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Can anybody help me in solving the following integral:

$$\int_0^\infty x\cdot e^{i \omega x-x^2}\,dx\quad?$$

Any help/hints will be highly appreciated.

Thanks

Jack D'Aurizio
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InvizibleSoul
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  • Are $j$ and $\omega$ constants? – 5xum Apr 08 '14 at 08:30
  • I supposed that $j$ is $\sqrt{-1}$ and $\omega$ is a constant. At least I hope ;-) – Umberto Apr 08 '14 at 08:33
  • yes exactly as Umberto has written it :) – InvizibleSoul Apr 08 '14 at 08:37
  • @Umberto : How can i split the integral ? As i understood from your hint is that $\int xe^{(j\omega x - x^2)} = \int xe^{ax}\int xe^{-x^2}$? How ? Please explain a little thanking in advance – InvizibleSoul Apr 08 '14 at 08:40
  • Sorry... Deleted my post. Was wrong :( my bad. I wrote that without using the brain. You cannot split the integral. But you can reduce it to the integral of the gaussian function (check here http://en.wikipedia.org/wiki/Gaussian_integral). Let me post a "correct" hint... Sorry for the mistake. – Umberto Apr 08 '14 at 08:46
  • @Umberto: no problem at all. Waiting for your valuable comments – InvizibleSoul Apr 08 '14 at 08:47

2 Answers2

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So let's call our integral $I$ and write it like this $$ I=\int_0^\infty xe^{ax-x^2} $$ where in our case $a=j\omega$. Now let's consider the following integral $K$ $$ K=\int_0^\infty (a-2x)e^{ax-x^2} $$ Now is easy to see (if there are no mistakes in my calculation) that $$ K=-2I+a\int_0^\infty e^{ax-x^2} \, \, \, (1) $$ now is easy to solve $K$ by doing the variable substitution $y=ax-x^2$. Then in (1) the last term is simply a generic gaussian integral (also should be easy to solve). Then from (1) one can get $I$. Sorry again for the mistaken answer.

Umberto
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  • Thanks I will look into it as it seems a bit complicated and will get back to you if i face difficulties. Thanks once again :) – InvizibleSoul Apr 08 '14 at 09:19
  • No problem. Let us know how it goes. – Umberto Apr 08 '14 at 09:45
  • I have followed your approach which seems interesting. Here is what I did. I slightly changed the integral $I$ and wrote it like this $$ I=\int_0^\infty xe^{ax-x^2} = \int_0^\infty xe^{-(x^2-ax)} $$ to help in limits of $K$. Then the integral $K$ becomes $$ K=\int_0^\infty (2x-a)e^{-(x^2-ax)} $$ Now as you mentioned $$ K=2I-a\int_0^\infty e^{-(x^2-ax)} , , , (1) $$ solving $K$ by doing the variable substitution $y=x^2-ax$, we get $K=1$. $$I = \frac{1}{2}[1 + a\int_0^\infty e^{-(x^2-ax)} dx]$$ The problem now is in the last term of (1). Guidence needed plz – InvizibleSoul Apr 08 '14 at 10:27
  • Well not so easy. You need to relate the integral to the error function $\text{erf}()$. Mainly what you can do is the following. You call $c=a/2$ then you can write $x^2-ax=x^2-2cx+c^2-c^2=(x-c)^2-c^2$ that means that you rewrite the last integral in $(1)$ as $e^{c^2} \int_0^\infty e^{-(x-c)^2}dx$ and that relates to the error function. Just to give you the result (check wolfram alpha) you will get $1/2 \sqrt \pi e^{a^2/4} ( \text{erf}(a/2)+1)$. You cannot solve the integral in closed form. If the extremes were $-\infty$ and $\infty$ then it would a total different story... – Umberto Apr 09 '14 at 07:55
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Assuming that $j$ is the immaginary unity and $\omega\in\mathbb{R}^+$ we have: $$\int_{0}^{+\infty}x\cdot e^{i\omega x - x^2}\,dx = e^{-\omega^2/4}\cdot\int_{0}^{+\infty}x\cdot e^{-(x-i\frac{\omega}{2})^2}\,dx$$ and: $$\int_{0}^{+\infty}x\cdot e^{-(x-i\frac{\omega}{2})^2}\,dx = \int_{0}^{+\infty}(x-i\omega/2)\cdot e^{-(x-i\frac{\omega}{2})^2}\,dx + \int_{0}^{+\infty}(i\omega/2)\cdot e^{-(x-i\frac{\omega}{2})^2}\,dx,$$ so: $$\int_{0}^{+\infty}x\cdot e^{-(x-i\frac{\omega}{2})^2}\,dx = \frac{1}{2}e^{\omega^2/4}+\frac{i\omega}{2}\int_{-i\omega/2}^{-i\omega/2+\infty}e^{-x^2}\,dx,$$ and by moving the integration path in the second integral we get: $$\int_{0}^{+\infty}x\cdot e^{i\omega x - x^2}\,dx = \frac{1}{2}+\frac{i\omega}{2}e^{-\omega^2/4}\left(\frac{\sqrt{\pi}}{2}+i\int_{0}^{\omega/2}e^{x^2}\,dx\right),$$ or:

$$\int_{0}^{+\infty}x\cdot e^{i\omega x - x^2}\,dx = \left(\frac{1}{2}-\frac{\omega\sqrt{\pi}}{4}\int_{0}^{\omega/2}e^{x^2-\omega^2/4}\,dx\right)+i\frac{\omega\sqrt{\pi}}{4}e^{-\omega^2/4}.$$

Notice that the real part depends on the non-elementary imaginary error function, or Dawson's integral.

Jack D'Aurizio
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