Evaluate the improper integral of x times e^(-x^2) dx from x = 0 to infinity.

Step-by-step solution

1. 1

   Rewrite the improper integral as a limit. An infinite upper limit is defined by a limit of proper integrals, and that definition is what makes the convergence question meaningful:

   $\int_0^{\infty}x\,e^{-x^2}dx=\lim_{b\to\infty}\int_0^{b}x\,e^{-x^2}dx$

2. 2

   Substitute $u=x^2$. The factor $x$ in front is exactly what the substitution needs — this is why $x\,e^{-x^2}$ is elementary while $e^{-x^2}$ alone is not:

   $u=x^2,\qquad du=2x\,dx,\qquad x\,dx=\frac{du}{2}$

   The limits transform as well: $x=0\Rightarrow u=0$ and $x=b\Rightarrow u=b^2$.

3. 3

   Integrate in $u$.

   $\int_0^{b}x\,e^{-x^2}dx=\frac12\int_0^{b^2}e^{-u}du=\frac12\left[-e^{-u}\right]_0^{b^2}=\frac12\left(1-e^{-b^2}\right)$

4. 4

   Take the limit. As $b\to\infty$, $b^2\to\infty$ and $e^{-b^2}\to 0$:

   $\int_0^{\infty}x\,e^{-x^2}dx=\lim_{b\to\infty}\frac12\left(1-e^{-b^2}\right)=\frac12$

   The integral converges, and the decay is extremely fast: at $b=3$ the remaining tail contributes only $\tfrac12 e^{-9}\approx 6\times 10^{-5}$.

5. 5

   Check against a related result. Because the integrand is odd, the same antiderivative gives $\int_{-\infty}^{\infty}x\,e^{-x^2}dx=0$, while the even companion $\int_0^{\infty}e^{-x^2}dx=\tfrac{\sqrt\pi}{2}\approx 0.886$ has no elementary antiderivative at all. Comparing the two shows how much the single factor of $x$ changes the problem: with it, an exact $\tfrac12$; without it, a value that needs the Gaussian trick.