SF10

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\int_0^\infty 3^{-4z^2}dz\,

Rewrite the problem as

\int_0^\infty e^{\ln \left[(3)^{-4z^2} \right]}dz = \int_0^\infty e^{-4z^2 \ln3}dz\,

Let x = -4z^2 \ln 3 \, so the last integral is now

\int_0^\infty e^{-x} d\left(\sqrt{\frac{x}{4\ln3}}\right) = \int_0^\infty e^{-x} \frac{\frac{1}{2}x^{-\frac{1}{2}}}{2\sqrt{\ln3}}dx = \frac{1}{4\sqrt{\ln3}}\int_0^\infty e^{-x} x^{-\frac{1}{2}}dx\,

=\frac{1}{4\sqrt{\ln3}}\Gamma \left(\frac{1}{2} \right) = \sqrt{\frac{\pi}{16\ln3}}\,

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Calculus

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