$\int_0^\infty 3^{-x^2} dx$ using gamma function

Question

I want to solv this problem: $$\int_0^\infty 3^{-x^2} dx,$$ using this substitution: $$a^{F(x)} \Longrightarrow F(x)\ln a=-t.$$

Answer 1

I/This is the definition of the gamma function: $$ \Gamma(z)=\int_0^\infty t^{z-1}e^{-t} dt, \qquad \Re(z)>0.$$ So first let's try to make appear the exponent in the original expression to see what is needed to be done.

$$\int_{0}^{\infty }3^{-x^2}dx=\int_{0}^{\infty }e^{\ln(3^{-x^2})}dx=\int_{0}^{\infty }e^{-x^2\ln(3)}dx.$$

II/Now: $t=x^2\ln(3)$ so $dx=\frac{dt}{2\sqrt{t\ln(3)}}$ (obviously the integration interval is the same)

$$\int_{0}^{\infty }e^{-t}\frac{dt}{2\sqrt{t\ln(3)}}dt=\frac{1}{2\sqrt{\ln(3)}}\int_{0}^{\infty }e^{-t}\frac{dt}{\sqrt{t}}=\frac{1}{2\sqrt{\ln(3)}}\int_{0}^{\infty }e^{-t}t^{1/2-1}dt\\ =\frac{1}{2\sqrt{\ln(3)}}\Gamma (1/2)=\frac{\sqrt{\pi}}{2\sqrt{\ln(3)}}.$$

Answer 2

$3^{-x^{2}}=e^{-x^{2}\ln(3)}$

${\int_0^\infty e^{-x^{2}\ln(3)}dx}$

using this substitution:

$$x^{2}\ln(3)=t$$

$x =\sqrt {\frac{t}{{\ln(3)}}}$

$dx={\frac{1}{{2}{\sqrt{{t}{\ln(3)}}}}}dt$

${\int_0^\infty {\frac{1}{{2}{\sqrt{{t} {\ln(3)}}}}}e^{-t} dt}$

$\frac{1}{{2}{\sqrt{\ln(3)}}}{\int_0^\infty t^{\frac{-1}{2}} e^{-t} dt}$

using gamma function:

${\int_0^\infty t^{\frac{-1}{2}} e^{-t} dt}=\Gamma(\frac{1}{2})={\sqrt\pi}$

so

$$\frac{1}{{2}{\sqrt{\ln(3)}}}{\int_0^\infty t^{\frac{-1}{2}} e^{-t} dt}=\frac{\sqrt\pi}{{2}{\sqrt{\ln(3)}}}$$