Compute integral $\displaystyle \int_{-\infty}^{\infty}e^{-k^2t+ikx}\, dk$.

Hint: Complete the square in the Exponent.

Okay, for the Exponent, we have $$ -k^2t+ikx=-t\cdot\left(k-\frac{ix}{2t}\right)^2-\frac{x^2}{4t}. $$

Now, is it easier to compute $$ \int_{-\infty}^{\infty}\exp\left(-t\cdot\left(k-\frac{ix}{2t}\right)^2-\frac{x^2}{4t}\right)\, dk? $$ Don't see the trick.

  • $\begingroup$ If integration is w.r.t. $k$ then what is $t$ and $x$ ? $\endgroup$ – Empty Apr 5 '17 at 14:14
  • $\begingroup$ They are constants, i.e. the Integrand is of the form $\exp(-c_1(k-c_2)^2-c_2^2)$, isn't it? $\endgroup$ – mathfemi Apr 5 '17 at 14:15
  • $\begingroup$ There's a standard trick for computing integrals of the form $\int_{-\infty}^\infty e^{Q(x)}\, dx$ where $Q(x)$ is a quadratic. Look up "Gaussian integral" if you need to. $\endgroup$ – Connor Harris Apr 5 '17 at 14:19
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    $\begingroup$ You can move the constant factor $e^{-x^2/4t}$ outside the integral. For the rest, the standard trick is to integrate around a rectangular contour in the complex plane, to reduse the integral to real form. Are you familiar with these sorts of manipulations? $\endgroup$ – Harald Hanche-Olsen Apr 5 '17 at 14:32

Well, we have:

$$\mathscr{I}_\text{n}\left(\text{t},\text{x}\right):=\int_{-\text{n}}^\text{n}\exp\left(\text{k}\cdot\text{x}\cdot i-\text{k}^2\cdot\text{x}\right)\space\text{d}\text{k}\tag1$$


$$\text{k}\cdot\text{x}\cdot i-\text{k}^2\cdot\text{x}=-\text{t}\cdot\left(\text{k}-\frac{\text{x}\cdot i}{2\cdot\text{t}}\right)^2-\frac{\text{x}^2}{4\cdot\text{t}}\tag2$$

So, we get for the integral:

$$\mathscr{I}_\text{n}\left(\text{t},\text{x}\right)=\exp\left(-\frac{\text{x}^2}{4\cdot\text{t}}\right)\cdot\int_{-\text{n}}^\text{n}\exp\left(-\text{t}\cdot\left(\text{k}-\frac{\text{x}\cdot i}{2\cdot\text{t}}\right)^2\right)\space\text{d}\text{k}\tag3$$


$$\text{u}:=\text{k}-\frac{\text{x}\cdot i}{2\cdot\text{t}}\tag4$$

So, the integral becomes:

$$\mathscr{I}_\text{n}\left(\text{t},\text{x}\right)=\exp\left(-\frac{\text{x}^2}{4\cdot\text{t}}\right)\cdot\int_{-\text{n}-\frac{\text{x}\cdot i}{2\cdot\text{t}}}^{\text{n}-\frac{\text{x}\cdot i}{2\cdot\text{t}}}\exp\left(-\text{t}\cdot\text{u}^2\right)\space\text{d}\text{u}\tag5$$

Now, using the error function we get:

$$\mathscr{I}_\text{n}\left(\text{t},\text{x}\right)=\frac{\sqrt{\pi}}{2\cdot\sqrt{t}}\cdot\exp\left(-\frac{\text{x}^2}{4\cdot\text{t}}\right)\cdot\left(\text{erf}\left(\left\{\text{n}-\frac{\text{x}\cdot i}{2\cdot\text{t}}\right\}\cdot\sqrt{t}\right)-\text{erf}\left(\left\{-\text{n}-\frac{\text{x}\cdot i}{2\cdot\text{t}} \right\}\cdot\sqrt{t}\right)\right)$$

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  • $\begingroup$ The integral is from $-\infty$ to $+\infty$, not from $-u$ to $+u$. Why mislead the OP into believing the function erf is even relevant here? $\endgroup$ – Did Apr 12 '17 at 9:31

There is a simpler way to finish what @JanEerland has started. Taking it from the quadratic, say

$$\int_{-\infty}^{\infty}e^{-k^2t+ikx}\, dk=e^{-x^2/4t}\int_{-\infty}^{\infty}e^{-t\left(k-\frac{ix}{2t}\right)^2}\,dk$$

Now, there is a theorem for Gaussian integrals with a complex offset that shows that

$$\int_{-\infty}^{\infty}e^{-p(t+c)^2}\,dt=\sqrt{\frac{\pi}{p}},\quad p,c\in\mathbb{C},\, \Re\{p\}>0$$

so that finally

$$\int_{-\infty}^{\infty}e^{-k^2t+ikx}\, dk=\sqrt{\frac{\pi}{t}}e^{-x^2/4t}$$

which is the same as you would get from @JanEerland if you take the limit as $n\to\infty$.

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