaling the other. – Daniel Berkowitz Nov 06 '16 at 17:27
  • What ? There is no rectangular contour here, but an infinite triangle with vertices $0,+\infty,z^{1/2}\infty$ – reuns Nov 06 '16 at 17:32
  • Ah got it use a infinite triangle as a contour got it. I'll try to work with that instead. – Daniel Berkowitz Nov 06 '16 at 17:39

1 Answers1

1

If you really meant $\int_{-\infty}^\infty e^{(ik-v^2) x^2}dx$, then let $$F(z) = \int_{-\infty}^\infty e^{-z x^2}dx$$ For $z > 0$ you can do the change of variable $y = z^{1/2}x$ to get $$F(z) = \frac{1}{z^{1/2}} \int_{-\infty}^\infty e^{- y^2}dy = \frac{\pi^{1/2}}{z^{1/2}}$$ Then for $z \in \mathbb{C}, Re(z) > 0$ note that $F(z) = \int_{-\infty}^\infty e^{-z x^2}dx$ and $\frac{\pi^{1/2}}{z^{1/2}}$ both are analytic in $z$,

hence by the identity theorem for analytic functions $F(z) = \frac{\pi^{1/2}}{z^{1/2}}$ stays true.

Overall : $$\int_{-\infty}^\infty e^{(ik-v^2) x^2}dx = \frac{\pi^{1/2}}{(v^2-ik )^{1/2}}$$ (where ${}^{1/2}$ is the branch of the square root analytic on $Re(z) > 0$ and such that $z^{1/2} > 0$ for $z > 0$)

reuns
  • 77,999
  • @Dr.MV ?? $\ \ \ $ I did the change of variable $y = z^{1/2}x$ only for $z > 0$, and extended the result to $Re(z) > 0$ by the identity theorem – reuns Nov 05 '16 at 05:30
  • I see that now. I had thought $z$ were complex. Well, I already up voted so ... – Mark Viola Nov 05 '16 at 05:39
  • @Dr.MV Of course you can use the Cauchy integral theorem, but I think the identity theorem is more direct/intuitive as a tool for extending a result from $z \in \mathbb{R}$ to $z \in \mathbb{C}$, which is a major tool in complex integration. – reuns Nov 05 '16 at 05:43
  • @user1952009 This is a very eloquent way of doing the problem but I think we must do it with contour integration. Can you give a hint to suggest a contour to use. I was thinking about a rectangular one but I don't know a priori what height I should use for the rectangle. – Daniel Berkowitz Nov 06 '16 at 03:06