$ f(z) = \int _{0} ^\infty e^{-z t^2} dt $ is holomorphic in $Re(z) > 0$

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I've been trying to find a reasonable way to approach this but nothing leads to a reasonable result.

With $z = a + ib$ the last thing I tried looking at was

$$| \frac{f(z) - f(z-h)}{h}| = |\frac{1}{h} \int_{0} ^{\infty} e^{-zt^2} (1 - e^{-ht^2}) dt | \leq \frac{1}{|h|} \int_{0} ^{\infty} |e^{-zt^2}| |1 - e^{-ht^2}| dt$$ I do cheat next since I asked wolfram alpha for $ \lim_{h\rightarrow0} \frac{|1 - e^{-ht^2}|}{|h|} = |t|$ and I'm not 100% on how to prove that.

$$ \leq \int_{0} ^{\infty} |e^{-at^2}| \frac{|1 - e^{-ht^2}|}{|h|} dt \rightarrow \int_{0} ^{\infty} |t^2e^{-at^2}|dt = \int_{0} ^{\infty} t^2 e^{-at^2} dt $$

I'm not sure how to then get something useful.

Any help would be greatly appreciated.