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

The integral of e^(-x^2) from negative infinity to infinity is equal to √π, that is, $\int_{-\infty}^\infty$exp(-x^2) dx = √π. Here we will learn how to integrate e^{-x^2} (e to the power minus x²) from -∞ to ∞.

Integral of exp(-x^2) from Minus Infinity to Infinity

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

Explanation:

The integral of e^(-x^2) from -∞ to ∞ will be computed using the following three properties of integrals. They are as follows:

1. $\int_{-a}^a f(x)_dx=2\int_{0}^a f(x)~dx$ if f(-x) = f(x).
2. $\Gamma(n)=\displaystyle \int_0^\infty e^{-x}x^{n-1}dx$
3. $\Gamma(\frac{1}{2})=\sqrt{\pi}$.

Step 1:

We have f(x) = $e^{-x^2}$.

So f(-x) = $e^{-(-x)^2}$ = $e^{-x^2}$ = f(x).

Therefore, the given integral can be written as

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

Step 2:

Let us put

x²=t.

Differentiating both sides, we have

2x dx=dt ⇒ dx = $\dfrac{dt}{2x}$ ⇒ dx = $\dfrac{dt}{2\sqrt{t}}$ as x2=t.

Now, change the limits.

x	0	∞
t	0	∞

Step 2:

Therefore, from Step 1, the integral of e^(-x^2) from negative infinity to positive infinity is equal to

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

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

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

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

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

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

= $\sqrt{\pi}$, since the value of Γ(1/2) is √π.

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

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

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

Related Articles:

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

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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