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In the WP article about propagators, there is an integral solved as:

$$K(x,x';t)=\frac{1}{2\pi}\int_{-\infty}^{+\infty}dk\,e^{ik(x-x')} e^{-\frac{i\hbar k^2 t}{2m}}=\left(\frac{m}{2\pi i\hbar t}\right)^{\frac{1}{2}}e^{-\frac{m(x-x')^2}{2i\hbar t}}$$

I was trying to find out how to evaluate the integral, using the method of "completing the squares" at the exponential. But at the end, I came up with $\gamma\int_{-\infty}^{\infty}e^{i\alpha (z-\beta)^2}dz$, where $\gamma$, $\alpha$ and $\beta$ are constants. The answer of the WP can be reached if it is solved as a gaussian integral: $= \gamma \sqrt{\frac{\pi}{i\alpha}}$

But is it valid to use the formula of the gaussian integral for a complex constant?

When I look at the main integral (of K), the exponentials are after all oscillating functions, that don't have necessarily to go to zero at infinity.

Claudio Saspinski
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  • See also here, and the linked Altland and Simons pages https://math.stackexchange.com/q/2667420/ – jklebes Mar 04 '21 at 20:08
  • Possible duplicate: https://physics.stackexchange.com/q/368186/2451 – Qmechanic Mar 04 '21 at 20:16
  • Thanks. I did not see the question. Testing the integral of $cos(x^2)$ in the excel, it converges quickly to a range around the expected limit $\sqrt{\frac{\pi}{2}}$. But that range of oscillations seems to decrease very slowly. – Claudio Saspinski Mar 05 '21 at 00:40

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