I want to calculate the integral using complex integration: $$ f(t) = \int_{-\infty}^\infty \dfrac{e^{-(z+iat)^2}}{4z^2+1} dz = \int_{-\infty}^\infty \dfrac{e^{-(z+iat)^2}}{(2z-i)(2z+i)} dz$$ where $a$ is real.

There are poles at $z=\pm i/2$. If I draw a semicircle anticlockwise contour on the upper half of the complex plane, the contour encloses the $i/2$ pole. Then by the method of residues,

$$f(t) = 2\pi i\mathrm{Res}_{z=i/2} = 2\pi i \left( \dfrac{1}{2i} e^{(1/2+at)^{2}}\right)$$

I double checked on Mathematica and this is not the correct answer.

I suspect I need to do something with the $e^{-(z+iat)^2}$ part, but I'm not sure what (I'm a novice at solving complex integrals). Can you please help?

  • $\begingroup$ Hi Medulla Oblongata, what are the signs of $a,t$? $\endgroup$ – Szeto Jul 24 '18 at 9:58
  • $\begingroup$ Hello again, Szeto. Both $a$ and $t$ are real, $t$ is positive and negative but $a$ can be restricted to being positive if necessary. $\endgroup$ – Medulla Oblongata Jul 24 '18 at 10:01

How did you get $-1+i$? The residue at $z=i/2$ is $\lim_{z\to i/2} (z-i/2) \frac {e^{-(z+iat)^{2}}} {(2z-i)(2z+i)}=\frac 1 {4i} e^{(1/2+at)^{2}}$ and you have to multiply this by $2 \pi i$

  • $\begingroup$ thanks, I fixed the error $\endgroup$ – Medulla Oblongata Jul 24 '18 at 8:49

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