Product of "Gaussian" integrals

Using the subsitution $u = t^{n}$ , we get that

$\int_{0}^{\infty} e^{-t^{n}} \ dt = \frac{1}{n} \int_{0}^{\infty} e^{-u}u^{\frac{1}{n}-1} \ du = \frac{1}{n} \Gamma(\frac{1}{n})$

where $\Gamma$ is the gamma function.

Thus, our function quickly collapses into

$f(z) = \left( \frac{z}{z-1} \right)^{2} \left( \frac{1}{z} \Gamma(\frac{1}{z}) \right) \left( \frac{z-1}{z} \Gamma(1-\frac{1}{z}) \right) = \frac{\Gamma(\frac{1}{z})\Gamma(1-\frac{1}{z})}{z-1}$

Using Euler's reflection formula for the gamma function, the function is reduced again to

$f(z) = \frac{\pi}{(z-1)\sin(\pi/z)}$

Looking at the denominator's zeroes, we can see that $z = \pm \frac{1}{1}, \pm \frac{1}{2}, \pm \frac{1}{3}, \ldots$ satisfy the requirement of singularity.

However, of all our $z$ , only for $z = 1$ do both limits go to $+\infty$ . Hence, our answer is then $z = \boxed{0.5}$ .