Question

For $\alpha>0$, the value of the integral $\displaystyle \int_{-\infty}^{+\infty} x e^{-\alpha x^2} \, dx$ is

• A. $\sqrt{\frac{\pi}{\alpha}}$
• B. $\infty$
• C. 0
• D. 1

Hint:

If $f(x)$ is an odd function, $\displaystyle \int_{-a}^{a} f(x)\,dx = 0$. Gaussian integrals involving odd powers of $x$ always vanish.

Answer 1

The Correct Option is C

Step 1: Recognize the function type.
The integrand $x e^{-\alpha x^2}$ is an odd function because: \[ f(-x) = (-x)e^{-\alpha (-x)^2} = -x e^{-\alpha x^2} = -f(x) \] Step 2: Apply property of definite integrals.
For any odd function integrated symmetrically about the origin: \[ \int_{-a}^{a} f(x)\, dx = 0 \] Step 3: Apply to limits from $-\infty$ to $\infty$.
\[ \int_{-\infty}^{+\infty} x e^{-\alpha x^2}\, dx = 0 \] Step 4: Conclusion.
Hence, the value of the integral is 0.