Q:

Definite Integral of $\int_{0}^{\infty}3^{-4(z^2)}dz$

Please help me  insolving this
$$\int_{0}^{\infty}3^{-4(z^2)}dz$$
I tried to do normal substitution but it didn'd work.... I wonder it is complex integration... I dont need the solution rather just need the approach to go to solution..Thanks in advance 

A:

Gaussian Integral (for $a\gt0$) is given as
$$I(a)=\int_{0}^{\infty}e^{-az^2}dz=\frac12\sqrt{\frac{\pi}{a}}$$
$$3^{-4z^2}=e^{-4\ln 3z^2 }\implies a=4\ln3$$

$$\large\int_{0}^{\infty}3^{-4z^2}dz=\frac14\sqrt{\frac{\pi}{\ln 3}}$$

