The integral you have brought up, $\int_0^\infty x^2e^{-x^2} dx$, is a famous problem in mathematics spanning statistics, physics, and probability. The exact answer is $\frac{\sqrt{\pi}}{4}$.

To solve it, we can use two incredibly elegant methods. Let's walk through both of them.

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Method 1: The "Split and Conquer" Approach (Integration by Parts)

The first elegant trick is realizing that while we cannot easily integrate $x^2 e^{-x^2}$, we can integrate $x e^{-x^2}$ using a simple simple substitution. Therefore, we should break $x^2$ apart.

Step 1: Set up Integration by Parts The formula for integration by parts is $\int u , dv = uv - \int v , du$. We separate our integrand $x^2 e^{-x^2}$ into $x \cdot (x e^{-x^2})$.

Let:

• $u = x$
• $dv = x e^{-x^2} dx$

Step 2: Differentiate and Integrate First, differentiate $u$:

• $du = dx$

Next, integrate $dv$. We can use $w$-substitution where $w = -x^2$, meaning $dw = -2x , dx$, or $x , dx = -\frac{1}{2} dw$:

• $v = \int x e^{-x^2} dx = \int -\frac{1}{2} e^w dw = -\frac{1}{2} e^{-x^2}$

Step 3: Apply the Formula Now, plug these into the integration by parts formula across our bounds from $0$ to $\infty$: $$\int_0^\infty x^2 e^{-x^2} dx = \left[ x \left(-\frac{1}{2} e^{-x^2}\right) \right]_0^\infty - \int_0^\infty \left(-\frac{1}{2} e^{-x^2}\right) dx$$

Step 4: Evaluate the Boundary Term (The First Trick) Let's look at the first term: $\left[ -\frac{x}{2e^{x^2}} \right]_0^\infty$.

• At the lower bound ($x = 0$), the numerator is $0$, so the term is $0$.
• At the upper bound ($x \to \infty$), we have $\infty / \infty$. However, the exponential function $e^{x^2}$ grows infinitely faster than the linear function $x$. Therefore, this limit approaches $0$.

Because both bounds evaluate to $0$, the entire first term gracefully vanishes! We are left with: $$\int_0^\infty x^2 e^{-x^2} dx = \frac{1}{2} \int_0^\infty e^{-x^2} dx$$

Step 5: The Gaussian Integral We are left with half of the standard Gaussian integral. Finding $\int_0^\infty e^{-x^2} dx$ requires one of the most famous elegant tricks in calculus (squaring the integral and converting to polar coordinates).

It is a known mathematical standard that the full area under the bell curve $\int_{-\infty}^\infty e^{-x^2} dx = \sqrt{\pi}$. Because the function is symmetrical, from $0$ to $\infty$, the area is exactly half: $$\int_0^\infty e^{-x^2} dx = \frac{\sqrt{\pi}}{2}$$

Step 6: Final Calculation Substitute this known value back into our equation: $$I = \frac{1}{2} \left( \frac{\sqrt{\pi}}{2} \right) = \mathbf{\frac{\sqrt{\pi}}{4}}$$

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Method 2: Feynman’s Trick (Differentiation Under the Integral Sign)

If you want to bypass Integration by Parts entirely, you can use an advanced, highly elegant trick popularized by physicist Richard Feynman. It turns integration into differentiation!

Step 1: Introduce a Parameter Instead of solving the specific integral, let's create a generalized function $I(\alpha)$ by introducing a dummy variable $\alpha$: $$I(\alpha) = \int_0^\infty e^{-\alpha x^2} dx$$

Step 2: Solve the Generalized Integral Using $u$-substitution where $u = \sqrt{\alpha}x$, we get $du = \sqrt{\alpha}dx$, which means $dx = \frac{du}{\sqrt{\alpha}}$. $$I(\alpha) = \int_0^\infty e^{-u^2} \frac{du}{\sqrt{\alpha}} = \frac{1}{\sqrt{\alpha}} \int_0^\infty e^{-u^2} du$$ Because we know the Gaussian integral equals $\frac{\sqrt{\pi}}{2}$, we can write: $$I(\alpha) = \frac{\sqrt{\pi}}{2} \alpha^{-1/2}$$

Step 3: The Magic Trick (Differentiate) Here is the elegant trick. We have an equation for $I(\alpha)$. Let's take the derivative of both sides with respect to $\alpha$.

