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

Question

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

Answer 1

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

$$3^{-4z^2}=e^{-4\ln 3z^2 }\implies a=4\ln3$$

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

Answer 2

Hint

Rewrite $$3^{-4z^2}=e^{-4z^2 \log(3)}$$ and make an appropriate change of variable to arrive to something looking like $e^{-x^2}$