Let $t \in \mathbb{R}_{>0}$ and $k \in \mathbb{R}$. I want to find $$\int_{-\infty}^\infty e^{-\frac{x^2}{2t}} e^{-ikx} \, \mathrm dx.$$ A hint told me to first determine $\int_{-\infty}^\infty e^{-\frac{x^2}{2}} \, \mathrm dx$ which I found to be equal to $\sqrt{2 \pi}$. Now I am told to compute the integral using Cauchy's formula for a convenient Cauchy contour. As I have not yet practically applied Cauchy's formula and I have no idea how it would be helpful in this case, I ask for a little help, a hint would be enough really. Thanks in advance.


After completing the square of the exponential argument, use the following contour to justify the change of variables, enter image description here

Here is what $c$ is $$\displaystyle \int_{-\infty}^\infty e^{-p(t+c)^2}dt = \sqrt{\frac{\pi}{p}}, \quad p,c\in {\bf C},\;\mathrm{Re}\displaystyle \left\{p\right\}>0$$

  • $\begingroup$ I don't quite understand. Could you maybe tell me what in my case would be $c$? What do I have to use as $f$ in Cauchy's formula? $\endgroup$ – studeth Oct 27 '12 at 13:09
  • $\begingroup$ Could I argue that the integral along this contour vanishes and as the paths left and right (in vertical direction) cancel each other, the integral along the complex path must equal the integral along the real axis and therefore for $T \to \infty$ the integral from $-\infty$ to $\infty$ equals the integral from $-i \infty$ to $i \infty$? $\endgroup$ – studeth Oct 27 '12 at 13:30
  • $\begingroup$ So I found this, thanks for your help I guess... $\endgroup$ – studeth Oct 27 '12 at 15:47
  • $\begingroup$ @studeth: You are welcome. $\endgroup$ – Mhenni Benghorbal Oct 27 '12 at 17:23
  • $\begingroup$ @studeth:It is a homework problem and you are supposed to do do some work. $\endgroup$ – Mhenni Benghorbal Oct 27 '12 at 17:31

Here is a method which circumvents complex analysis. As DonAntonio pointed out, we have

$$ \exp\left\{ -\frac{x^2}{2t} - ikx \right\} = \exp\left\{ -\frac{1}{2t}\left( x + ikt \right)^2 - \frac{k^2t}{2} \right\} $$


$$ \frac{d}{du} \exp\left\{ -\frac{1}{2t}\left( x + iut \right)^2 \right\} = -i(x+iut) \exp\left\{ -\frac{1}{2t}\left( x + iut \right)^2 \right\}, $$

we have

$$ \begin{align*} & \int_{-\infty}^{\infty} \exp\left\{ -\frac{1}{2t}\left( x + ikt \right)^2 \right\} \, dx - \int_{-\infty}^{\infty} \exp\left\{ -\frac{x^2}{2t} \right\} \, dx \\ &= \int_{-\infty}^{\infty} \left[ \frac{d}{du} \exp\left\{ -\frac{1}{2t}\left( x + iut \right)^2 \right\} \right]_{u=0}^{u=k} \, dx \\ &= -i \int_{-\infty}^{\infty} \int_{0}^{k} (x+iut) \exp\left\{ -\frac{1}{2t}\left( x + iut \right)^2 \right\} \, dudx. \end{align*}$$

Since the integrand is Lebesgue integrable on $(x,u) \in \Bbb{R} \times [0, k]$, we can apply Fubini's theorem and we have

$$ \begin{align*} & \int_{-\infty}^{\infty} \exp\left\{ -\frac{1}{2t}\left( x + ikt \right)^2 \right\} \, dx - \int_{-\infty}^{\infty} \exp\left\{ -\frac{x^2}{2t} \right\} \, dx \\ &= -i \int_{0}^{k} \int_{-\infty}^{\infty} (x+iut) \exp\left\{ -\frac{1}{2t}\left( x + iut \right)^2 \right\} \, dxdu. \\ &= \int_{0}^{k} \left[ it \exp\left\{ -\frac{1}{2t}\left( x + iut \right)^2 \right\} \right]_{x=-\infty}^{x=\infty} \, du = 0. \end{align*}$$


$$ \begin{align*} \int_{-\infty}^{\infty} \exp\left\{ -\frac{x^2}{2t} - ikx \right\} \, dx &= \int_{-\infty}^{\infty} \exp\left\{ -\frac{1}{2t}\left( x + ikt \right)^2 - \frac{k^2t}{2} \right\} \, dx \\ &= \exp\left\{ - \frac{k^2t}{2} \right\} \int_{-\infty}^{\infty} \exp\left\{ -\frac{x^2}{2t} \right\} \, dx \\ &= \sqrt{2\pi t} \exp\left\{ - \frac{k^2t}{2} \right\}. \end{align*}$$


Complete the square:


$$\Longrightarrow e^{-\frac{x^2}{2t}-ikx}=e^{-\frac{1}{2t}(x+tik)^2}\,e^{-\frac{1}{2}tk^2}$$

Now, substituting

$$u:=\frac{1}{\sqrt{2t}}(x+tik)\Longrightarrow du=\frac{1}{\sqrt{2t}}dx\Longrightarrow$$

$$\Longrightarrow \int_{-\infty}^\infty e^{-\frac{1}{2t}(x+tik)^2}dx=\sqrt{2t}\int_{-\infty}^\infty e^{-u^2}du=\sqrt {2t\pi}$$

End now the exercise.

  • 4
    $\begingroup$ Your answer may be right, but the last step isn't since $u$ is complex. How did you get the domain of integration in the last integral? $\endgroup$ – Mercy King Oct 27 '12 at 11:53
  • $\begingroup$ Well, I assume the OP knows what to do with integration limits of the form $\,\pm\infty +ia\,\,,\,a+i\infty\,$ and etc., but I can see your point. I'll try to add some explanation later. Thanx. $\endgroup$ – DonAntonio Oct 27 '12 at 12:05
  • $\begingroup$ @Mercy: Deformation of the infinite coutour of integration is justified by easy estimates on the function. $\endgroup$ – Alexander Shamov Oct 27 '12 at 12:08
  • $\begingroup$ Perhaps so, @AlexanderShamov, yet I think Mercy meant that in my answer there must be a little more info that it shows and, perhaps, the OP won't be able to apply this as he write he's hardly used Cauchy formula and etc. I've no time right now but I'll try to come up with some idea later to help the OP out, if in the meantime someone doesn't fill up this hole. $\endgroup$ – DonAntonio Oct 27 '12 at 12:12
  • $\begingroup$ @AlexanderShamov I don't get your point. What do you mean exactly? $\endgroup$ – Mercy King Oct 27 '12 at 12:12

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