Evaluating $\int^\infty _{-\infty} \frac{e^{-i p x / h}}{x^2 + a^2}\,\mathrm{d}x $

Question

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

Answer 1

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 $(6)$

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

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

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