Gaussian integral $\int_{0}^{\infty} e^{-2x^2} dx$

Question

How can I evaluate the Gaussian integral

$$\int_{0}^{\infty} e^{-2x^2} dx$$

The result that I'm getting (which is wrong) is:

Let's say

$$ I = \int_{0}^{\infty} e^{-2x^2} dx $$

Then

$$ I^2 = \left( \int_{0}^{\infty} e^{-2x^2} dx \right)^2 = \int_{0}^{\infty} \int_{0}^{\infty} e^{-2(x^2+y^2)} dxdy = \int_{0}^{2 \pi} \int_{0}^{\infty} e^{-2(r^2)} r dr d\theta $$

Substitution:

$$ u = -2r^2\\ - \frac{du}{4} = r dr $$

Therefore

$$ - \frac{1}{4} \lim_{a \to - \infty} \int_{0}^{2 \pi} \int_{0}^{-2a^2} e^{u} du d\theta $$

Since

$$ \lim_{a \to - \infty} \int_{0}^{-2a^2} e^{u} du = -1 $$

Then

$$ - \frac{1}{4} \int_{0}^{2 \pi} \int_{0}^{\infty} e^{u} du d\theta = \frac{1}{4} \int_{0}^{2 \pi} d\theta = \frac{1}{4} \cdot 2\pi $$

So

$$I = \sqrt{\frac{\pi}{2}}$$

The correct result should be

$$I = \frac{1}{4} \sqrt{2 \pi}$$

Where did I go wrong?

Thank you.

Answer 1

Another Method to handle this problem (too long for a comment)

$$I=\int _{ 0 }^{ \infty }{ { e }^{ -2{ x }^{ 2 } }dx } $$

Substitute: $x\rightarrow \sqrt { \frac { x }{ 2 } } $

$$I=\frac { 1 }{ 2\sqrt { 2 } } \int _{ 0 }^{ \infty }{ { x }^{ \frac { -1 }{ 2 } }{ e }^{ -x }dx } $$

By the definition of Gamma Function, we get: $$I=\frac { \Gamma \left( \frac { 1 }{ 2 } \right) }{ 2\sqrt { 2 } } $$

Hence,

$$\boxed{\int _{ 0 }^{ \infty }{ { e }^{ -2{ x }^{ 2 } }dx }=\frac { \sqrt { \pi } }{ 2\sqrt { 2 } } }$$