On the left side, we move the derivative inside the integral: $$\frac{d}{d\alpha} I(\alpha) = \int_0^\infty \frac{\partial}{\partial \alpha} \left( e^{-\alpha x^2} \right) dx$$ By the chain rule, taking the derivative of $e^{-\alpha x^2}$ with respect to $\alpha$ brings down a factor of $-x^2$: $$\frac{d}{d\alpha} I(\alpha) = \int_0^\infty -x^2 e^{-\alpha x^2} dx$$

On the right side, we just use the power rule on $\frac{\sqrt{\pi}}{2} \alpha^{-1/2}$: $$\frac{d}{d\alpha} \left( \frac{\sqrt{\pi}}{2} \alpha^{-1/2} \right) = \frac{\sqrt{\pi}}{2} \left(-\frac{1}{2} \alpha^{-3/2}\right) = -\frac{\sqrt{\pi}}{4} \alpha^{-3/2}$$

Step 4: Set $\alpha = 1$ Equating the two sides, we now have: $$\int_0^\infty -x^2 e^{-\alpha x^2} dx = -\frac{\sqrt{\pi}}{4} \alpha^{-3/2}$$

Multiply both sides by $-1$ to remove the negatives, and set $\alpha = 1$ to return to our original problem: $$\int_0^\infty x^2 e^{-(1) x^2} dx = \frac{\sqrt{\pi}}{4} (1)^{-3/2}$$

$$\int_0^\infty x^2 e^{-x^2} dx = \mathbf{\frac{\sqrt{\pi}}{4}}$$

Both methods yield the exact same satisfying result!

Evaluating ∫₀^∞ x²e^(-x²) dx

Strategy Overview

This integral doesn't yield to direct antidifferentiation, so I'll use two connected techniques: integration by parts and the Gaussian integral.

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Step 1: Establish the Gaussian Integral

First, I need the foundational result:

$$I_0 = \int_0^\infty e^{-x^2}\, dx = \frac{\sqrt{\pi}}{2}$$

This is the classic result (proved via polar coordinates / squaring the integral). I'll build on it.

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Step 2: The Elegant Trick — Differentiation Under the Integral Sign

Define a parameter-dependent integral:

$$I(\alpha) = \int_0^\infty e^{-\alpha x^2}\, dx \quad \text{for } \alpha > 0$$

Evaluate it by substitution $u = \sqrt{\alpha},x$, so $du = \sqrt{\alpha},dx$:

$$I(\alpha) = \frac{1}{\sqrt{\alpha}}\int_0^\infty e^{-u^2}\, du = \frac{\sqrt{\pi}}{2\alpha^{1/2}} = \frac{\sqrt{\pi}}{2}\,\alpha^{-1/2}$$

Now differentiate both sides with respect to α:

$$\frac{d}{d\alpha}I(\alpha) = \frac{d}{d\alpha}\int_0^\infty e^{-\alpha x^2}\, dx = \int_0^\infty \frac{\partial}{\partial\alpha}\left(e^{-\alpha x^2}\right)dx$$

$$\boxed{-\int_0^\infty x^2 e^{-\alpha x^2}\, dx = \frac{\sqrt{\pi}}{2}\cdot\left(-\frac{1}{2}\right)\alpha^{-3/2}}$$

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Step 3: Extract the Target Integral

Canceling the negatives:

$$\int_0^\infty x^2 e^{-\alpha x^2}\, dx = \frac{\sqrt{\pi}}{4}\,\alpha^{-3/2}$$

Set α = 1:

$$\int_0^\infty x^2 e^{-x^2}\, dx = \frac{\sqrt{\pi}}{4}$$

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Verification via Integration by Parts

As a satisfying cross-check, integrating $\int x^2 e^{-x^2}dx$ with:

• $u = x$, $\quad dv = x,e^{-x^2}dx$
• $du = dx$, $\quad v = -\tfrac{1}{2}e^{-x^2}$

$$\int_0^\infty x^2 e^{-x^2}\,dx = \left[-\frac{x}{2}e^{-x^2}\right]_0^\infty + \frac{1}{2}\int_0^\infty e^{-x^2}\,dx$$

The boundary term vanishes (exponential beats polynomial), leaving:

$$= 0 + \frac{1}{2}\cdot\frac{\sqrt{\pi}}{2} = \frac{\sqrt{\pi}}{4} \checkmark$$

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Result

$$\int_0^\infty x^2 e^{-x^2}\, dx = \boxed{\dfrac{\sqrt{\pi}}{4}}$$

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Why This is Elegant

Feature	What makes it beautiful
Differentiation under integral sign	Converts a hard integral into a derivative of an easy one
Parameter trick	Links an entire family of integrals through one formula
Integration by parts check	Two completely different methods yield the same answer
Deeper pattern	This is $\Gamma(3/2) / 2$, connecting to the Gamma function

The deeper truth: $\int_0^\infty x^{2n}e^{-x^2}dx = \frac{(2n-1)!!}{2^{n+1}}\sqrt{\pi}$, showing this result sits inside a beautiful infinite family.