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Integration in complex analysis does not always seem as obvious at in the real plane, with reasons that are not obvious to me. In particular, consider $$\int_{- \infty}^{+ \infty} e^{- \pi x^2 z} dx$$

where $z$ is a complex number of positive real part, so that the integral converges. The value of this integral is $$z^{-1/2} \int_{- \infty}^{+ \infty} e^{-\pi x^2} dx = z^{-1/2}$$

and that seems to be like a standard change of variables as it would be for $z$ real. However, it is not and the argument I read for computing this integral involves Cauchy's contour theorem and turning the path of integration by an angle of $-arg(z)/2$. I do not understand:

  • why isn't the change of variable (which is a "linear" one) possible, even if $z$ is complex ?
  • how does the contour changing work ? (I understand it in the case of horizontal translation, but changing the angle could create convergence issues, isn't it?)
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  • $\begingroup$ The original integral is over the real line. If you make the change of variable $y=x\sqrt z$ the new variable is not a real variable. $\endgroup$ – Kavi Rama Murthy Feb 25 at 0:27
  • $\begingroup$ Complex integrals are path integrals so a change of variables needs to preserve the path and in your example the straight change moves from the reals to a different line in the complex plane (for real integrals there is only one path so to speak up to orientation so a change of variable is just a reparametrization) $\endgroup$ – Conrad Feb 25 at 0:39
  • $\begingroup$ Contour integration uses that the integral of an analytic function on a loop is zero, so you make a loop from the reals to the complex line obtained by turning by $z$ - technically you truncate at a large $R$ and join the segments by another curve, generally the arc of radius $R$ and show that the integral on that arc goes to zero as $R$ goes to infinity. Of course there are many variations and the ingenuity is in finding both the right analytic function and the right closed curve $\endgroup$ – Conrad Feb 25 at 0:45
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Actually, there's another way to prove this. Suppose you have a formula for an integral, say $$F(z) = \int_{-\infty}^\infty f(x,z)\; dx $$ which is true for $z$ in some interval $J$, and there is a connected open subset $U$ of
$\mathbb C$ containing $J$ such that both $F(z)$ and $\int_{-\infty}^\infty f(x,z)\;dx$ are analytic in $U$. Then the formula must be true throughout $U$.

In your example, $z^{-1/2}$ is analytic in the right half plane, while the integral is also analytic there because of locally uniform absolute convergence.

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