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How can I find the the following integral:

$$\mathcal{I}\left(\text{k},\text{z}\right)=\int_\mathbb{R}\exp\left[\frac{\pi\left(2x\text{z}-\text{k}\right)^2i}{2\text{z}}\right]\space\text{d}x=\int_{-\infty}^\infty\exp\left[\frac{\pi\left(2x\text{z}-\text{k}\right)^2i}{2\text{z}}\right]\space\text{d}x$$

Where $\text{k}$ is an integer number and $\text{z}$ is a complex number.

For the undefinte integral we can write using the substitution $\text{u}=2x\text{z}-\text{k}$:

$$\int\exp\left[\frac{\pi\left(2x\text{z}-\text{k}\right)^2i}{2\text{z}}\right]\space\text{d}x=\frac{1}{2\text{z}}\int\exp\left[\frac{\pi\text{u}^2i}{2\text{z}}\right]\space\text{d}\text{u}$$

Looking back at the definite integral the boundaries will become realy 'weird' after the use of the substitution. When I look at the lower bound it will something be, like this:

$$\text{u}=2\left(-\infty\right)\text{z}-\text{k}$$

So, how can I compute $\mathcal{I}\left(\text{k},\text{z}\right)$?

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  • $\begingroup$ Doesn't $z$ have a part that varies over the $\mathbb{R}$? $\endgroup$ – Michael McGovern Jan 8 '17 at 19:54
  • $\begingroup$ @MichaelMcGovern Yes, the real part of $\text{z}$, say: $\Re\left(\text{z}\right)$ $\endgroup$ – user403351 Jan 8 '17 at 20:23
  • $\begingroup$ Then why did you factor out $\frac{1}{2z}$? Isn't $\mathfrak{I}(z)i$ what remains constant? $\endgroup$ – Michael McGovern Jan 8 '17 at 20:57
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It is a clever way to use the well known result $$\int_0^\infty e^{-x^2}\cos 2bx \,dx=\frac{\sqrt{\pi}}{2}e^{-b^2},\quad \text{or}\,\,\int_0^\infty e^{-ax^2}\cos 2k\pi x \,dx=\frac{\sqrt{\pi}}{2\sqrt{a}}\,e^{-\frac{k^2\pi^2}{a}}\quad (a>0). $$ We rewrite $\mathcal{I}\left(k,z\right)$ : \begin{align} \mathcal{I}\left(k,z\right)&=\int_{-\infty}^\infty\exp\left[\frac{\pi\left(2xz-k\right)^2i}{2z}\right]dx =\exp\left(\frac{k^2\pi^2}{2z}i\right)\int_{-\infty}^\infty\exp\left(2\pi izx^2-2k\pi ix\right)dx\\ &=\exp\left(\frac{k^2\pi^2}{2z}i\right)\int_{-\infty}^\infty\exp\left(-ax^2-2k\pi ix\right)dx, \end{align} where $a=-2\pi iz.$ The integral converges for $a$ with $\operatorname{Re}a>0$ , in other words for $z$ with $\operatorname{Im} z>0$.

First we consider the case $a$ is real and positive. Then \begin{align} \int_{-\infty}^\infty e^{-ax^2-2k\pi ix}dx&=\int_{-\infty}^0 e^{-ax^2-2k\pi ix}dx +\int_0^\infty e^{-ax^2-2k\pi ix}dx\\ &=-\int_{\infty}^0 e^{-at^2+2k\pi it}dt+\int_0^\infty e^{-ax^2-2k\pi ix}dx\\ &=\int_0^\infty e^{-ax^2}\left(e^{2k\pi ix}+e^{-2k\pi ix}\right)dx\\ &=2\int_0^\infty e^{-ax^2}\cos 2k\pi x\,dx\\ &=\frac{\sqrt{\pi}}{\sqrt{a}}\exp\left(-{\frac{k^2\pi^2}{a}}\right). \end{align} Thus we have \begin{align} \mathcal{I}\left(k,z\right)=\frac{1}{\sqrt{-2iz}}\tag{1} \end{align} for $z$ with $z=iv,$ $v>0.$
By Identity Theorem $(1)$ holds for all $z$ with $\operatorname{Im} z>0$.

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