Question

Determine whether the improper integral diverges or converges. Evaluate the integral if it converges. $$ \int_{-\infty}^{\infty} 2 x e^{-3 x^{2}} d x $$

Short answer

The improper integral \(\int_{-\infty}^{\infty} 2x e^{-3x^2}\) dx converges and its value is 0.

Step-by-step solution

Split the integral into two parts

We can rewrite the original integral as the sum of two integrals: \(\int_{-\infty}^{0} 2x e^{-3x^2} dx + \int_{0}^{\infty} 2x e^{-3x^2} dx\). This allows us to deal with each part of the integral separately and also allows us to use different techniques of integration for each integral if necessary.

Perform the integration

For both integrals, we can perform a substitution \(u = -3x^2\), \(du = -6x dx\), and \(x dx = - du / 6\). We then re-rewrite and integrate: \(\int_{-\infty}^{0} 2x e^{-3x^2} dx = -\int_{\infty}^{0} e^u du / 3\), \(\int_{0}^{\infty} 2x e^{-3x^2} dx = -\int_{0}^{\infty} e^u du / 3\). Both integrals will then evaluate to \(-[-e^u / 3]_{\infty}^{0}\) and \(-[-e^u / 3]_{0}^{\infty}\) respectively.

Evaluate the limits and check convergence

Evaluating the upper and lower limits of each integral gives us \(-[0+e^0 / 3] = -1/3\) for the first integral and \(-[-e^0 / 3 + 0] = 1/3\) for the second integral. Adding these two values gives us \(-1/3 + 1/3 = 0\). So the original improper integral converges to a value of 0.