Doubt in substitution for a Gamma Function

Question

My Doubt : I just want to know what kind of substitution should one take in this problem, after that I think I'll manage this

Answer 1

The integral can be transformed into the gamma function with some algebraic manipulation and a straightforward substitution. First, recall that the gamma function is defined as

$$\Gamma(n+1)=\int\limits_0^{+\infty}x^ne^{-x}\,\mathrm dx$$

Our integrand has an exponential term with a base of $3$ and in order to convert our integral into the form above, it must take on a base of $e$. Therefore, rewrite the power as

$$3^{-4x^2}=e^{-4x^2\log 3}\qquad\implies\qquad I=\int\limits_0^{+\infty}e^{-4x^2\log 3}\,\mathrm dx$$

Next, make the substitution $t=4x^2\log 3$ so

$$x=\frac 12\sqrt{\frac t{\log 3}}\qquad\implies\qquad\mathrm dx=\frac {\mathrm dt}{4\sqrt{t\log 3}}$$

The integral $I$ becomes

\begin{align*}I & =\frac 1{4\sqrt{\log 3}}\int\limits_0^{+\infty}t^{-1/2}e^{-t}\,\mathrm dt\\ & =\frac 1{4\sqrt{\log 3}}\Gamma\left(\frac 12\right)\\ & =\frac {\sqrt {\pi}}{4\sqrt {\log 3}}\end{align*}

Answer 2

That integral is not so related with gamma function, but Gauss error function. Taking $u=2x$ you get the integral $$\int_{0}^{\infty}\frac{1}{2\cdot 3^{u^2}}du=\frac{\sqrt{\pi}}{4\sqrt{\log 3}}\int_0^{\infty}\frac{2\sqrt{\log 3 }}{\sqrt{\pi}3^{u^2}}du$$ where $$\int_0^{\infty}\frac{2\sqrt{\log 3 }}{\sqrt{\pi}3^{u^2}}du=erf(\sqrt{\log 3}u)$$ So $$\int_{0}^{\infty}3^{-4x^2}dx=\frac{\sqrt{\pi}}{4\sqrt{\log 3}}(\lim_{u\to\infty} erf(\sqrt{\log 3}u)-erf(0))=\frac{\sqrt{\pi}}{4\sqrt{\log 3}}$$