Integral of e^(-x^2) from 0 to Infinity

The integral of e^(-x^2) from 0 to infinity is equal to √π/2, that is, ∫exp(-x^2) dx = √π/2. Here we will learn how to integrate e^{-x^2} (e to the power -x²) from 0 to ∞.

Integration of exp(-x^2) from 0 to infinity

Answer: The integral of $e^{-x^2}$ from 0 to infinity is equal to √π/2. So, $\displaystyle \int_0^\infty e^{-x^2} dx =\dfrac{\sqrt{\pi}}{2}$.

Explanation:

To find the integral of e^(-x^2) from 0 to infinity, we will use following two properties of gamma function:

1. $\Gamma(n)=\displaystyle \int_0^\infty e^{-x}x^{n-1}dx$
2. $\Gamma(\frac{1}{2})=\sqrt{\pi}$.

Now, the integration of e^(-x^2) from 0 to infinity is given by

$\displaystyle \int_0^\infty e^{-x^2} dx$.

Put x²=t.

Differentiating, 2x dx=dt

dx = $\dfrac{dt}{2x}$

dx = $\dfrac{dt}{2\sqrt{t}}$ as x²=t.

Now, change the limits.

x	0	∞
t	0	∞

Therefore, $\displaystyle \int_0^\infty e^{-x^2} dx$.

= $\displaystyle \int_0^\infty e^{-t} \dfrac{dt}{2\sqrt{t}}$.

= $\dfrac{1}{2} \displaystyle \int_0^\infty e^{-t} t^{-\frac{1}{2}}~dt$

= $\dfrac{1}{2} \displaystyle \int_0^\infty e^{-t} t^{\frac{1}{2}-1}~dt$

= $\dfrac{1}{2} \Gamma \left(\frac{1}{2}\right)$ by the above definition of Γ(n).

= $\dfrac{1}{2} \times \sqrt{\pi}$ as we know Γ(1/2)=√π.

= $\dfrac{\sqrt{\pi}}{2}$.

So the integral of e^(-x^2) from 0 to infinity is equal to √π/2, that is, ∫₀^∞ $e^{-x^2}$ dx =√π/2.

Conclusion: The value of the integral of e^(x^2) from 0 to infinity is given by:

$\boxed{\displaystyle \int_0^\infty e^{-x^2} dx = \dfrac{\sqrt{\pi}}{2}}$

Read Also:

Integral of log(sinx) from 0 to pi/2

Integral of e^-x from 0 to Infinity: $\int_0^\infty e^{-x}dx$

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