# $\int_{-\infty}^{\infty}e^{-zt^2 + 2wt} \ dt = e^{w^2 / z} \sqrt{\frac{\pi}{z}}$ for $z,w \in \mathbb{C}, Re(z) > 0$

I'm doing complex analysis out of Gamelin, and I'm stuck on a problem in section V.7 on zeros of analytic functions.

Problem: Show that $\int_{-\infty}^{\infty}e^{-zt^2 + 2wt} \ dt = e^{w^2 / z} \sqrt{\frac{\pi}{z}}$ for $z,w \in \mathbb{C}, Re(z) > 0$.

Gamelin gives a hint to start with $z$ as a real positive number, and I was able to show this is true when $z$ is a positive real number. But I do not see how to extend this result to general $z$.

• Fix $w$ and recall two holomorphic functions on an open set that agree on a subset with at least one accumulation point are equal – shdp Mar 13 '17 at 23:51

First we complete the square and write $-zt^2+2wt=-z(t-w/z)^2+w^2/z$. Thus, we see that

$$\int_{-\infty}^\infty e^{-zt^2+2wt}\,dt=e^{w^2/z}\int_{-\infty}^\infty e^{-z(t-w/z)^2}\,dt \tag 1$$

Enforcing the substitution $t\to t/\sqrt{z}+w/z$ into $(1)$ yields

$$\int_{-\infty}^\infty e^{-zt^2+2wt}\,dt=\frac{e^{w^2/z}}{\sqrt z}\lim_{L\to \infty}\int_{-\sqrt{z}(L+w/z)}^{\sqrt{z}(L-w/z)} e^{-t^2}\,dt\tag2$$

Denote $\text{Re}(\sqrt{z})=a>0$. From Cauchy's Integral Theorem, we have

\begin{align} 0&=\oint_C e^{-t^2}\,dt\\\\ &=\int_{-\sqrt{z}(L+w/z)}^{\sqrt{z}(L-w/z)}e^{-t^2}\,dt+\int_{\sqrt{z}(L-w/z)}^{aL} e^{-t^2}\,dt+\int_{aL}^{-aL}e^{-t^2}\,dt+\int_{-aL}^{-\sqrt{z}(L+w/z)}e^{-t^2}\tag3 \end{align}

As $L\to \infty$, the second and fourth integrals on the right-hand side of $(3)$ approach $0$ where we have tacitly used the fact that since $\text{Re}(z)>0$, then $\text{Im}(\sqrt{z})<\text{Re}(\sqrt{z})$. Therefore,

$$\lim_{L\to \infty}\int_{-\sqrt{z}(L+w/z)}^{\sqrt{z}(L-w/z)} e^{-t^2}\,dt=\int_{-\infty}^\infty e^{-t^2}\,dt=\sqrt \pi\tag4$$

Substituting $(4)$ into $(2)$ yields

$$\int_{-\infty}^\infty e^{-zt^2+2wt}\,dt=\sqrt{\frac{\pi}{z}}e^{w^2/z}$$

as was to be shown!