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I'm trying to figure out this integral but cannot figure out the right substitution $$ \int^\infty _{-\infty} \frac{e^{-i p x / h}}{x^2 + a^2}dx $$

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  • $\begingroup$ I think you can find the value by consider the contour integral over a semi-circle living in $\{y\ge 0\}$ with radius $r$ and let $r\rightarrow \infty$ $\endgroup$ – Ben Nov 21 '15 at 20:34
  • $\begingroup$ It looks like you're trying to find a wave function in the momentum representation. I believe you'll need to consider the two cases $p/h > 0$ and $p/h < 0$ separately if you're doing contour integration. $\endgroup$ – John Barber Nov 21 '15 at 20:44
  • $\begingroup$ What is $i$, in your integral? $\endgroup$ – Jan Nov 21 '15 at 20:46
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    $\begingroup$ Mathematica gives the following: $$\int^\infty _{-\infty} \frac{e^{-i p x / h}}{x^2 + a^2}\space\text{d}x=\frac{\pi e^{-\frac{ap}{h}}}{a}$$ With $\frac{p}{h}\in\mathbb{R}\space\space\text{&&}\space\space\Re(a)\ne 0\space\space\text{&&}\space\space\left(\Re(a^2)\ge 0\space||\space a^2\notin \mathbb{R}\right)$ $\endgroup$ – Jan Nov 21 '15 at 20:54
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    $\begingroup$ A related question. $\endgroup$ – Lucian Nov 21 '15 at 22:49
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Let $$ \begin{align} I(\alpha,k) &= \int^\infty _{-\infty} \frac{e^{-i \alpha x}}{x^2 + k^2}\,\mathrm{d}x \tag{1} \\ &= \int^{\infty} _{-\infty} \frac{e^{i \alpha x}}{x^2 + k^2}\,\mathrm{d}x \tag{2} \\ &= \frac12\int^{\infty} _{-\infty} \frac{e^{i \alpha x} + e^{-i \alpha x}}{x^2 + k^2}\,\mathrm{d}x \tag{3} \\ &= \int^{\infty} _{-\infty} \frac{\cos \alpha x}{x^2 + k^2}\,\mathrm{d}x \tag{4} \\ &= 2\int^{\infty} _{0} \frac{\cos \alpha x}{x^2 + k^2}\,\mathrm{d}x \tag{5} \\ I(\alpha,k) &= \frac\pi {|k|} e^{-|\alpha|| k|} \tag{6} \end{align}$$

Putting $\alpha = \dfrac ph$ and $k = a$ in $(6)$

$$I\left(\frac ph,a\right) = \int^\infty _{-\infty} \frac{e^{-i p x /h}}{x^2 + a^2}\,\mathrm{d}x = \frac \pi {|a|} e^{-|a||p/h|}$$


$\text{Explanation (2) :}$ Set $x \rightarrow -x$

$\text{Explanation (6) :}$ Using $$\int_{0}^{\infty} \frac{\cos \alpha x}{x^2 + k^2} dx = \frac{\pi e^{-k\alpha}}{2k}$$

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  • $\begingroup$ Why are you able to set $x → -x$ and what are you doing to get (3)? $\endgroup$ – TheStrangeQuark Nov 23 '15 at 0:35
  • $\begingroup$ I believe Eq. (6) here is incorrect if $\alpha$ or $k$ are negative. There should be some absolute values taken. The reason I harp on this is that this looks like an attempt to calculate a quantum mechanical wave function in the momentum basis, and it will not be normalizable without the absolute values. $\endgroup$ – John Barber Nov 23 '15 at 1:14

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