# Changing variable in complex variable integrals

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?)
• 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. – Kavi Rama Murthy Feb 25 at 0:27
• 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) – Conrad Feb 25 at 0:39
• 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 – Conrad Feb 25 at 0:45

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.